solutions functions of one complex variable i, second edition by john b. conway

Solutions Manual forFunctions of One Complex Variable I, Second Edition1

© Copyright by Andreas Kleefeld, 2009All Rights Reserved2

1by John B. Conway2Last updated on January, 7th 2013

PREFACE

Most of the exercises I solved were assigned homeworks in thegraduate courses Math 713 and Math 714”Complex Analysis I and II” at the University of Wisconsin – Milwaukee taught by Professor Dashan Fanin Fall 2008 and Spring 2009.The solutions manual is intented for all students taking a graduate level Complex Analysis course. Studentscan check their answers to homework problems assigned from the excellent book “Functions of One Com-plex Variable I”, Second Edition by John B. Conway. Furthermore students can prepare for quizzes, tests,exams and final exams by solving additional exercises and check their results. Maybe students even studyfor preliminary exams for their doctoral studies.However, I have to warn you not to copy straight of this book and turn in your homework, because thiswould violate the purpose of homeworks. Of course, that is upto you.I strongly encourage you to send me solutions that are still missing [email protected](LATEXpreferredbut not mandatory) in order to complete this solutions manual. Think about the contribution you will giveto other students.If you find typing errors or mathematical mistakes pop an email to [email protected]. The recentversion of this solutions manual can be found athttp: //www.math.tu-cottbus.de/INSTITUT /kleefeld/Files/Solution.html.The goal of this project is to give solutions to all of the 452 exercises.

CONTRIBUTION

I thank (without special order)

Christopher T. AlvinMartin J. MichaelDavid Perkins

for contributions to this book.

2

Contents

1 The Complex Number System 11.1 The real numbers. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 The field of complex numbers. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.3 The complex plane. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.4 Polar representation and roots of complex numbers. . . . . . . . . . . . . . . . . . . . . 51.5 Lines and half planes in the complex plane. . . . . . . . . . . . . . . . . . . . . . . . . . 71.6 The extended plane and its spherical representation. . . . . . . . . . . . . . . . . . . . . 7

2 Metric Spaces and the Topology ofC 92.1 Definitions and examples of metric spaces. . . . . . . . . . . . . . . . . . . . . . . . . . 92.2 Connectedness. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.3 Sequences and completeness. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.4 Compactness. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.5 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.6 Uniform convergence. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

3 Elementary Properties and Examples of Analytic Functions 213.1 Power series. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213.2 Analytic functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243.3 Analytic functions as mappings. Möbius transformations . . . . . . . . . . . . . . . . . . 31

4 Complex Integration 424.1 Riemann-Stieltjes integrals. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 424.2 Power series representation of analytic functions. . . . . . . . . . . . . . . . . . . . . . . 484.3 Zeros of an analytic function. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 584.4 The index of a closed curve. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 604.5 Cauchy’s Theorem and Integral Formula. . . . . . . . . . . . . . . . . . . . . . . . . . . 614.6 The homotopic version of Cauchy’s Theorem and simple connectivity . . . . . . . . . . . 634.7 Counting zeros; the Open Mapping Theorem. . . . . . . . . . . . . . . . . . . . . . . . 664.8 Goursat’s Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

5 Singularities 685.1 Classification of singularities. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 685.2 Residues. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 755.3 The Argument Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

3

6 The Maximum Modulus Theorem 846.1 The Maximum Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 846.2 Schwarz’s Lemma. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 866.3 Convex functions and Hadamard’s Three Circles Theorem. . . . . . . . . . . . . . . . . 886.4 The Phragmén-Lindelöf Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

7 Compactness and Convergence in the Space of Analytic Functions 937.1 The space of continuous functionsC(G,Ω) . . . . . . . . . . . . . . . . . . . . . . . . . 937.2 Spaces of analytic functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 977.3 Spaces of meromorphic functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1007.4 The Riemann Mapping Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1037.5 The Weierstrass Factorization Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . 1067.6 Factorization of the sine function. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1107.7 The gamma function. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1127.8 The Riemann zeta function. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120

8 Runge’s Theorem 1238.1 Runge’s Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1238.2 Simple connectedness. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1248.3 Mittag-Leffler’s Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

9 Analytic Continuation and Riemann Surfaces 1309.1 Schwarz Reflection Principle. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1309.2 Analytic Continuation Along a Path. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1319.3 Monodromy Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1329.4 Topological Spaces and Neighborhood Systems. . . . . . . . . . . . . . . . . . . . . . . 1329.5 The Sheaf of Germs of Analytic Functions on an Open Set. . . . . . . . . . . . . . . . . 1339.6 Analytic Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1349.7 Covering spaces. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

10 Harmonic Functions 13710.1 Basic properties of harmonic functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . 13710.2 Harmonic functions on a disk. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14110.3 Subharmonic and superharmonic functions. . . . . . . . . . . . . . . . . . . . . . . . . . 14410.4 The Dirichlet Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14810.5 Green’s Function. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148

11 Entire Functions 15111.1 Jensen’s Formula. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15111.2 The genus and order of an entire function. . . . . . . . . . . . . . . . . . . . . . . . . . 15311.3 Hadamard Factorization Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158

12 The Range of an Analytic Function 16112.1 Bloch’s Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16112.2 The Little Picard Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16212.3 Schottky’s Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16212.4 The Great Picard Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162

4

Chapter 1

The Complex Number System

1.1 The real numbers

No exercises are assigned in this section.

1.2 The field of complex numbers

Exercise 1. Find the real and imaginary parts of the following:

1z

;z− az+ a

(a ∈ R); z3;3+ 5i7i + 1

;

−1+ i√

32

3

;

−1− i√

32

6

; in;

(

1+ i√2

)n

for 2 ≤ n ≤ 8.

Solution. Let z= x+ iy. Thena)

Re

(

1z

)

=x

x2 + y2=

Re(z)|z|2

Im

(

1z

)

= − yx2 + y2

= − Im(z)|z|2

b)

Re(z− az+ a

)

=x2+ y2 − a2

x2 + y2 + 2ax+ a2=

|z|2 − a2

|z|2 + 2aRe(z) + a2

Im(z− az+ a

)

=2ya

x2 + y2 + 2xa+ a2=

2Im(z)a|z|2 + 2aRe(z) + a2

c)

Re(

z3)

= x3 − 3xy2= Re(z)3 − 3Re(z)Im(z)2

Im(

z3)

= 3x2y− y3= 3Re(z)2Im(z) − Im(z)3

1

d)

Re

(

3+ 5i7i + 1

)

=1925

Im

(

3+ 5i7i + 1

)

= − 825

e)

Re

−1+ i√

32

3= 1

Im

−1+ i√

32

3= 0

f)

Re

−1− i√

32

6= 1

Im

−1− i√

32

6= 0

g)

Re(in) =

0, n is odd

1, n ∈ 4k : k ∈ Z−1, n ∈ 2+ 4k : k ∈ Z

Im (in) =

0, n is even

1, n ∈ 1+ 4k : k ∈ Z−1, n ∈ 3+ 4k : k ∈ Z

2

h)

Re

((

1+ i√2

)n)

=

0, n = 2

−√

22 , n = 3

−1, n = 4

−√

22 , n = 5

0, n = 6√2

2 , n = 7

1, n = 8

Im

((

1+ i√2

)n)

=

1, n = 2√2

2 , n = 3

0, n = 4

−√

22 , n = 5

−1, n = 6

−√

22 , n = 7

0, n = 8

Exercise 2. Find the absolute value and conjugate of each of the following:

−2+ i;−3; (2+ i)(4+ 3i);3− i√2+ 3i

;i

i + 3; (1+ i)6; i17.

Solution. It is easy to calculate:a)

z= −2+ i, |z| =√

5, z= −2− i

b)z= −3, |z| = 3, z= −3

c)z= (2+ i)(4+ 3i) = 5+ 10i, |z| = 5

√5, z= 5− 10i

d

z=3− i√2+ 3i

, |z| = 111

√110, z=

3+ i√2− 3i

e)

z=i

i + 3=

110+

310

i, |z| = 110

√10, z=

110− 3

10i

f)z= (1+ i)6

= −8i, |z| = 8, z= 8i

g)i17= i, |z| = 1, z= −i

Exercise 3. Show that z is a real number if and only if z= z.

3

Solution. Let z= x+ iy.⇒: If z is a real number, then z= x (y= 0). This impliesz= x and therefore z= z.⇔: If z = z, then we must have x+ iy = x− iy for all x, y ∈ R. This implies y= −y which is true if y= 0 andtherefore z= x. This means that z is a real number.

Exercise 4. If z and w are complex numbers, prove the following equations:

|z+ w|2 = |z|2 + 2Re(zw) + |w|2.|z− w|2 = |z|2 − 2Re(zw) + |w|2.|z+ w|2 + |z− w|2 = 2

(

|z|2 + |w|2)

.

Solution. We can easily verify thatz= z. Thus

|z+ w|2 = (z+ w)(z+ w) = (z+ w)(z+ w) = zz+ zw+ wz+ ww

= |z|2 + |w|2 + zw+ zw= |z|2 + |w|2 + zw+ z ¯w

= |z|2 + |w|2 + zw+ zw = |z|2 + |w|2 + 2zw+ zw

2= |z|2 + |w|2 + 2Re(zw) = |z|2 + 2Re(zw) + |w|2.

|z− w|2 = (z− w)(z− w) = (z− w)(z− w) = zz− zw− wz+ ww

= |z|2 + |w|2 − zw− zw= |z|2 + |w|2 − zw− z ¯w

= |z|2 + |w|2 − zw− zw = |z|2 + |w|2 − 2zw+ zw

2= |z|2 + |w|2 − 2Re(zw) = |z|2 − 2Re(zw) + |w|2.

|z+ w|2 + |z− w|2 = |z|2 + Re(zw) + |w|2 + |z|2 − Re(zw) + |w|2

= |z|2 + |w|2 + |z|2 + |w|2 = 2|z|2 + 2|w|2 = 2(

|z|2 + |w|2)

.

Exercise 5. Use induction to prove that for z= z1 + . . . + zn; w = w1w2 . . .wn:

|w| = |w1| . . . |wn|; z= z1 + . . . + zn; w = w1 . . . wn.

Solution. Not available.

Exercise 6. Let R(z) be a rational function of z. Show thatR(z) = R(z) if all the coefficients in R(z) are real.

Solution. Let R(z) be a rational function of z, that is

R(z) =anzn+ an−1zn−1

+ . . .a0

bmzm + bm−1zm−1 + . . .b0

where n,m are nonnegative integers. Let all coefficients of R(z) be real, that is

a0,a1, . . . ,an,b0,b1, . . . ,bm ∈ R.Then

R(z) =anzn + an−1zn−1 + . . . a0

bmzm + bm−1zm−1 + . . . b0=

anzn + an−1zn−1 + . . . a0

bmzm + bm−1zm−1 + . . . b0

=anzn + an−1zn−1 + . . . a0

bmzm + bm−1zm−1 + . . . b0

=anzn+ an−1zn−1

+ . . . a0

bmzm + bm−1zm−1 + . . . b0= R(z).

4

1.3 The complex plane

Exercise 1. Prove (3.4) and give necessary and sufficient conditions for equality.

Solution. Let z and w be complex numbers. Then

||z| − |w|| = ||z− w+ w| − |w||≤ ||z− w| + |w| − |w||= ||z− w||= |z− w|

Notice that|z| and |w| is the distance from z and w, respectively, to the origin while |z− w| is the distancebetween z and w. Considering the construction of the impliedtriangle, in order to guarantee equality, it isnecessary and sufficient that

||z| − |w|| = |z− w|⇐⇒ (|z| − |w|)2

= |z− w|2

⇐⇒ (|z| − |w|)2= |z|2 − 2Re(zw) + |w|2

⇐⇒ |z|2 − 2|z||w| + |w|2 = |z|2 − 2Re(zw) + |w|2

⇐⇒ |z||w| = Re(zw)

⇐⇒ |zw| = Re(zw)

Equivalently, this is zw ≥ 0. Multiplying this by ww, we get zw · w

w = |w|2 ·zw ≥ 0 if w , 0. If

t = zw =

(1|w|2

)

· |w|2 · zw. Then t≥ 0 and z= tw.

Exercise 2. Show that equality occurs in (3.3) if and only if zk/zl ≥ 0 for any integers k and l,1 ≤ k, l ≤ n,for which zl , 0.


Exercise 3. Let a∈ R and c> 0 be fixed. Describe the set of points z satisfying

|z− a| − |z+ a| = 2c

for every possible choice of a and c. Now let a be any complex number and, using a rotation of the plane,describe the locus of points satisfying the above equation.


1.4 Polar representation and roots of complex numbers

Exercise 1. Find the sixth roots of unity.

Solution. Start with z6 = 1 and z= rcis(θ), therefore r6cis(6θ) = 1. Hence r= 1 and θ = 2kπ6 with k ∈

−3,−2,−1,0,1,2. The following table gives a list of principle values of arguments and the correspondingvalue of the root of the equation z6

= 1.θ0 = 0 z0 = 1θ1 =

π3 z1 = cis(π3)

θ2 =2π3 z2 = cis(2π

3 )θ3 = π z3 = cis(π) = −1θ4 =

−2π3 z4 = cis(−2π

3 )θ5 =

−π3 z5 = cis(−π3 )

5

Exercise 2. Calculate the following:

a) the square roots of i

b) the cube roots of i

c) the square roots of√

3+ 3i

Solution. c) The square roots of√

3+ 3i.

Let z=√

3+ 3i. Then r= |z| =√

(√3)2+ 32 =

√12 andα = tan−1

(

3√3

)

=π3 . So, the2 distinct roots of z

are given by 2√

r(

cosα+2kπn + i sin α+2kπ

n

)

where k= 0,1. Specifically,

√z=

4√12

(

cosπ3 + 2kπ

2+ i sin

π3 + 2kπ

2

)

.

Therefore, the square roots of z, zk, are given by

z0 =4√12

(

cosπ6 + i sin π6

)

=4√12

( √3

2 +12 i)

z1 =4√12

(

cos7π6 + i sin 7π

6

)

=4√12

(

−√

32 − 1

2 i)

.

So, in rectangular form, the second roots of z are given by(

4√1082 ,

4√122

)

and(

−4√

1082 ,−

4√122

)

.

Exercise 3. A primitive nth root of unity is a complex number a such that1,a,a2, ...,an−1 are distinct nthroots of unity. Show that if a and b are primitive nth and mth roots of unity, respectively, then ab is a kth rootof unity for some integer k. What is the smallest value of k? What can be said if a and b are nonprimitiveroots of unity?


Exercise 4. Use the binomial equation

(a+ b)n=

n∑

k=0

(

nk

)

an−kbk,

where (

nk

)

=n!

k!(n− k)!,

and compare the real and imaginary parts of each side of de Moivre’s formula to obtain the formulas:

cosnθ = cosn θ −(

n2

)

cosn−2 θ sin2 θ +

(

n4

)

cosn−4 θ sin4 θ − . . .

sinnθ =

(

n1

)

cosn−1 θ sinθ −(

n3

)

cosn−3 θ sin3 θ + . . .


Exercise 5. Let z= cis2πn for an integer n≥ 2. Show that1+ z+ . . . + zn−1

= 0.

6

Solution. The summation of the finite geometric sequence1, z, z2, . . . , zn−1 can be calculated as∑n

j=1 zj−1=

zn−1z−1 . We want to show that zn is an nth root of unity. So, using de Moivre’s formula, zn

=

(

cis(

2πn

))n=

cis(

n · 2πn

)

= cis(2π) = 1. It follows that1+ z+ z2+ ... + zn−1

=zn−1z−1 =

1−1z−1 = 0 as required.

Exercise 6.Show thatϕ(t) = cis t is a group homomorphism of the additive groupR onto the multiplicativegroup T= z : |z| = 1.


Exercise 7. If z ∈ C and Re(zn) ≤ 0 for every positive integer n, show that z is a non-negative real number.

Solution. Let n be an arbitrary but fixed positive integer and let z∈ C and Re(zn) ≥ 0. Since zn =rn(cos(nθ) + i sin(nθ)), we have

Re(zn) = rn cos(nθ) ≥ 0.

If z = 0, then we are done, since r= 0 and Re(zn) = 0. Therefore, assume z, 0, then r> 0. Thus

Re(zn) = rn cos(nθ) ≥ 0

implies cos(nθ) ≥ 0 for all n. This impliesθ = 0 as we will show next. Clearly,θ < [π/2,3π/2]. Ifθ ∈ (0, π/2), then there exists a k∈ 2,3, . . . such that π

k+1 ≤ θ < πk . If we choose n= k+ 1, we have

π ≤ nθ <(k+ 1)π

k

which is impossible sincecos(nθ) ≥ 0. Similarly, we can derive a contradiction if we assumeθ ∈ (3π/2,2π).Then2π − π/k ≤ θ < 2π − π/(k+ 1) for some k∈ 2,3, . . ..

1.5 Lines and half planes in the complex plane

Exercise 1. Let C be the circlez : |z− c| = r, r > 0; let a = c+ rcisα and put

Lβ =

z : Im(z− a

b

)

= 0

where b= cisβ. Find necessary and sufficient conditions in terms ofβ that Lβ be tangent to C at a.


1.6 The extended plane and its spherical representation

Exercise 1. Give the details in the derivation of (6.7) and (6.8).


Exercise 2. For each of the following points inC, give the corresponding point of S :0,1+ i,3+ 2i.

Solution. We have

Φ(z) =

(

2x|z|2 + 1

,2y|z|2 + 1

,|z|2 − 1|z|2 + 1

)

.

If z1 = 0, then|z1| = 0 and thereforeΦ(z1) = (0,0,−1).

7

Thus, z1 = 0 corresponds to the point(0,0,−1) on the sphere S .If z2 = 1+ i, then|z2| =

√2 and therefore

Φ(z2) =

(

23,23,13

)

.

Thus, z2 = 1+ i corresponds to the point(

23 ,

23 ,

13

)

on the sphere S .

If z3 = 3+ 2i, then|z3| =√

13and therefore

Φ(z3) =

(

37,27,67

)

.

Thus, z3 = 3+ 2i corresponds to the point(

37 ,

27 ,

67

)

on the sphere S .

Exercise 3. Which subsets of S correspond to the real and imaginary axes inC.

Solution. If z is on the real axes, then z= x which implies|z|2 = x2. Thus

Φ(z) =

(

2xx2 + 1

,0,x2 − 1x2 + 1

)

.

Therefore the set(

2xx2 + 1

,0,x2 − 1x2 + 1

)

|x ∈ R

⊂ S

corresponds to the real axes inC. That means the unit circle x2+ z2= 1 lying in the xz−plane corresponds

to the real axes inC.If z is on the imaginary axes, then z= iy which implies|z|2 = y2. Thus

Φ(z) =

(

0,2y

y2 + 1,y2 − 1y2 + 1

)

.

Therefore the set(

0,2y

y2 + 1,y2 − 1y2 + 1

)

|y ∈ R

⊂ S

corresponds to the imaginary axes inC. That means the unit circle y2+ z2

= 1 lying in the yz−planecorresponds to the imaginary axes inC.

Exercise 4. LetΛ be a circle lying in S . Then there is a unique plane P inR3 such that P∩ S = Λ. Recallfrom analytic geometry that

P = (x1, x2, x3) : x1β1 + x2β2 + x3β3 = l

where(β1, β2, β3) is a vector orthogonal to P and l is some real number. It can be assumed thatβ21+β

22+β

23 =

1. Use this information to show that ifΛ contains the point N then its projection onC is a straight line.Otherwise,Λ projects onto a circle inC.


Exercise 5. Let Z and Z′ be points on S corresponding to z and z′ respectively. Let W be the point on Scorresponding to z+ z′. Find the coordinates of W in terms of the coordinates of Z andZ′.


8

Chapter 2

Metric Spaces and the Topology ofC

2.1 Definitions and examples of metric spaces

Exercise 1. Show that each of the examples of metric spaces given in (1.2)-(1.6) is, indeed, a metric space.Example (1.6) is the only one likely to give any difficulty. Also, describe B(x; r) for each of these examples.


Exercise 2. Which of the following subsets ofC are open and which are closed: (a)z : |z| < 1; (b) thereal axis; (c)z : zn

= 1 for some integer n≥ 1; (d) z ∈ C is real and0 ≤ z< 1; (e) z ∈ C : z is real and0 ≤ z≤ 1?

Solution. We have

(a) A := z ∈ C : |z| < 1Let z∈ A and setεz =

1−|z|2 , then B(z, εz) ⊂ A is open and therefore A=

⋃

z∈A B(z, εz) is open also. Acannot be closed, otherwise A andC − A were both closed and open sets yetC is connected.

(b) B := z ∈ C : z= x+ iy, y = 0 (the real axis)

Let z ∈ C − B, thenImz , 0. Setεz =|Imz|

2 , then B(z, εz) ⊂ C − B.Hence B is closed since itscomplement is open.

For any real x and anyε > 0 the point x+ i ε2 ∈ B(x, ε) but x+ i ε2 ∈ C − R. Thus B is not open.

(c) C := z ∈ C : zn= 1 for some integer n≥ 1

Claim: C is neither closed nor open.

C is not open because if zn= 1 then z= rcis(θ) with r = 1 and anyε-ball around z would contain an

element z′ := (1+ ε4)cis(θ).

To show that C cannot be closed, note that forpq ∈ Q with p ∈ Z and q ∈ N the number z:=

cis(

pq2π

)

∈ C sincezq= cis(p2π) = cos(p2π) + i sin(p2π) = 1.

Now fix x∈ R − Q and letxnn be a rational sequence that converges to x. Let m be any naturalnumber. Now zm = 1 implies thatsin(mx2π) = 0 which in turn means that mx∈ Z contradicting thechoice x∈ R − Q. We have constructed a sequence of elements of C that converges to a point that isnot element of C. Hence C is not closed.

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(d) D := z ∈ C : z is real and0 ≤ z< 1The set cannot be open by the argument given in2.1) and it is not closed because zn := 1 − 1

n is asequence in D that converges to a point outside of D.

(e) E := z ∈ C : z is real and0 ≤ z≤ 1This set E is not open by the observation in2.1) and it is closed because its complement is open: Ifz , E and z real, then B(z, min|x|,|x−1|

2 ) is contained in the complement of E, if z is imaginary then

B(z, |Imy|2 ) is completely contained in the complement of E.

Exercise 3. If (X,d) is any metric space show that every open ball is, in fact, an open set. Also, show thatevery closed ball is a closed set.


Exercise 4. Give the details of the proof of (1.9c).


Exercise 5. Prove Proposition 1.11.


Exercise 6. Prove that a set G⊂ X is open if and only if X−G is closed.


Exercise 7. Show that(C∞,d) where d is given (I. 6.7) and (I. 6.8) is a metric space.

Solution. To show(C∞,d) with d(z, z′) = 2|z−z′ |[(1+|z|2)(1+|z′ |2)]1/2 for z, z′ ∈ C and d(z,∞) = 2

(1+|z|2)1/2 , z ∈ C is ametric space.i) d(z, z′) ≥ 0.Since|z−z′| ≥ 0, we have d(z, z′) = 2|z−z′ |

[(1+|z|2)(1+|z′ |2)]1/2 ≥ 0 for all z, z′ ∈ C (the denominator is always positive).

Obviously, d(z,∞) = 2(1+|z|2)1/2 ≥ 0 for all z ∈ C.

ii) d(z, z′) = 0 iff z= z′.We have2|z − z′| = 0 iff z = z′. Therefore, d(z, z′) = 2|z−z′ |

[(1+|z|2)(1+|z′ |2)]1/2 = 0 iff z = z′. d(∞,∞) =

limz→∞2

(1+|z|2)1/2 = 0. Thus, d(∞,0) = 0 iff z= ∞.iii) d (z, z′) = d(z′, z).We have d(z, z′) = 2|z−z′ |

[(1+|z|2)(1+|z′ |2)]1/2 =2|z′−z|

[(1+|z′ |2)(1+|z|2)]1/2 = d(z′, z) for all z, z′ ∈ C. Also d(z,∞) = 2(1+|z|2)1/2 =

d(∞, z) by the symmetry of d(z, z′) in general.iv) d(z, z′) ≤ d(z, x) + d(x, z′).We have

d(z, z′) =2|z− z′|

[(1 + |z|2)(1+ |z′|2)]1/2

≤ 2|z− x|[(1 + |z|2)(1+ |x|2)]1/2

+2|x− z′|

[(1 + |x|2)(1+ |z′|2)]1/2

= d(z, x) + d(x, z′)

for all x, z, z′ ∈ C. The first inequality will be shown next.First, we need the following equation

(z− z′)(1+ xx) = (z− x)(1+ xz′) + (x− z′)(1+ xz), (2.1)

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which follows from a simple computation

(z− x)(1+ xz′) + (x− z′)(1+ xz) = z− x+ xz′z+ xxz′ + x− z′ + xxz− xz′z

= z+ xxz− z′ − xxz′

= (z− z′)(1+ xx).

Using (2.1) and taking norms gives

|(z− z′)(1+ xx)| = |(z− z′)(1+ |x|2)| = |(z− z′)|(1+ |x|2)

= |(z− x)(1+ xz′) + (x− z′)(1+ xz)|≤ |z− x|(1+ |x|2)1/2(1+ |z′|2)1/2

+ |x− z′|(1+ |x|2)1/2(1+ |z|2)1/2 (2.2)

where the last inequality follows by

|z+ xz′| ≤ (1+ |x|2)1/2(1+ |z′|2)1/2

⇔ (1+ xz′)(1+ xz′) ≤ (1+ |x|2)(1+ |z′|2)

and

|z+ xz| ≤ (1+ |x|2)1/2(1+ |z|2)1/2

⇔ (1+ xz)(1+ xz) ≤ (1+ |x|2)(1+ |z|2)

since

(1+ xz)(1+ xz) ≤ (1+ |x|2)(1+ |z|2)

⇔ xz+ xz≤ |x|2 + |y|2

⇔ 2Re(xz) ≤ |x|2 + |y|2

which is true by Exercise 4 part 2 on page 3. Thus, dividing (2.2) by (1+ |x|2)1/2 yields

|z− z′|(1+ |x|2)1/2 ≤ |z− x|(1+ |z′|2)1/2+ |x− z′|(1+ |z|2)1/2.

Multiplying this by2

(1+ |x|2)1/2(1+ |z′|2)1/2(1+ |z|2)1/2

gives the assertion above.

Exercise 8. Let (X,d) be a metric space and Y⊂ X. Suppose G⊂ X is open; show that G∩ Y is open in(Y,d). Conversely, show that if G1 ⊂ Y is open in(Y,d), there is an open set G⊂ X such that G1 = G∩ Y.

Solution. Set G1 = G ∩ Y, let G be open in(X,d) and Y⊂ X. In order to show that G1 is open in(Y,d),pick an arbitrary point p∈ G1. Then p∈ G and since G is open, there exists anε > 0 such that

BX(p; ε) ⊂ G.

But thenBY(p; ε) = BX(p; ε) ∩ Y ⊂ G∩ Y = G1

which proves that p is an interior point of G1 in the metric d. Thus G1 is open in Y (Proposition 1.13a).

Let G1 be an open set in Y. Then, for every p∈ G1, there exists anε−ball

BY(p : ε) ⊂ G1.

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ThusG1 =

⋃

p∈G1

BY(p; ε).

Since we can writeBY(p; ε) = BX(p; ε) ∩ Y,

we getG1 =

⋃

p∈G1

BY(p; ε) =⋃

p∈G1

(

BX(p; ε) ∩ Y)

=

⋃

p∈G1

BX(p; ε) ∩ Y = G∩ Y

where G=⋃

p∈G1BX(p; ε) is open in(X,d). (Proposition 1.9c since each BX(p; ε) is open).

Exercise 9. Do Exercise 8 with “closed” in place of “open.”


Exercise 10.Prove Proposition 1.13.


Exercise 11. Show thatcis k : k a non-negative integer is dense in T= z ∈ C : |z| = 1. For whichvalues ofθ is cis(kθ) : k a non-negative integer dense in T?


2.2 Connectedness

Exercise 1. The purpose of this exercise is to show that a connected subset ofR is an interval.

(a) Show that a set A⊂ R is an interval iff for any two points a and b in A with a< b, the interval[a,b] ⊂ A.

(b) Use part (a) to show that if a set A⊂ R is connected then it is an interval.


Exercise 2. Show that the sets S and T in the proof of Theorem 2.3 are open.


Exercise 3. Which of the following subsets X ofC are connected; if X is not connected, what are itscomponents: (a) X= z : |z| ≤ 1 ∪ z : |z− 2| < 1. (b) X = [0,1)∩

1+ 1n : n ≥ 1

. (c) X = C − (A∩ B)where A= [0,∞) and B= z= r cis θ : r = θ,0 ≤ θ ≤ ∞?

Solution. a) Define X= z : |z| ≤ 1∪ z : |z−2| < 1 := A∪B. It suffices to show that X is path-connected.Obviously A is path-connected and B is path-connected. Next, we will show that X is path-connected.Recall that a space is path-connected if for any two points x and y there exists a continuous function f fromthe interval[0,1] to X with f(0) = x and f(1) = y (this function f is called the path from x to y).Let x∈ A and y∈ B and define the function f(t) : [0,1]→ X by

f (t) =

(1− 3t)x+ 3tRe(x), 0 ≤ t ≤ 13

(2− 3t)Re(x) + (3t − 1)Re(y), 13 < t ≤ 2

3

(3− 3t)Re(y) + (3t − 2)y, 23 < t ≤ 1.

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This function is obviously continuous, since f(1/3) = limt→ 13− f (t) = limt→ 1

3+ f (t) = Re(x) ∈ X and

f (2/3) = limt→ 23− f (t) = limt→ 2

3+ f (t) = Re(y) ∈ X. In addition, we have f(0) = x and f(1) = y. Therefore

X is path-connected and hence X is connected.b) There is no way to connect2 and 32. Therefore X is not connected. The components are[0,1), 2, 32, 43, . . ., 1+ 1

n.c) X is not connected, since there is no way to connect(2,1) and(1,−2). The k− th component is given byC − A∩ B where

A = [2πk− 2π,2πk)

andB = z= r cis θ : r = θ,2πk− 2π ≤ θ < 2πk, k ∈ 1,2,3, . . ..

Exercise 4. Prove the following generalization of Lemma 2.6. IfD j : j ∈ J is a collection of connectedsubsets of X and if for each j and k in J we have Dj ∩ Dk , then D=

⋃D j : j ∈ J is connected.

Solution. Let D=⋃

j∈J D j andC = D j : j ∈ J. If D is connected, we could write D as the disjoint unionA∪ B where A and B are nonempty subsets of X. Thus, for each C∈ C either C⊂ A or C ⊂ B.We have C⊂ A ∀C ∈ C or C ⊂ B ∀C ∈ C. If not, then there exist E, F ∈ C such that E⊂ A and F⊂ B.But, we assume A∪ B is disjoint and thus E∪ F is disjoint which contradicts the assumption E∩ F , ∅.Therefore, all members ofC are contained in either A or all B. Thus, either D= A and B= ∅ or D = B andA = ∅. Both contradicting the fact that A, B are assumed to be nonempty. Hence,

D =⋃

j∈JD j

is connected.

Exercise 5. Show that if F⊂ X is closed and connected then for every pair of points a, b in Fand eachε > 0 there are points z0, z1, . . . , zn in F with z0 = a, zn = b and d(zk−1, zk) < ε for 1 ≤ k ≤ n. Is thehypothesis that F be closed needed? If F is a set which satisfies this property then F is not necessarilyconnected, even if F is closed. Give an example to illustratethis.


2.3 Sequences and completeness


Solution. a) A set is closed iff it contains all its limit points.Let S⊂ X be a set.“⇐”: Assume S contains all its limit points. We have to show thatS is closed or that Sc is open. Letx ∈ Sc. By assumption x is not a limit point and hence there exists anopenε−ball around x, B(x; ε) suchthat B(x; ε) ∩ S = ∅ (negation of Proposition 1.13f). So B(x; ε) ⊂ Sc and therefore Sc is open.“⇒”: Let S be closed and x be a limit point. We claim x∈ S . If not, Sc (open) would be an openneighborhood of x, that does not intersect S . (Proposition 1.13f again) which contradicts the fact that x isa limit point of S .b) If A ⊂ X, then A− = A∪ x : x is a limit point of A := A∪ A′.Let A⊂ X be a set.“ ⊆”: Let x ∈ A. We want to show x∈ A∪ A′. If x ∈ A, then obviously x∈ A∪ A′. Suppose x< A. Sincex ∈ A, we have (Proposition 1.13f) that for everyε > 0, B(x; ε) ∩ A , ∅. Because x< A, B(x; ε) must

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intersect A in a point different from x. In particular, for every integer n there is a point xn in B(x; 1n) ∩ A.

Thus d(x, xn) < 1n which implies xn→ x (see Proof of Proposition 3.2). Then x∈ A′, so x∈ A∪ A′.

“ ⊇”: To show A∪ A′ ⊆ A−. Let x∈ A∪ A′. If x ∈ A, then x∈ A− (A ⊂ A−). Now, assume x∈ A′ but notin A. Then there existsxn ⊂ A with limn→∞ xn = x. It follows,∀ε > 0 B(x; ε) ∩ A , ∅ sincexn ⊂ A. ByProposition 1.13f we get x∈ A−.

Exercise 2. Furnish the details of the proof of Proposition 3.8.


Exercise 3. Show that diam A= diam A−.


Exercise 4. Let zn, z be points inC and let d be the metric onC∞. Show that|zn − z| → 0 if and only ifd(zn, z)→ 0. Also show that if|zn| → ∞ thenzn is Cauchy inC∞. (Mustzn converge inC∞?)

Solution. First assume that|zn − z| → 0, then

d(zn, z) =2

√

(1+ |zn|2)(1+ |z|2)|zn − z| → 0

since the denominator√

(1+ |zn|2)(1+ |z|2) ≥ 1 is bounded below away from 0. To see the converse, letd(zn, z) → 0 or equivalently d2(zn, z) → 0. We need to show that if zn → z in the d-norm, then|zn| 9 ∞because otherwise the denominator grows without bounds. Infact we will show that if|zn| → ∞, thend(zn, z)9 0 for any z∈ C. Then

d2(zn, z) =4(

|zn|2 − znz− znz+ |z|2)

(

1+ |zn|2) (

1+ |z|2)

=

4(

|zn|2 − 2Re(znz) + |z|2)

(

1+ |z|2) |zn|2 + 1+ |z|2

=

4(

1− 2Re(znz)|zn|2 +

|z|2|zn|2

)

1+ |z|2 + 1+|z|2|zn|2

if |zn| , 0

in particular if |zn| → ∞, d(zn, z) → 41+|z|2 , 0. This shows that the denominator remains bounded as

d(zn, z)→ 0 and therefore the numerator2|zn − z| → 0. Hence convergence in d-norm implies convergencein |·|-norm for numbers zn, z ∈ C. Next assume that|zn| → ∞. Then clearly also

√

1+ |zn|2 → ∞ andtherefore d(zn,∞) = 2√

1+|zn|2→ 0. The last thing to show is that if zn → ∞ in (C∞,d), also |zn| → ∞. But

d(zn,∞)→ 0 implies√

1+ |zn|2→ ∞ which is equivalent to|zn| → ∞.

Exercise 5. Show that every convergent sequence in(X,d) is a Cauchy sequence.

Solution. Let xn be a convergent sequence with limit x. That is, givenε > 0 ∃N such that d(xn, x) < ε2 if

n > N and d(x, xm) < ε2 if m > N. Thus

d(xn, xm) ≤ d(xn, x) + d(x, xm) <ε

2+ε

2= ε, ∀n,m≥ N

and thereforexn is a Cauchy sequence.

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Exercise 6. Give three examples of non complete metric spaces.

Solution. Example 1: Let X= C[−1,1] and the metric d( f ,g) =√

∫ 1

−1[ f (x) − g(x)]2 dx, f,g ∈ X. Consider

the Cauchy sequence

fn(x) =

0, −1 ≤ x ≤ 0

nx, 0 < x ≤ 1n

1, 1n < x ≤ 1

.

It is obvious that the limit function f is discontinuous. Hence, the metric space(X,d) is not complete.Example 2: Let X= (0,1] with metric d(x, y) = |x− y|, x, y ∈ X. The sequence 1n is Cauchy, but convergesto 0, which is not in the space. Thus, the metric space is not complete.Example 3: Let X= Q, the rationals, with metric d(x, y) = |x−y|, x, y ∈ X. The sequence defined by x1 = 1,xn+1 =

xn

2 +1xn

is a Cauchy sequence of rational numbers. The limit√

2 is not a rational number. Therefore,the metric space is not complete.

Exercise 7. Put a metric d onR such that|xn − x| → 0 if and only if d(xn, x)→ 0, but thatxn is a Cauchysequence in(R,d) when|xn| → ∞. (Hint: Take inspiration fromC∞.)


Exercise 8. Supposexn is a Cauchy sequence andxnk is a subsequence that is convergent. Show thatxn must be convergent.

Solution. Sincexnk is convergent, there is a x such that xnk → x as k→ ∞. We have to show that xn→ xas n→ x. Letε > 0. Then we have∃N ∈ N such that d(xn, xm) < ε

2 ∀n,m ≥ N sincexn is Cauchy and∃M ∈ N such that d(xnk , x) < ε

2 ∀nk ≥ M since xnk → x as k→ ∞. Now, fix nk0 > M + N, then

d(xn, x) ≤ d(xn, xnk0) + d(xnk0

, x) <ε

2+ε

2= ε, ∀n ≥ N.

Thus, xn→ x as n→ ∞ and thereforexn is convergent.

2.4 Compactness

Exercise 1. Finish the proof of Proposition 4.4.


Exercise 2. Let p = (p1, . . . , pn) and q = (q1, . . . ,qn) be points inRn with pk < qk for each k. LetR= [p1,q1] × . . . × [pn,qn] and show that

diam R= d(p,q) =

n∑

k=1

(qk − pk)2

12

.

Solution. By definitiondiam(R) = sup

x∈R,y∈Rd(x, y).

Obviously, R is compact, so we have

diam(R) = maxx∈R,y∈R

d(x, y).

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Let x = (x1, . . . , xn) ∈ R and y= (y1, . . . , yn) ∈ R. Then clearly, pi ≤ xi ≤ qi and pi ≤ yi ≤ qi for alli = 1, . . . ,n. We also have

(yi − xi)2 ≤ (qi − pi)

2, ∀i = 1, . . . ,n. (2.3)

Therefore, by (2.3) we obtain

d(x, y) =

√√n∑

i=1

(yi − xi)2 ≤

√√n∑

i=1

(qi − pi)2 = d(p,q).

Note that we get equality if for example xi = pi and yi = qi for all i = 1, . . . ,n. Thus the maximum distanceis obtained, so

diam(R) = supx∈R,y∈R

d(x, y) =

√√n∑

i=1

(qi − pi)2 = d(p,q).

Exercise 3. Let F = [a1,b1] × . . . × [an,bn] ⊂ Rn and letε > 0; use Exercise 2 to show that there arerectangles R1, . . . ,Rm such that F=

⋃mk=1 Rk and diam Rk < ε for each k. If xk ∈ Rk then it follows that

Rk ⊂ B(xk; ε).


Exercise 4. Show that the union of a finite number of compact sets is compact.

Solution. Let K =⋃n

i=1 Ki be a finite union of compact sets. LetGλλ∈Γ be an open cover of K, that is

K ⊂⋃

λ∈ΓGλ.

Of course,Gλλ∈Γ is an open cover of each Ki , i = 1, . . . ,n, that is

Ki ⊂⋃

λ∈ΓGλ, ∀i = 1, . . . ,n.

Since each Ki is compact,

Ki ⊂ki⋃

j=1

Gi, j

for each i= 1, . . . ,n. Thus

K =n⋃

j=1

Ki ⊂n⋃

i=1

ki⋃

j=1

Gi, j

which is a finite union and therefore K is compact.

Exercise 5. Let X be the set of all bounded sequences of complex numbers. That is,xn ∈ X iff sup|xn| :n ≥ 1 < ∞. If x = xn and y= yn, define d(x, y) = sup|xn − yn| : n ≥ 1. Show that for each x in X andε > 0, B(x; ε) is not totally bounded although it is complete. (Hint: you might have an easier time of it ifyou first show that you can assume x= (0,0, . . .).)


Exercise 6. Show that the closure of a totally bounded set is totally bounded.

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Solution. Let (X,d) be a given metric space and let S⊂ X be totally bounded. Letε > 0. Since S is totallybounded, we have by Theorem 4.9 d) p. 22 that there exist a finite number of points x1, . . . , xn ∈ S such that

S ⊆n⋃

k=1

B(xk; ε/2).

Taking the closure, gives the desired result

S ⊆n⋃

k=1

B(xk; ε).

2.5 Continuity



Exercise 2. Show that if f and g are uniformly continuous (Lipschitz) functions from X intoC then so isf + g.

Solution. The distance inC is ρ(x, y) = |x − y|. The distance in X is d(x, y). Let f : X → C be Lipschitz,that is, there is a constant M> 0 such that

ρ( f (x), f (y)) = | f (x) − f (y)| ≤ Md(x, y), ∀x, y ∈ X

and g: X→ C be Lipschitz, that is, there is a constant N> 0 such that

ρ(g(x),g(y)) = |g(x) − g(y)| ≤ Nd(x, y), ∀x, y ∈ X.

We have

ρ( f (x) + g(x), f (y) + g(y)) = | f (x) + g(x) − f (y) − g(y)| = | f (x) − f (y) + g(x) − g(y)|≤ | f (x) − f (y)| + |g(x) − g(y)|≤ Md(x, y) + Nd(x, y) = (M + N)d(x, y), ∀x, y ∈ X.

Since there is a constant K= M + N > 0 such that

ρ( f (x) + g(x), f (y) + g(y)) ≤ Kd(x, y), ∀x, y ∈ X,

we have that f+ g is Lipschitz.

Now, let f,g be both uniformly continuous, that is,

∀ε1 > 0 ∃δ1 > 0 such thatρ( f (x), f (y)) = | f (x) − f (y)| < ε1 whenever d(x, y) < δ1

and∀ε2 > 0 ∃δ2 > 0 such thatρ( f (x), f (y)) = | f (x) − f (y)| < ε2 whenever d(x, y) < δ2.

We have

ρ( f (x) + g(x), f (y) + g(y)) = | f (x) + g(x) − f (y) − g(y)| = | f (x) − f (y) + g(x) − g(y)|≤ | f (x) − f (y)| + |g(x) − g(y)|≤ ε1 + ε2,

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whenever d(x, y) < δ1 and d(x, y) < δ2, that is, whenever d(x, y) < min(δ1, δ2). So, choosingε = ε1+ ε2 andδ = min(δ1, δ2), we have shown that

∀ε > 0 ∃δ > 0 such thatρ( f (x), f (y)) = | f (x) − f (y)| < ε whenever d(x, y) < δ.

Thus f+ g is uniformly continuous.

Exercise 3. We say that f: X → C is bounded if there is a constant M> 0 with | f (x)| ≤ M for all x in X.Show that if f and g are bounded uniformly continuous (Lipschitz) functions from X intoC then so is f g.

Solution. Let f be bounded, that is, there exists a constant M1 > 0 with | f (x)| < M1 for all x ∈ X and let gbe bounded, that is, there exists a constant M2 > 0 with |g(x)| < M2 for all x ∈ X. Obviously f g is bounded,because

| f (x)g(x)| ≤ | f (x)| |g(x)| ≤ M1·M2 ∀x ∈ X.

So, there exists a constant M= M1M2 with

| f (x)g(x)| ≤ M ∀x ∈ X.

Let f be Lipschitz, that is, there exists a constant N1 > 0 such that

ρ( f (x), f (y)) ≤ N1d(x, y) ∀x, y ∈ X

and let g be Lipschitz, that is, there exists a constant N2 > 0 such that

ρ(g(x),g(y)) ≤ N1d(x, y) ∀x, y ∈ X.

Now,

ρ( f (x)g(x), f (y)g(y)) = | f (x)g(x) − f (y)g(y)| == | f (x)g(x) − f (x)g(y) + f (x)g(y) − f (y)g(y)|≤ | f (x)g(x) − f (x)g(y)| + | f (x)g(y) − f (y)g(y)|≤ | f (x)| |g(x) − g(y)| + | f (x)| |g(y) − g(y)|≤ M1N2d(x, y) + M2N1d(x, y) = (M1N2 + M2N1)d(x, y) ∀x, y ∈ X.

Thus, there exists a constant K= M1N2 + M2N1 > 0 with

ρ( f (x)g(x), f (y)g(y)) ≤ Kd(x, y) ∀x, y ∈ X,

so f g is Lipschitz and bounded.

Now, let f and g be both bounded and uniformly continuous, that is

∃ M1 such that| f (x)| ≤ M1 ∀x ∈ X

∃ M2 such that|g(x)| ≤ M2 ∀x ∈ X

∀ ε1 > 0 ∃δ1 > 0 such thatρ( f (x), f (y)) = | f (x) − f (y)| < ε1 whenever d(x, y) < δ1

∀ ε2 > 0 ∃δ2 > 0 such thatρ( f (x), f (y)) = | f (x) − f (y)| < ε2 whenever d(x, y) < δ2.

18

It remains to verify that f g is uniformly continuous, since we have already shown that f g is bounded. Wehave

ρ( f (x)g(x), f (y)g(y)) = | f (x)g(x) − f (y)g(y)| == | f (x)g(x) − f (x)g(y) + f (x)g(y) − f (y)g(y)|≤ | f (x)g(x) − f (x)g(y)| + | f (x)g(y) − f (y)g(y)|≤ | f (x)| |g(x) − g(y)| + | f (x)| |g(y) − g(y)|≤ M1ε2 + M2ε1,

whenever d(x, y) < min(δ1, δ2). So choosingε = M1ε2 + M2ε1 andδ = min(δ1, δ2), we have∀ε > 0 ∃δ > 0such that

| f (x)g(x) − f (y)g(y)| < εwhenever d(x, y) < δ. Thus, f g is uniformly continuous and bounded.

Exercise 4. Is the composition of two uniformly continuous (Lipschitz)functions again uniformly continu-ous (Lipschitz)?


Exercise 5. Suppose f: X→ Ω is uniformly continuous; show that ifxn is a Cauchy sequence in X then f (xn) is a Cauchy sequence inΩ. Is this still true if we only assume that f is continuous? (Prove or givea counterexample.)

Solution. Assume f: X → Ω is uniformly continuous, that is, for everyε > 0 there existsδ > 0 such thatρ( f (x), f (y)) < ε if d(x, y) < δ. If xn is a Cauchy sequence in X, we have, for everyε1 > 0 there existsN ∈ N such that d(xn, xm) < ε1 for all n,m≥ N. But then, by the uniform continuity, we have that

ρ( f (xn), f (xm)) < ε ∀n,m≥ N

whenever d(xn, xm) < δ which tells us that f (xn) is a Cauchy sequence inΩ.If f is continuous, the statement is nottrue. Here is a counterexample: Let f(x) = 1

x which is continuouson (0,1). The sequence xn =

1n is apparently convergent and therefore a Cauchy sequence inX. But

f (xn) = f ( 1n) = n is obviously not Cauchy. Note that f(x) = 1

x is not uniformly continuous on(0,1).To see that pickε = 1. Then there is noδ > 0 such that| f (x)− f (y)| < 1 whenever|x− y| < δ. Assume thereexists such aδ. WLOG assumeδ < 1 since the interval(0,1) is considered. Let y= x+ δ/2 and set x= δ/2,then

| f (x) − f (y)| =∣∣∣∣∣

1x− 1

y

∣∣∣∣∣=

y− xxy=

δ/2δ/2· δ =

1δ> 1,

that is no matter whatδ < 1 we choose, we always obtain| f (x) − f (y)| > 1. Therefore f(x) = 1x cannot be

uniformly continuous.

Exercise 6. Recall the definition of a dense set (1.14). Suppose thatΩ is a complete metric space and thatf : (D,d) → (Ω; ρ) is uniformly continuous, where D is dense in(X,d). Use Exercise 5 to show that thereis a uniformly continuous function g: X→ Ω with g(x) = f (x) for every x in D.


Exercise 7. Let G be an open subset ofC and let P be a polygon in G from a to b. Use Theorems 5.15and 5.17 to show that there is a polygon Q⊂ G from a to b which is composed of line segments which areparallel to either the real or imaginary axes.

19


Exercise 8. Use Lebesgue’s Covering Lemma (4.8) to give another proof ofTheorem 5.15.

Solution. Suppose f: X → Ω is continuous and X is compact. To show f is uniformly continuous. Letε > 0. Since f is continuous, we have for all x∈ X there is aδx > 0 such thatρ( f (x), f (y)) < ε/2 wheneverd(x, y) < δx. In addition,

X =⋃

x∈XB(x; δx)

is an open cover of X. Since X is by assumption compact (it is also sequentially compact as stated inTheorem 4.9 p. 22), we can use Lebesgue’s Covering Lemma 4.8 p. 21 to obtain aδ > 0 such that x∈ Ximplies that B(x, δ) ⊂ B(z; δz) for some z∈ X. More precisely, x, y ∈ B(z; δz) and therefore

ρ( f (x), f (z)) ≤ ρ( f (x), f (z)) + ρ( f (z), f (y)) <ε

2+ε

2= ε

and hence f is uniformly continuous on X.

Exercise 9. Prove the following converse to Exercise 2.5. Suppose(X,d) is a compact metric space havingthe property that for everyε > 0 and for any points a, b in X, there are points z0, z1, . . . , zn in X with z0 = a,zn = b, and d(zk−l , zk) < ε for 1 ≤ k ≤ n. Then(X,d) is connected. (Hint: Use Theorem 5.17.)


Exercise 10. Let f and g be continuous functions from(X,d) to (Ω, p) and let D be a dense subset of X.Prove that if f(x) = g(x) for x in D then f= g. Use this to show that the function g obtained in Exercise 6is unique.


2.6 Uniform convergence

Exercise 1. Let fn be a sequence of uniformly continuous functions from(X,d) into (Ω, ρ) and supposethat f = u − − lim fn exists. Prove that f is uniformly continuous. If each fn is a Lipschitz function withconstant Mn andsupMn < ∞, show that f is a Lipschitz function. IfsupMn = ∞, show that f may fail tobe Lipschitz.


20

Chapter 3

Elementary Properties and Examples ofAnalytic Functions

3.1 Power series



Exercise 2. Give the details of the proof of Proposition 1.6.


Exercise 3.Prove thatlim sup(an+bn) ≤ lim supan+ lim supbn andlim inf(an+bn) ≥ lim inf an+ lim inf bn

for bounded sequences of real numbersan andbn.

Solution. Let r > lim supn→∞ an (we know there are only finitely many by definition) and let s> lim supn→∞(same here, there are only finitely many by definition). Then r+ s > an + bn for all but finitely many n’s.This however, implies that

r + s≥ lim supn→∞

(an + bn).

Since this holds for any r> lim supn→∞ an and s> lim supn→∞ bn, we have

lim supn→∞

(an + bn) ≤ lim supn→∞

an + lim supn→∞

bn.

Let r < lim inf n→∞ an (we know there are only finitely many by definition) and let s< lim inf n→∞ (samehere, there are only finitely many by definition). Then r+ s < an + bn for all but finitely many n’s. Thishowever, implies that

r + s≤ lim infn→∞

(an + bn).

Since this holds for any r< lim inf n→∞ an and s< lim inf n→∞ bn, we have

lim infn→∞

(an + bn) ≥ lim infn→∞

an + lim infn→∞

bn.

Exercise 4. Show thatlim inf an ≤ lim supan for any sequence inR.

21

Solution. Let m = lim inf n→∞ an and bn = inf an,an+1, . . .. Let M = lim supn→∞ an. Take any s> M.Then, by definition of thelim supn→∞ an = M, we obtain that an < s for infinitely many n’s which impliesthat bn < s for all n and hencelim supn→∞ bn = m < s. This holds for all s> M. But the infimum of allthese s’s is M. Therefore m≤ M which is

lim infn→∞

an ≤ lim supn→∞

an.

Exercise 5. If an is a convergent sequence inR and a= lim an, show that a= lim inf an = lim supan.

Solution. Suppose thatan is a convergent sequence inR with limit a = limn→∞ an. Then by definition, wehave:∀ε > 0 ∃N > 0 such that∀n ≥ N, we have|an − a| ≤ ε, that is a− ε ≤ an ≤ a+ ε. This means that allbut finitely many an’s are≤ a+ ε and≥ a− ε. This shows that

a− ε ≤ lim infn→∞

an︸︷︷︸

=:m

≤ a+ ε

anda− ε ≤ lim sup

n→∞an

︸︷︷︸

=:M

≤ a+ ε.

By the previous Exercise 4, we also have

a− ε ≤ m≤ M ≤ a+ ε.

Hence,0 ≤ M −m≤ 2ε.

Sinceε > 0 is arbitrary, we obtain m= M and further

a− ε ≤ lim infn→∞

an ≤ lim supn→∞

an ≤ a+ ε,

we obtainlim inf

n→∞an = lim sup

n→∞an = a.

Exercise 6. Find the radius of convergence for each of the following power series: (a)∑∞

n=0 anzn, a ∈ C;(b)

∑∞n=0 an2

zn, a ∈ C; (c)∑∞

n=0 knzn, k an integer, 0; (d)∑∞

n=0 zn! .

Solution. a) We have∑∞

n=0 anzn=

∑∞k=0 bkzn with bk = ak, a ∈ C. We also have,

lim supk→∞

|bk|1/k = lim supk→∞

|ak|1/k = lim supk→∞

|a| = |a|.

Therefore, R= 1/|a|, so

R=

1|a| , a , 0

∞, a = 0.

b) In this case, bn = an2where a∈ C.

R = limn→∞

∣∣∣∣∣

bn

bn+1

∣∣∣∣∣= lim

n→∞

∣∣∣∣∣∣

an2

a(n+1)2

∣∣∣∣∣∣= lim

n→∞

∣∣∣∣∣∣

an2

an2+2n+1

∣∣∣∣∣∣= lim

n→∞

∣∣∣∣∣

1a2n+1

∣∣∣∣∣

= limn→∞

1|a|2n+1

=

0, |a| > 1

1, |a| = 1

∞, |a| < 1

.

22

c) Now, bn = kn, k is an integer, 0. We have

R= lim supn→∞

|bn|1/n = lim supn→∞

|kn|1/n = lim supn→∞

|k| = |k|.

So

R=1|k| =

1k , k > 0, k integer

− 1k , k < 0, k integer

.

d) We can write∑∞

n=0 zn!=

∑∞k=0 akzk where

ak =

0, k = 0

2, k = 1

1, k = n!,n ∈ N,n > 1

0, otherwise

Thus,lim sup

k→∞|ak|1/k = lim sup

k→∞|1|1/k!

= 1.

Therefore1/R= 1 which implies R= 1.

Exercise 7. Show that the radius of convergence of the power series∞∑

n=1

(−1)n

nzn(n+1)

is 1, and discuss convergence for z= 1,−1, and i. (Hint: The nth coefficient of this series is not(−1)n/n.)

Solution. Rewrite the power series in standard form, then

∞∑

n=1

(−1)n

nzn(n+1)

=

∞∑

n=1

akzk with ak =

(−1)n

n if ∃ n ∈ N s.t. k= n(n+ 1)

0 else.

To find the radius of convergence we use the root criterion andtherefore need the estimates

1 ≤ n(n+1)√

n ≤ n√

n for n ∈ N.

The first inequality is immediate from the fact that n≥ 1 and hence n1

n(n+1) ≥ 1. For the second inequalitynote that

n ≤ nn+1

⇔ n1

n(n+1) ≤ n1n

⇔ n(n+1)√

n ≤ n√

n .

Using this one obtains

n(n+1)

√

| (−1)n

n| = 1

n(n+1)√

n≤ 1

and

n(n+1)

√

| (−1)n

n| = 1

n(n+1)√

n≥ 1

n√

n.

23

Vague memories of calculus classes tell me thatn√

n→ 1, thus 1R = lim sup n(n+1)

√an = 1, i.e. R= 1.

If z = 1 the series reduces to∑∞

n=1(−1)n

n which converges with the Leibniz Criterion.If z = −1 we note that the exponents n(n + 1) are always even integers and therefore the series is the

same as in the previous case of z= 1.Now let z= i. The expression in(n+1) will always be real, so if the series converges at z= i, it converges

to a real number. We also note that formally

∞∑

n=1

(−1)n

nin(n+1)

=

∞∑

n=1

cn with

1n ifn mod 4∈ 0,1− 1

n if n mod 4∈ 2,3.

Define the partial sums Sk :=∑k

n=0 ck. We claim that the following chain of inequalities holds

0 ≤ S4k+3a)< S4k

b)< S4k+4

c)< S4k+2

d)< S4k+1 ≤ 1.

To verify this, note that

S4k+3 − S4k = c4k+1 + c4k+2 + c4k+3 = −16k2

+ 8k− 1(4k+ 1)(4k+ 2)(4k+ 3)

< 0, hence a)

S4k+4 − S4k = c4k+3 + c4k+4 = −1

4k+ 3+

14k+ 4

< 0, hence c)

S4k+2 − S4k+1 = c4k+2 < 0, hence d).

Relation b) is obvious and so are the upper bound c1 = 1 and the non-negativity constraint. We remark thatS4k+lk≥1, l ∈ 0,1,2,3 describe bounded and monotone subsequences that converge to some point. Nowthat |cn| 0 the difference between S4k+l and S4k+m, l,m ∈ 0,1,2,3 tends to zero, i.e. all subsequencesconverge to the same limit. Therefore the power series converges also in the case of z= i.

3.2 Analytic functions

Exercise 1. Show that f(z) = |z|2 = x2+ y2 has a derivative only at the origin.

Solution. The derivative of f at z is given by

f ′(z) = limh→0

f (z+ h) − f (z)h

, h ∈ C

provided the limit exist. We have

f (z+ h) − f (z)h

=|z+ h|2 − |z|2

h=

(z+ h)(z+ h) − zzh

=zz+ hz+ zh+ hh− zz

h

= z+ h+ zhh=: D.

If the limit of D exists, it may be found by letting the point h= (x, y) approach the origin (0,0) in the complexplaneC in any manner.1.) Take the path along the real axes, that is y= 0. Thenh = h and thus

D = z+ h+ zhh= z+ h+ z

24

and therefore, if the limit of D exists, its value has to bez+ z.2.) Take the path along the imaginary axes, that is x= 0. Thenh = −h and thus

D = z− h− zhh= z− h− z

and therefore, if the limit of D exists, its value has to bez− z.Because of the uniqueness of the limit of D, we must have

z+ z= z− z ⇐⇒ z= −z ⇐⇒ z= 0,

if the limit of D exists. It remains to show that the limit of D exists at z= 0. Since z= 0, we have that D= hand thus the limit of D is 0.In summary, the function f(z) = |z|2 = x2

+ y2 has a derivative only at the origin with value 0.

Exercise 2. Prove that if bn, an are real and positive and0 < b = lim bn < ∞, a = lim supan thenab= lim sup(anbn). Docs this remain true if the requirement of positivity is dropped?

Solution. Let a= lim supn→∞

an < ∞. Then there exists a monotonic subsequenceank of an that converges

to a. Sincelimk→∞

bnk = b, limn→∞

ankbnk = ab. Hence,ankbnk is a subsequence of anbn that converges to ab. So

ab≤ lim supn→∞

anbn. Hence,lim supn→∞

anbn ≥ b lim supn→∞

an.

Now, let a= lim supn→∞

an = ∞. Then there exists a subsequenceank of an such thatlimk→∞

ank = a > 0.

And, sincelimn→∞

bn > 0, limk→∞

ankbnk = ∞. Hencelim supn→∞

anbn = ∞. In this second case,lim supn→∞

anbn ≥b lim sup

n→∞an.

In both cases, we have established that ab≤ lim supn→∞

anbn. Now, since for all n∈ N, an > 0 and bn > 0,

consider limn→∞

1bn=

1b

. Applying the inequality we have established replacing bn with 1bn

and an replaced

with anbn:

lim supn→∞

an = lim supn→∞

1bn

(anbn) ≥ 1b

lim supn→∞

anbn.

Rearranging,

lim supn→∞

anbn ≤ b lim supn→∞

an.

It follows that ab= lim supn→∞

anbn as required.

Now, consider the case where we drop the positivity requirement. Let bn = 0,− 12 ,0,− 1

3 ,0,− 14 , . . . and

note0 = b = limn→∞

bn < ∞. Also let an = 0,−2,0,−3,0,−4 . . . and notelim supn→∞

an = a = 0. In this case,

ab= 0 , 1 = lim supn→∞

anbn.

Exercise 3. Show thatlim n1/n= 1.

25

Solution. Let n∈ N. Also let a= n1/n. Then

a = n1/n

⇐⇒ loga = logn1/n

⇐⇒ loga =logn

n

⇐⇒ limn→∞

loga = limn→∞

lognn

Now, let f(x) =log(x)

x. Then lim

n→∞f (x) = 0 by L’Hopital’s Rule. Thus,lim

n→∞logn

n= lim

n→∞f (n) = 0 also. So

limn→∞

a = limn→∞

n1/n= 1.

Exercise 4. Show that(cosz)′ = − sinz and(sinz)′ = cosz.

Solution. We have by definition

(cosz)′ =

(

12

(

eiz+ e−iz

))′=

i2

(

eiz − e−iz)

= − 12i

(

eiz − e−iz)

= − sinz

and similarly

(sinz)′ =

(

12i

(

eiz − e−iz))′=

i2i

(

eiz+ e−iz

)

=12

(

eiz+ e−iz

)

= cosz.

Exercise 5. Derive formulas (2.14).


Exercise 6. Describe the following sets:z : ez= i, z : ez

= −1, z : ez= −i, z : cosz = 0,

z : sinz= 0.

Solution. Using the definition we obtain

z : ez= i =

(

12+ 2k

)

πi

, z : ez= −1 = (1+ 2k) πi ,

z : ez= −i =

(

−12+ 2k

)

πi

, z : cosz= 0 =(

12+ k

)

π

,

andz : sinz= 0 = kπ

where k∈ Z.

Exercise 7. Prove formulas forcos(z+ w) andsin(z+ w).

Solution. We have

cos(z) =eiz+ e−iz

2

sin(z) =eiz − e−iz

2i.

26

1. Claim: cos(z) cos(w) − sin(z) sin(w) = cos(z+ w).Proof:

cos(z) cos(w) − sin(z) sin(w) =eiz+ e−iz

2eiw+ e−iw

2− eiz − e−iz

2ieiw − e−iw

2i

=14

(

eizeiw+ e−ize−iw

)

+14

(


)

=12

eizeiw+

12

e−ize−iw

=12

(


)

= cos(z+ w).

2. Claim: sin(z) cos(w) + cos(z) sin(w) = sin(z+ w).Proof:

sin(z) cos(w) + cos(z) sin(w) =eiz − e−iz

2ieiw+ e−iw

2+

eiz+ e−iz

2eiw − e−iw

2i

=14i

(

eizeiw − e−ize−iw)

+14i

(


=12i

eizeiw − 12

e−ize−iw

=12i

(


= sin(z+ w).

Exercise 8. Definetanz= sinzcosz; where is this function defined and analytic?

Solution. Since bothsinz andcosz are analytic in the entire complex plane, it follows from the discussionin the text following Definition 2.3 thattanz = sinz

cosz is analytic wherevercosz , 0. Now,cosz = 0 impliesthat z is real and equal to an odd multiple ofπ

2 . Thus let

G ≡

(2k+ 1)π2

∣∣∣∣ k ∈ Z

.

Thentanz is defined and analytic onC−G. If z∈ G, thencosz= 0 sotanz is undefined on the non-extendedcomplex plane.

Exercise 9. Suppose that zn, z ∈ G = C − z : z ≤ 0 and zn = rneiθn, z = reiθ where−π < θ, θn < π. Provethat if zn→ z thenθn→ θ and rn→ r.


Exercise 10.Prove the following generalization of Proposition 2.20. Let G andΩ be open inC and supposef and h are functions defined on G, g: Ω → C and suppose that f(G) ⊂ Ω. Suppose that g and h areanalytic, g′(ω) , 0 for anyω, that f is continuous, h is one-one, and that they satisfy h(z) = g( f (z)) for z inG. Show that f is analytic. Give a formula for f′(z).


Exercise 11. Suppose that f: G → C is a branch of the logarithm and that n is an integer. Prove thatzn= exp(n f(z)) for all z in G.

27

Solution. Let f(z) be a branch of the logarithm so that ef (z)= z. Let n∈ Z and consider en f(z). In the

following cases we shall apply several of the properties of the complex exponential function developed inthe discussion on page 38 in the text.CASE 1: Assume n> 0. Then we note that

en f(z)= ef (z)+ f (z)+···+ f (z) (n times)

= ef (z)ef (z) · · · ef (z) (n times)

=

(

ef (z))n

= xn.

CASE 2: Assume n< 0. Then let m= −n so that m> 0 and

en f(z)=

1em f(z)

=1zm= zn,

the middle step following from Case 1.CASE 3: Assume n= 0. Then en f(z)

= e0= 1 = z0

= zn.

Exercise 12.Show that the real part of the function z12 is always positive.

Solution. We know that we can write z= reiθ, 0, −π < θ < π and Log(z) = ln(r) + iθ. Thus

z12 = eLogz

12= e

12Logz

= e12 (lnr+iθ)

= e12 lnre

12θi = e

12 lnr

(

cos(θ

2

)

+ i sin(θ

2

))

.

Hence,

Re(z) = e12 lnr cos

(θ

2

)

> 0,

since e12 lnr > 0 andcos

(θ2

)

> 0 since−π < θ < π. Thus, the real part of the function z12 is always positive.

Exercise 13. Let G = C − z ∈ R : z ≤ 0 and let n be a positive integer. Find all analytic functionsf : G→ C such that z= ( f (z))n for all z ∈ G.

Solution. Let Log(z) be the principal branch, thenlog(z) = Log(z) + 2kπi for some k∈ Z. Thus, we canwrite

f (z) = z1/n= elog(z)/n

= e(Log(z)+2kπi)/n= eLog(z)/n·e2kπi/n.

We know that the latter factor are the n-th roots of unity and depend only on k and n. They correspond tothe n distinct powers of the expressionζ = e2πi/n. Therefore, the branches of z1/n on the set U are given by

f (z) = ζk·eLog(z)/n,

where k= 0, . . . ,n− 1 and therefore they are all constant multiples of each other.

Exercise 14. Suppose f: G→ C is analytic and that G is connected. Show that if f(z) is real for all z inG then f is constant.

Solution. First of all, we can write f: G→ C as

f (z) = u(z) + iv(z)

28

where u, v are real-valued functions. Since f: G → C is analytic, that is f is continuously differentiable(Definition 2.3), we have that u and v have continuous partialderivatives. By Theorem 2.29, this impliesthat u, v satisfy the Cauchy-Riemann equations. That is,

∂u∂x=∂v∂y

and∂u∂y= −∂v

∂x. (3.1)

Since f(z) is real∀z ∈ G, this impliesv(z) ≡ 0

and therefore f(z) = u(z). So, since v(z) = 0, we have

∂v∂x=∂v∂y= 0

and by (3.1) we obtain∂u∂x=∂u∂y= 0

and thus u′(z) = 0 (see reasoning of equation 2.22 and 2.23 on page 41). Hence f′(z) = 0. Since G isconnected and f: G→ C is differentiable with f′(z) = 0 ∀z ∈ G, we have that f is constant.

Exercise 15.For r > 0 1et A=

ω : ω = exp(

1z

)

where0 < |z| < r

; determine the set A.

Solution. Define the set S= z : 0 < |z| < r where r> 0. The image of this set under1/z is clearly

T =

z :1r< |z|

.

To find the image of A is the same as finding the image of T under ez.Claim: The image of A isC − 0 (thus not depending on r).To proof the claim, we need to show that for w, 0, the equation ez = w has a solution z∈ T. Using polarcoordinates we can write w= |w|eiθ. We have to find a complex number z= x+ iy such that x2+ y2 > 1

r andexeiy

= w = |w|eiθ. We want ex = |w| and y= θ + 2kπ, for some k∈ Z. Using x= log |w| and k 0 suchthat (log |w|)2

+ (θ + 2kπ)2 > 1r , then we found z= x+ iy.

Exercise 16. Find an open connected set G⊂ C and two continuous functions f and g defined on G suchthat f(z)2

= g(z)2= 1− z2 for all z in G. Can you make G maximal? Are f and g analytic?


Exercise 17.Give the principal branch of√

1− z.


Exercise 18.Let f : G→ C and g: G→ C be branches of za and zb respectively. Show that f g is a branchof za+b and f/g is a branch of za−b. Suppose that f(G) ⊂ G and g(G) ⊂ G and prove that both f g andg f are branches of zab.


Exercise 19. Let G be a region and define G∗ = z : z ∈ G. If f : G → C is analytic prove thatf ∗ : G∗ → C, defined by f∗(z) = f (x), is also analytic.

29

Solution. Let z = x + iy and let f(z) = u(x, y) + iv(x, y). By assumption f is analytic and thereforeu and v have continuous partial derivatives. In addition theCauchy-Riemann Equations ux = vy anduy = −vx are satisfied. Since f∗(z) = f (x), we have f∗(z) = u∗(x, y) + iv∗(x, y) where u∗(x, y) = u(x,−y)and v∗(x, y) = −v(x,−y). Hence, we have u∗x(x, y) = ux(x,−y) = vy(x,−y), u∗y(x, y) = −uy(x,−y) = vx(x,−y),v∗x(x, y) = −vx(x,−y) and v∗y(x, y) = vy(x,−y) and therefore u∗x = v∗y and u∗y = −v∗x so f∗ is analytic.

Exercise 20. Let z1, z2, . . . , zn be complex numbers such that Re zk > 0 and Re(zl . . . zk) > 0 for 1 ≤ k ≤ n.Show thatlog(z1 . . . zn) = logz1 + . . . + logzn, wherelogz is the principal branch of the logarithm. If therestrictions on the zk are removed, does the formula remain valid?

Solution. Let z1, . . . , zn ∈ C such that Re(zj) > 0 and Re(z1 · · · zj) > 0 for 1 ≤ j ≤ n. The proof will beby induction. Consider first the case where n= 2. Let z1, z2 ∈ C as above. Then− π2 < argz1 <

π2 , − π2 <

argz2 <π2 and− π2 < arg(z1z2) < π

2 . But notearg(z1z2) = argz1+argz2 implies that−π < argz1+argz2 < π.Now, recall

log(z1z2) = ln |z1z2| + i arg(z1z2)

= ln |z1| + ln |z2| + i arg(z1) + i arg(z2)

= log(z1) + log(z2)

Assume this formula is true for n= k− 1. Let z1, . . . , zk ∈ C as above. Then

log(z1 . . . zk) = log((z1 · · · zk−1)zk)

= log(z1 · · · zk−1) + logzk by the case when n= 2

= logz1 + logz2 + . . . + logzk−1 + logzk by the case when n= k− 1

Hence, this is true for all n such that zi , 1 ≤ i ≤ n, satisfy the restrictions.If the restriction on the zk are removed, does the formula remain valid? Consider the complex numbers

z1 = −1+ 2i, z2 = −2+ i. Then z1z2 = 0− 5i. Clearly, z1, z2, and z1z2 do not meet the restrictions as statedabove. Now,

log(z1) = ln |z1| + i argz1

log(z2) = ln |z2| + i argz2

Thuslogz1 + logz2 = ln√

5 + ln√

5 + i( 3π2 ) = ln 5 + i( 3π

2 ), but note that3π2 < (−π, π) and log(z1z2) =ln 5+ i(− π2). Thuslogz1 + logz2 , log(z1z2) wherelogz is the principal branch of the logarithm. Hence,the formula is invalid.

Exercise 21.Prove that there is no branch of the logarithm defined on G= C − 0. (Hint: Suppose such abranch exists and compare this with the principal branch.)

Solution. Define the subsetG of G byG = C − z ∈ R : z ≤ 0. We use the notation Log for the principalpart of thelog onG, that is

Log(z) = log |z| + i arg(z)

wherearg(z) ∈ (−π, π). We will prove the statement above by contradiction.Assume f(z) is a branch of the logarithm defined on G. Then restricting f toG gives us a branch of thelogonG. Therefore its only difference to the principal branch is2πik for some k∈ Z. This yields

f (z) = log |z| + i arg(z) + 2πik

where z∈ G. Since f is analytic in G, it is continuous at−1. But we can check that this is not the case, thuswe have derived a contradiction

30

3.3 Analytic functions as mappings. Möbius transformations

Exercise 1. 1. Find the image ofz : Re z< 0, |Im z| < π under the exponential function.

Solution. We have

z : Re z< 0, |Im z| < π = z= x+ iy : x < 0,−π < y < π.The image ofz= x+ iy : x < 0,−π < y < π under the exponential function is given by

ex+iy : x < 0,−π < y < π = exeiy : x < 0,−π < y < π= ex(cos(y) + i sin(y)) : x < 0,−π < y < π= r(cos(y) + i sin(y)) : 0 < r < 1,−π < y < π.

If 0 < r < 1 is fixed, then r(cos(y) + i sin(y)) describes a circle with radius r without the point(−r,0)centered at(0,0).Since r varies between 0 and 1, we get that the image is a solid circle with radius 1, where the boundarydoes not belong to it, the negative x-axis does not belong to it and the origin does not belong to is.

Exercise 2. Do exercise 1 for the setz : |Im z| < π/2.Solution. We have

z : |Im z| < π/2 = z= x+ iy : x ∈ R,−π2< y <

π

2.

The image ofz= x+ iy : x ∈ R,− π2 < y < π2 under the exponential function is given by

ex+iy : x ∈ R,−π2< y <

π

2 = ex(cos(y) + i sin(y)) : x ∈ R,−π

2< y <

π

2

= r(cos(y) + i sin(y)) : r > 0,−π2< y <

π

2.

If r > 0 is fixed, then r(cos(y) + i sin(y)) describes a half circle with radius r centered at(0,0) lying in theright half plane not touching the imaginary axis.Since r varies between 0 and infinity, we get that the image is the right half plane without the imaginaryaxis.

Exercise 3. Discuss the mapping properties ofcosz andsinz.


Exercise 4. Discuss the mapping properties of zn and z1/n for n ≥ 2. (Hint: use polar coordinates.)


Exercise 5. Find the fixed points of a dilation, a translation and the inversion onC∞.

Solution. In general, we have

S(z) =az+ bcz+ d

.

To get a dilation S(z) = az, we have that b= 0, c = 0,d = 1 and a> 0. To find a fixed point, we have to findall z such that S(z) = z. In this case az= z. Obviously z= 0 is a fixed point. Also z= ∞ is a fixed point,since a· ∞ = ∞(a > 0) or S(∞) = a

c =a0 = ∞.

To get a translation S(z) = z+ b, we have that a= 1, c = 0,d = 1 and b∈ R. To find a fixed point, we haveto find all z such that S(z) = z. In this case z+b = z, which is true if z= ∞. To see that S(∞) = a

c =10 = ∞.

To get a inversion S(z) = 1z, we have that a= 0,b = 1, c = 1 and d= 0. To find a fixed point, we have to

find all z such that S(z) = z. In this case1z = z which is equivalent to z2

= 1 and thus z= 1 and z= −1 arefixed points.

31

Exercise 6. Evaluate the following cross ratios: (a)(7+ i,1,0,∞) (b) (2,1− i,1,1+ i) (c) (0,1, i,−1) (d)(i − 1,∞,1+ i,0).

Solution. We have

S(z) =z− z3

z− z4/

z2 − z3

z2 − z4, if z2, z3, z4 ∈ C (3.2)

S(z) =z− z3

z− z4, if z2 = ∞ (3.3)

S(z) =z− z3

z2 − z3, if z4 = ∞. (3.4)

a) By (3.4) we get

(7+ i,1,0,∞) =7+ i − 0

1− 0= 7+ i.

b) By (3.2) we get

(2,1− i,1,1+ i) =2− 1

2− 1− i/

1− i − 11− i − 1− i

=1

1− i/−i−2i=

21− i

= 21+ i

(1− i)(i + 1)= 2

1+ i2= 1+ i.

c) By (3.2) we get

(0,1, i,−1) =0− i0+ 1

/1− i1+ 1

= −i/1− i

2= −i

21− i

1+ i1+ i

= −i22

(1+ i) = −i − i2 = 1− i.

d) By (3.3) we get

(i − 1,∞,1+ i,0) =i − 1− 1− i

i − 1− 0=−2

i − 1i + 1i + 1

=−2−2

(i + 1) = 1+ i.

Exercise 7. If Tz= az+bcz+d , find z2, z3, z4 (in terms of a, b, c, d) such that Tz= (z, z2, z3, z4).

Solution. The inverse of T is given by

T−1(z) =dz− b−cz+ a

as shown on p. 47. We have T−1(1) = d−ba−c , T−1(0) = − b

a and T−1(∞) = − dc . Set z2 = d−b

a−c , z3 = − ba and

z4 = − dc to obtain Tz= (z, z2, z3, z4).

Exercise 8. If Tz= az+bcz+d show that T(R∞) = R∞ iff we can choose a,b, c,d to be real numbers.

Solution. Let Tz=az+ bcz+ d

. Assume T(R∞) = R∞. Then let z0 ∈ R∞ such that Tz0 = 0. Observe that this

implies az0 = −b, so−ba∈ R∞ and r1 ≡

ba∈ R∞ as well. Likewise, if z∞ ∈ R∞ such that Tz∞ = ∞, then

32

r2 ≡dc∈ R∞. Now let z1 ∈ R∞ such that Tz1 = 1. Then

az1 + bcz1 + d

= 1

az1 + b = cz1 + d

z1

(

1− ca

)

=da− b

az1

c− z1

a− r2

a+

r1

c= 0

z1 + r1

c=

z1 + r2

az1 + r1

z1 + r2=

ca∈ R∞.

Let r3 =ca

. Thenda=

dc× c

a= r2r3 ∈ R∞. Thus

Tz=az+ bcz+ d

=z+ b

acaz+ d

a

=z+ r1

r3z+ r2r3

and we have thus found real coefficients for T .Now to prove the converse, assume T(R∞) , R∞. Then recognizing thatR∞ is a circle in C∞, by

Theorem 3.14 we conclude that T(R∞) is some other circle inC∞. In particular, this means that T(R∞) ∩(C∞ − R∞) , ∅. In other words, there must be some value zc ∈ R∞ for which Tzc < R∞.

Now, suppose by way of contradiction that there is some representation of T in which a, b, c, and d areall real. Then

Tzc =azc + bczc + d

is clearly an element ofR∞, contradicting the observation that Tzc is not real. We see, therefore, that thereis a real choice for a, b, c, and d simultaneously if and only ifT(R∞) = R∞.

Exercise 9. If Tz = az+bcz+d , find necessary and sufficient conditions that T(Γ) = Γ whereΓ is the unit circle

z : |z| = 1.

Solution. We want if zz= 1, then T(z)T(z) = 1.

T(z)T(z) = 1 ⇐⇒ az+ bcz+ d

az+ b

cz+ d= 1 ⇐⇒ (az+ b)(az+ b)

(cz+ d)(cz+ d)= 1

⇐⇒ azaz+ baz+ azb+ bb = czcz+ dcz+ dcz+ dd

⇐⇒ zz(aa− cc) + z(ab− cd) + z(ba+ dc) − dd = 0.

This is the same as the equation zz− 1 = 0 if

ab− cd = 0

ba− dc = 0

and

aa− cc = 1

bb− dd = −1

33

which is equivalent to|a|2 − |c|2 = 1 and1 = |d|2 − |b|2. Hence, we have the two conditions

ab− cd = 0 and |a|2 + |b|2 = |c|2 + |d|2.

These are the sufficient conditions. Let c= λb, then ab− cd = 0 yields ab− λbd = 0 iff a = λd iff d = aλ.

Insert this into|c|2 + |d|2 = |a|2 + |b|2 yields

|λ|2|b|2 + |a|2

|λ|2 = |a|2+ |b|2,

and therefore we can take|λ| = 1. Then1λ= λ, and so the form of the Möbius transformation is

T(z) =az+ b

λ(bz+ a), where|λ| = 1

or

T(z) = λaz+ b

bz+ a, where|λ| = 1.

Takeλ = e−iθ, then

T(z) = eiθ az+ b

bz+ a, for someθ.

This mapping transforms|z| = 1 into |T(z)| = 1.

Exercise 10.Let D= z : |z| < 1 and find all Möbius transformations T such that T(D) = D.

Solution. Take anα ∈ D = z : |z| < 1 such that T(α) = 0. Its symmetric point with respect to the unitcircle is

α∗ =1α

since

z∗ − a =R2

z− a

(see page 51) with a= 0 and R= 1 (the unit circle). Therefore T(α∗) = ∞. Thus T looks like

T(z) = Kz− ααz− 1

where K is a constant. (It is easy to check that T(α) = 0 and T(α∗) = T( 1α) = ∞). Finally, we are going to

choose the constant K in such a way that|T(z0)| = 1 where z0 = eiθ. We have

T(z0) = Keiθ − ααeiθ − 1

and therefore

1 = |T(z0)| = |K| |eiθ − α||αeiθ − 1| = |K|

(

eiθ − α) (

e−iθ − α)

|eiθ |· |α − e−iθ | = |K|(

eiθ − α) (

e−iθ − α)

(

α − e−iθ) (

α − eiθ) = |K|.

So|K| = 1 implies K= eiθ for some realθ. We arrive at

T(z) = eiθ z− ααz− 1

, for some realθ.

34

Exercise 11.Show that the definition of symmetry (3.17) does not depend onthe choice of points z2, z3, z4.That is, show that ifω2, ω3, ω4 are also inΓ then equation (3.18) is satisfied iff (z∗, ω2, ω3, ω4) = (z, ω2, ω3, ω4).(Hint: Use Exercise 8.)

Solution. LetΓ be a circle inC∞ containing points z2, z3, z4, w2, w3, and w4, with all zi distinct and all wi

distinct. Let z∈ C∞ and consider z∗, as defined in the text and established by the points zi . We know that

(z∗, z2, z3, z4) = (z, z2, z3, z4)

Recall that the above left-hand cross-ratio implies a unique Möbius transformation T for which Tz2 = 1,Tz3 = 0, and Tz4 = ∞. By Proposition 3.10, T mapsΓ to R∞. In fact, the definition of symmetry may berewritten as

Tz∗ = Tz.

orz∗ = T−1Tz (3.5)

Likewise, there is some transformation S for which S w2 = 1, S w3 = 0, and S w4 = ∞. Again, S mapsΓ toR∞. We wish to show that

z∗ = S−1S z. (3.6)

Now, we proceed by showing that the right hand sides of (3.5) and (3.6) must be equal:

T−1Tz= T−1TS−1S z

Observe here that TS−1 must have real coefficients by Exercise 8 becauseR∞ 7→ R∞ under TS−1. Thus weobserve thatTS−1 = TS−1 so

T−1Tz= T−1TS−1S z= S−1S z.

Hence we have the desired result.

Exercise 12.Prove Theorem 3.4.


Exercise 13.Give a discussion of the mapping f(z) = 12(z+ 1/z).

Solution. The function f(z) = 12

(

z+ 1z

)

=z2+1

2z can be defined for all z∈ C − 0 and therefore also in thepunctured disk0 < |z| < 1. To see that in this domain the function is injective, let z1, z2 be two numbers inthe domain of f with the same image, then

0 = f (z1) − f (z2) =z2

1 + 1

z1−

z22 + 1

z2=

z21z2 + z2 − z1z2

2 − z1

z1z2=

(z1z2 − 1)(z1 − z2)z1z2

.

With the assumption0 < |zi | < 1, i = 1,2 the factor z1z2−1, is always nonzero and we conclude that z1 = z2;hence f(z) is injective.The range of the function isC − z ∈ C | < z ∈ [−1,1] andIm z = 0. To see this, write z= reiθ in polarcoordinates and let f(z) = w = a+ ib,a,b ∈ R.

f (z) = f (reiθ) =12

(

reiθ+

1r

e−iθ

)

=12

[(

r +1r

)

cosθ + i

(

r − 1r

)

sinθ

]

.

35

For the real and imaginary part of w the following equations must hold

a =12

(

r + 1r

)

cosθ

b =12

(

r − 1r

)

sinθ.(3.7)

If f (z) = w has imaginary part b= 0 then sinθ = 0 and |cosθ| = 1. Therefore points of the forma+ ib,a ∈ [−1,1],b = 0 cannot be in the range of f . For all other points the equations(3.7) can be solvedfor r and θ uniquely (after restricting the argument to[−π, π)).Given any value of r∈ (0,1), the graph of f(reiθ) as a function ofθ looks like an ellipse. In fact from

formulas (3.7) we see that(

a12 (r+ 1

r )

)2+

(

b12 (1− 1

r )

)2= 1.

If we fix the argumentθ and let r vary in(0,1) it follows from from equation (3.7) that the graph of f(reiθ)is a hyperbola and it degenerates to rays if z is purely real orimaginary. In the caseθ ∈ (2k+ 1)π | k ∈ Zthe graph of f in dependence on r is on the imaginary axis and for θ ∈ 2kπ | k ∈ Z the graph of f(reiθ) is

either(−∞,−1) or (1,∞). If cosθ , 0 andsinθ , 0 then(

acosθ

)2 −(

bsinθ

)2= 1.

Exercise 14. Suppose that one circle is contained inside another and thatthey are tangent at the point a.Let G be the region between the two circles and map G conformally onto the open unit disk. (Hint: first try(z− a)−1.)

Solution. Using the hint, define the Möbius transformation T(z) = (z− a)−1 which sends the region Gbetween two lines. Afterward applying a rotation followed by a translation, it is possible to send this regionto any region bounded by two parallel lines we want. Hence, choose S(z) = cz+ d where|c| = 1 such that

S(T(G)) =

x+ iy : 0 < y <π

2

.

Applying the exponential function to this region yields theright half plane

exp(S(T(G))) = x+ iy : x > 0.

Finally, the Möbius transformation

R(z) =z− 1z+ 1

maps the right half plane onto the unit disk (see page 53). Hence the function f defined by R(exp(S(T(z))))maps G onto D and is the desired conformal mapping ( f is a composition of conformal mappings). Doingsome simplifications, we obtain

f (z) =e

cz−a+d − 1

ec

z−a+d+ 1

where the constants c and d will depend on the circle location.

Exercise 15.Can you map the open unit disk conformally ontoz : 0 < |z| < 1?


Exercise 16. Map G = C − z : −1 ≤ z ≤ 1 onto the open unit disk by an analytic function f . Can f beone-one?


Exercise 17. Let G be a region and suppose that f: G → C is analytic such that f(G) is a subset of acircle. Show that f is constant.

36


Exercise 18.Let−∞ < a < b < ∞ and put Mz= z−iaz−ib . Define the lines L1 = z : Im z= b, L2 = z : Im z=

a and L3 = z : Re z= 0. Determine which of the regions A, B,C,D,E, F in Figure 1, are mapped by Monto the regions U,V,W,X,Y,Z in Figure 2.

Solution. We easily see that we have M(ia) = 0. Therefore the region B,C,E and F which touch the lineia are mapped somehow to the region U,V,X and Y which touch0. Similarly we have M(ib) = ∞ andtherefore the region B and E which touch the line ib are mappedsomehow to the region U and X whichtouch∞. Thus we conclude that C or F goes to either V or Y. Let us find out. Let x, y be small positivereal numbers such that the poin z= x + iy + ia ∈ E. Thus, the imaginary part of Mz is a positive numbermultiplied by x(b− a) and therefore also positive. Therefore we conclude that M maps E to U and B to X.Because B and C meet at the line ia, we conclude that X and M(C) do, too. Hence, M maps C to Y and Fto V. By a similar argument, we obtain that M maps A to Z and D to W.

Exercise 19.Let a, b, and M be as in Exercise 18 and letlog be the principal branch of the logarithm.(a) Show thatlog(Mz) is defined for all z except z= ic, a ≤ c ≤ b; and if h(z) = Im [log Mz] then0 < h(z) < π for Re z> 0.(b) Show thatlog(z− ic) is defined for Re z> 0 and any real number c; also prove that|Im log(z− ic)| < π

2if Re z> 0.(c) Let h be as in (a) and prove that h(z) = Im [log(z− ia) − log(z− ib)].(d) Show that

∫ b

a

dtz− it

= i[log(z− ib) − log(z− ia)]

(Hint: Use the Fundamental Theorem of Calculus.)(e) Combine (c) and (d) to get that

h(x+ iy) =∫ b

a

xx2 + (y− t)2

dt = arctan(y− a

x

)

− arctan

(

y− bx

)

(f) Interpret part (e) geometrically and show that for Re z> 0, h(z) is the angle depicted in the figure.


Exercise 20.Let S z= az+bcz+d and Tz= αz+β

γz+δ , show that S= T iff there is a non zero complex numberλ suchthatα = λa, β = λb, γ = λc, δ = λd.

Solution. ⇐: Letλ , 0 be a complex number such that

α = λa

β = λb

γ = λc

δ = λd.

Then

T(z) =αz+ βγz+ δ

=λaz+ λbλcz+ λd

=λ(az+ b)λ(cz+ d)

=(az+ b)(cz+ d)

= S(z).

Thus, S= T.⇒: Let S = T, that is S(z) = T(z). Then S(0) = T(0), S(1) = T(1) and S(∞) = T(∞) which is equivalent

37

to

bd=

β

δ= λ1 ⇐⇒ b = λ1d, β = λ1δ (3.8)

a+ bc+ d

=α + β

γ + δ(3.9)

ac=

α

γ= λ3 ⇐⇒ a = λ3c, α = λ3γ. (3.10)

Insert b= λ1d, β = λ1δ, a = λ3c andα = λ3γ into (3.9), then

λ3c+ λ1dc+ d

=λ3γ + λ1δ

γ + δ

⇐⇒ λ3cγ + λ1dγ + λ3cδ + λ1dδ = λ3γc+ λ1δc+ λ3γd + λ1δd

⇐⇒ λ1dγ + λ3cδ − λ1δc− λ3γd = 0

⇐⇒ λ1(dγ − δc) − λ3(dγ − δc) = 0

⇐⇒ (λ1 − λ3)(dγ − δc) = 0.

Thus, eitherλ1 − λ3 = 0 or dγ − δc = 0.If λ1− λ3 = 0, then from (3.8) and (3.9), we getλ3 =

ac =

bd = λ1 which implies ad− bc= 0 a contradiction

(Not possible, otherwise we do not have a Möbius transformation).Thus, dγ − δc = 0 ⇐⇒ c

d =γ

δ⇐⇒ γ

c =δd := λ, λ , 0 or

γ = cλ (3.11)

δ = dλ. (3.12)

Insert (3.11) and (3.12) into (3.8) to obtain

bd=β

δ⇒ b

d=

β

dλ⇒ β = λb.

Insert (3.11) and (3.12) into (3.10) to obtain

ac=α

γ⇒ a

c=α

cλ⇒ α = λa.

Therefore, we haveα = λa, β = λb, γ = λc, δ = λd.

Exercise 21.Let T be a Möbius transformation with fixed points z1 and z2. If S is a Möbius transformationshow that S−1TS has fixed points S−1z1, and S−1z2.

Solution. To showS−1TS(S−1z1) = S−1z1.

We haveS−1TS(S−1z1) = S−1TS S−1z1 = S−1Tz1 = S−1z1,

where the first step follows by the associativity of compositions, the second step since S S−1= id and the

last step by Tz1 = z1 since z1 is a fixed point of T .To show

S−1TS(S−1z2) = S−1z2.

38

We haveS−1TS(S−1z2) = S−1TS S−1z2 = S−1Tz2 = S−1z2,

where the first step follows by the associativity of compositions, the second step since S S−1= id and the

last step by Tz2 = z2 since z2 is a fixed point of T .Note that S−1TS is a well defined Möbius transformation, since S and T are Möbius transformation. Thecomposition of Möbius transformation is a Möbius transformation.

Exercise 22. (a) Show that a Möbius transformation has0 and∞ as its only fixed points iff it is a dilation,but not the identity.(b) Show that a Möbius transformation has∞ as its only fixed point iff it is a translation, but not the identity.

Solution. a) Let M be a Möbius transformation with exactly two fixed points, at0 and∞. We know that

Mz=az+ bcz+ d

with a,b, c,d ∈ C. We know that M(0) = 0 sobd= 0, whence we conclude that b= 0. But we also

know that M(∞) = ∞ soac= ∞, meaning c= 0. Thus Mz=

azd= αz for someα ∈ C. Since both

b and c were0, we know that both a and d are nonzero soα is likewise nonzero and finite. Supposeby way of contradiction thatα = 1. Then for anyz ∈ C∞, Mz = z and M has infinitely many fixedpoints, a contradiction. Thusα , 1 and M is a (complex) dilation not equal to the identity.

On the other hand, assume M is a dilation not equal to the identity. In this case, we are free to expressM as Mz= αz withα , 1. It is clear from this representation that M has fixed points at 0 and∞ and,of course, at no other points.

b) Let M be a Möbius transformation with exactly one fixed point at∞. Again, we know that

Mz=az+ bcz+ d

and we can see that c= 0 as before. This time, however, we also examine the lack of a second fixedpoint. Recall that ifz is a fixed point of M, thenz satisfies cz2

+ (d − a)z− b = 0, which in the case

of c = 0 reduces to(d − a)z− b = 0 or z =b

d − a. We note that b, 0 because0 is not a fixed point

of M. Suppose that d, a. In such a case, there is a finite value ofz that is a fixed point of M, acontradiction. Thus, it must be the case that d= a and

Mz= z+bd.

Thus M is a translation.

Now assume M is a translation. Then we can express M as Mz= z+ β. Here, in the convention ofthe text, we have a= 1, b = β, c = 0, and d= 1. The fixed pointsz of M satisfy cz2

+ (d− a)z− b = 0,but no finitez satisfies this equation for the given coefficients. Clearly, however, M has∞ as a fixedpoint, so∞ is the only fixed point of M.

Exercise 23. Show that a Möbius transformation T satisfies T(0) = ∞ and T(∞) = 0 iff Tz = az−1 forsome a inC.


39

Exercise 24. Let T be a Möbius transformation, T, the identity. Show that a Möbius transformation Scommutes with T if S and T have the same fixed points. (Hint: UseExercises 21 and 22.)

Solution. Let T and S have the same fixed points. To show TS= S T, T, id.

? : Suppose T and S have two fixed points, say z1 and z2. Let M be a Möbius transformation with M(z1) = 0and M(z2) = ∞. Then

MS M−1(0) = MS M−1Mz1 = MS z1 = Mz1 = 0

andMS M−1(∞) = MS M−1Mz2 = MS z2 = Mz2 = ∞.

Thus MS M−1 is a dilation by exercise 22 a) since MS M−1 has 0 and∞ as its only fixed points. Similar,we obtain MT M−1(0) = 0 and MT M−1(∞) = ∞ and therefore is also a dilation. It is easy to check thatdilations commute (define C(z) = az, a> 0 and D(z) = bz, b> 0, then CD(z) = abz= baz= DC(z)), thus

(MT M−1)(MS M−1) = (MS M−1)(MT M−1)

⇒ MTS M−1= MS T M−1

⇒ TS = S T.

? : Suppose T and S have one fixed points, say z. Let M be a Möbius transformation with M(z) = ∞. Then

MS M−1(∞) = MS M−1Mz= MS z= Mz= ∞.Thus MS M−1 is a translation by exercise 22 b) since MS M−1 has∞ as its only fixed point. Similar, weobtain MT M−1(∞) = ∞ and therefore is also a translation. It is easy to check that translations commute(define C(z) = z+ 1 and D(z) = z+ b, b> 0, then CD(z) = z+ a+ b = z+ b+ a = DC(z)), thus

(MT M−1)(MS M−1) = (MS M−1)(MT M−1)

⇒ MTS M−1= MS T M−1

⇒ TS = S T.

Exercise 25.Find all the abelian subgroups of the group of Möbius transformations.


Exercise 26. 26. (a) Let GL2(C) = all invertible 2× 2 matrices with entries inC and letM be the group

of Möbius transformations. Defineϕ : GL2(C) → M by ϕ

(

a bc d

)

=az+bcz+d . Show thatϕ is a group

homomorphism of GL2(C) ontoM. Find the kernel ofϕ.(b) Let S L2(C) be the subgroup of GL2(C) consisting of all matrices of determinant 1. Show that the imageof S L2(C) underϕ is all ofM. What part of the kernel ofϕ is in S L2(C)?

Solution. a) We have to check that if

A =

(

a bc d

)

and B=

(

a bc d

)

thenϕ(AB) = ϕ(A) ϕ(B). A simple calculation shows that this is true. To find the kernel of the grouphomomorphism we have to find all z such thataz+b

cz+d = z. This is equivalent to az+ b = cz2+ dz and by

comparing coefficients we obtain b= c = 0 and a= d. Therefore, the kernel is given by N= ker(ϕ) = λI :λ ∈ C×. Note that the kernel is a normal subgroup of GL2(C).b) Restrictingϕ to S L2(C) still yields a surjective map since for any matrix A∈ GL2(C) both A andthe modification M= 1√

detAA have the same image and the modification matrix M has by construction

determinant 1. The kernel of the restriction is simply N∩ S L2(C) = ±I .

40

Exercise 27.If G is a group andN is a subgroup thenN is said to be a normal subgroup ofG if S−1TS ∈ Nwhenever T∈ N and S∈ G. G is a simple group if the only normal subgroups ofG are I (I = the identityofG) andG itself. Prove that the groupM of Möbius transformations is a simple group.


Exercise 28.Discuss the mapping properties of(1− z)i .


Exercise 29.For complex numbersα andβ with |α|2 + |β|2 = 1

uα,β =αz− ββz+ α

and let U= uα,β : |α|2 + |β|2 = 1

(a) Show that U is a group under composition.(b) If S U2 is the set of all unitary matrices with determinant 1, show that S U2 is a group under matrixmultiplication and that for each A in S U2 there are unique complex numbersα andβ with |α|2 + |β|2 = 1and

A =

(

α β

−β α

)

(c) Show that

(

α β

−β α

)

7→ uα,β is an isomorphism of the group S U2 onto U. What is its kernel?

(d) If l ∈ 0, 12 ,1,

32 , . . . let Hl = all the polynomials of degree≤ 2l. For uα,β = u in U define T(l)u : Hl → Hl ,

by (T(l)u f )(z) = (βz+ α)2l f (u(z)). Show that T(l)u is an invertible linear transformation on Hl , and u 7→ T(l)

u

is an injective homomorphism of U into the group of invertible linear transformations of Hl , onto Hl .


Exercise 30.For |z| < l define f(z) by

f (z) = exp

−i log

[

i

(

1+ z1− z

)]1/2

(a) Show that f maps D= z : |z| < 1 conformally onto an annulus G.(b) Find all Möbius transformations S(z) that map D onto D and such that f(S(z)) = f (z) when|z| < 1.


41

Chapter 4

Complex Integration

4.1 Riemann-Stieltjes integrals

Exercise 1. Let γ : [a,b] → R be non decreasing. Show thatγ is of bounded variation and V(γ) =γ(b) − γ(a).

Solution. Let γ : [a,b] → R be a monotone, non-decreasing function. For a given partition of [a,b],a = t0 < t1 < . . . < tm = b, for all n ∈ N, γ(tn) ≥ γ(tn−1). Therefore,|γ(tn) − γ(tn−1)| = γ(tn) − γ(tn−1) ≥ 0.Let P= a = t0 < t1 < . . . < tm = b, any partition of[a,b]. Then,

m∑

n=1

|γ(tn) − γ(tn−1)| = [γ(tm) − γ(tm−1)] + [γ(tm−1) − γ(tm−2)] + . . . + [γ(t2) − γ(t1)] + [γ(t1) − γ(t0)]

= [γ(b) − γ(tm−1)] + [γ(tm−1) − γ(tm−2)] + . . . + [γ(t2) − γ(t1)] + [γ(t1) − γ(a)]

= γ(b) − γ(a)

Sinceγ(a), γ(b) ∈ R, γ(b) − γ(a) < ∞. It follows thatγ is of bounded variation for any partition P. Andalso v(γ; P) = supv(γ; P) : P a partition of[a,b] = V(γ) = γ(b) − γ(a).


Solution. Letγ : [a,b] → C be of bounded variation.(a) If P and Q are partitions of[a,b] and P⊂ Q then v(γ; P) ≤ v(γ; Q).Proof will be by induction. Let P= a = t0 < t1 < . . . < tm = b. Suppose Q= P∪ x, where tk < x < tk+1.Then,

v(γ; P) =∑

i,k+1

|γ(ti) − γ(ti−1)| + |γ(tk+1) − γ(tk)|

≤∑

i,k+1

|γ(ti) − γ(ti−1)| + |γ(x) − γ(tk)| + |γ(tk+1) − γ(x)|

= v(γ; Q)

Now consider the general case on the number of elements of Q\ P. The smallest partition set for Q\ Pcontains the trivial partition composed ofa,b. Using the case defined above, we have v(γ; Q) ≥ v(γ; P) ≥v(γ; Q \ P) ≥ |γ(b) − γ(a)|, whenever P⊂ Q.(b) If σ : [a,b] → C is also of bounded variation andα, β ∈ C, thenαγ + βσ is of bounded variation and

42

V(αγ + βσ) ≤ |α|V(γ) + |β|V(σ).Let P= a = t0 < t1 < . . . < tm = b be a partition. Sinceγ andσ are of bounded variation for[a,b] ⊂ R,

there exists constants Mγ > 0 and Mσ > 0, respectively, such that v(γ; P) =m∑

k=1

|γ(tk) − γ(tk−1)| ≤ Mγ and

v(γ; P) =m∑

k=1

|σ(tk) − σ(tk−1)| ≤ Mσ. Then,

v(αγ + βσ; P) =m∑

k=1

|(αγ + βσ)(tk) − (αγ + βσ)(tk−1)|

=

m∑

k=1

|αγ(tk) + βσ(tk) − αγ(tk−1) − βσ(tk−1)|

=

m∑

k=1

|(αγ(tk) − αγ(tk−1)) + (βσ(tk) − βσ(tk−1))|

≤m∑

k=1

|αγ(tk) − αγ(tk−1)| + |βσ(tk) − βσ(tk−1)|

=

m∑

k=1

|αγ(tk) − αγ(tk−1)| +m∑

k=1

|βσ(tk) − βσ(tk−1)|

=

m∑

k=1

|α| |(γ(tk) − γ(tk−1))| +m∑

k=1

|β| |(σ(tk) − σ(tk−1))|

= |α|m∑

k=1

|γ(tk) − γ(tk−1)| + |β|m∑

k=1

|σ(tk) − σ(tk−1)|

= |α|v(γ; P) + |β|v(σ; P)

≤ |α|supv(γ; P) : P a partition of[a,b] + |β|supv(σ; P) : P a partition of[a,b]= |α|V(γ) + |β|V(σ)

< ∞

The result is two-fold, V(αγ + βσ) ≤ |α|V(γ) + |β|V(σ) and it followsαγ + βσ is of bounded variation by|α|V(γ) + |β|V(σ).



Exercise 4. Prove Proposition 1.8 (Use induction).


Exercise 5. Let γ(t) = exp((−1+ i)/t−1) for 0 < t ≤ 1 andγ(0) = 0. Show thatγ is a rectifiable path andfind V(γ). Give a rough sketch of the trace ofγ.


Exercise 6. Show that ifγ : [a,b] → C is a Lipschitz function thenγ is of bounded variation.

43

Solution. Letγ : [a,b] → C be Lipschitz, that is∃ a constant C> 0 such that

|γ(x) − γ(y)| ≤ C|x− y|, ∀x, y ∈ [a,b].

Then, for any partition P= a = t0 < t1 < . . . < tm = b of [a,b], we have

v(γ; P) =m∑

k=1

|γ(tk) − γ(tk−1)| ≤m∑

k=1

C|tk − tk−1| = Cm∑

k=1

|tk − tk−1| = C(b− a) =: M > 0.

Hence, the functionγ : [a,b] → C, for [a,b] ⊂ R, is of bounded variation, since there is a constant M> 0such that for any partition P= a = t0 < t1 < . . . < tm = b of [a,b]

v(γ; P) =m∑

k=1

|γ(tk) − γ(tk−1)| ≤ M.

Exercise 7. Show thatγ : [0,1] → C, defined byγ(t) = t + it sin 1t for t , 0 andγ(0) = 0, is a path but is

not rectifiable. Sketch this path.


Exercise 8. Letγ andσ be the two polygons[1, i] and[1,1+ i, i]. Expressγ andσ as paths and calculate∫

γf and

∫

σf where f(z) = |z|2.

Solution. For γ we haveγ(t) = (1− t) + it for 0 ≤ t ≤ 1. Soγ′(t) = −1+ i. Therefore,

∫

γ

|z|2dz=∫ 1

0((1− t)2

+ t2)(−1+ i)dt

= (−1+ i)∫ 1

0(2t2 − 2t + 1)dt

= (−1+ i)

[

23

t3 − t2 + t

]1

0

= (−1+ i)

(

23− 1+ 1

)

= −23+ i

23.

Sinceσ is composed of multiple lines, the path is given byγ1(t) = 1+ it : 0 ≤ t ≤ 1;γ2(t) = (1− t) + i :

44

0 ≤ t ≤ 1. Soγ′1(t) = i andγ′2(t) = −1. Therefore,

∫

γ

f =∫

γ1

f +∫

γ2

f

=

∫

γ1

|z|2dz+∫

γ2

|z|2dz

=

∫ 1

0(t2 + 1)(i)dt+

∫ 1

0((1− t)2

+ 1)(−1)dt

= i∫ 1

0(t2 + 1)dt−

∫ 1

0(t2 − 2t + 2)dt

= i

[

t3

3+ t

]1

0

−[

t3

3− t2 + 2t

]1

0

=43

i −(

13− 1+ 2

)

=43

i − 43

=43

(−1+ i)

Exercise 9. Defineγ : [0,2π] → C byγ(t) = exp(int) where n is some integer (positive, negative, or zero).Show that

∫

γ

1z dz= 2πin.

Solution. Clearly,γ(t) = eint is continuous and smooth on[0,2π]. Thus

∫

γ

z−1 dz=∫ 2π

0e−intineint dt =

∫ 2π

0in dt = in(2π − 0) = 2πin.

Exercise 10.Defineγ(t) = eit for 0 ≤ t ≤ 2π and find∫

γzn dz for every integer n.

Solution. Clearly,γ(t) = eint is continuous and smooth on[0,2π] (It is the unit circle).Case 1: n= −1

∫

γ

z−1 dz=∫ 2π

0e−it ieit dt = i

∫ 2π

0dt = 2πi.

Case 2: n, −1

∫

γ

zn dz =

∫ 2π

0eintieit dt = i

∫ 2π

0ei(n+1)t dt = t

[

ei(n+1)t

i(n+ 1)

]2π

0

=1

n+ 1

[

ei(n+1)t]2π

0=

1n+ 1

[

ei(n+1)2π − 1]

=1

n+ 1[cos((n+ 1)2π) − i sin((n+ 1)2π)1] =

1n+ 1

[1− i·0− 1]

= 0.

Exercise 11.Letγ be the closed polygon[1 − i,1+ i,−1+ i,−1− i,1− i]. Find∫

γ

1z dz.

45

Solution. Define

γ1(t) = 1+ it, t from − 1 to 1 → γ′1(t) = iγ2(t) = t + i, t from 1 to − 1 → γ′2(t) = 1γ3(t) = −1+ it, t from 1 to − 1 → γ′3(t) = iγ4(t) = t − i, t from − 1 to 1 → γ′4(t) = 1.

Thus∫

γ

1z

dz

=

∫ 1

−1

11+ it

i dt +∫ −1

1

1t + i

dt+∫ −1

1

1−1+ it

i dt +∫ 1

−1

1t − i

dt

=

∫ 1

−1

( −1t + i+

1t − i

)

dt+ i∫ 1

−1

(

11+ it

− 1−1+ it

)

dt

=

∫ 1

−1

−t + i + t + i(t + i)(t − i)

dt+ i∫ 1

−1

−1+ it − 1− it(1+ it)(−1+ it)

dt

= 2i∫ 1

−1

1t2 + 1

dt− 2i∫ 1

−1

1−1− t2

dt

= 2i∫ 1

−1

1t2 + 1

dt+ 2i∫ 1

−1

1t2 + 1

dt

= 4i∫ 1

−1

1t2 + 1

dt = 4i [arctan(z)]1−1 = 4i

[π

4−

(

−π4

)]

= 4iπ

2= 2πi.

Exercise 12.Let I(r) =∫

γ

eiz

z dz whereγ : [0, π] → C is defined byγ(t) = reit . Show thatlimr→∞ I (r) = 0.

Solution. To showlimr→∞ I (r) = 0 which is equivalent to∀ε > 0, ∃n ∈ N such that if r> N, then|I (r)| < ε.Let ε > 0, then

|I (r) − 0| = |I (r)| =∣∣∣∣∣∣

∫

γ

eiz

zdz

∣∣∣∣∣∣=

∣∣∣∣∣∣

∫ π

0

eireit

reitrieit dt

∣∣∣∣∣∣=

∣∣∣∣∣i∫ π

0eireit

dt∣∣∣∣∣≤

∫ π

0

∣∣∣∣e

ireit∣∣∣∣ dt.

Now, recall that|ez| = eRe(z) (p. 38 eq. 2.13), so∣∣∣∣e

ireit∣∣∣∣ = eRe(ireit )

= eRe(ir cos(t)+ir ·i sin(t))= eRe(ir cos(t)−r sin(t))

= e−r sin(t).

Hence,∫ π

0

∣∣∣∣e

ireit∣∣∣∣ dt =

∫ π

0e−r sin(t) dt =

∫ δ1

0e−r sin(t) dt+

∫ π−δ2

δ1

e−r sin(t) dt+∫ π

π−δ2

e−r sin(t) dt

for δ1 > 0 andδ2 > 0. Since e−r sin(t) is continuous, soδ1 > 0 can be chosen such that∫ δ1

0e−r sin(t) dt <

ε

3, ∀r > 0.

Similar,δ2 > 0 can be chosen such that∫ π

π−δ2

e−r sin(t) dt <ε

3, ∀r > 0.

46

Finally, because of the Lebesgue’s dominated convergence Theorem

limr→∞

∫ π−δ2

δ1

e−r sin(t) dt =∫ π−δ2

δ1

limr→∞

e−r sin(t) dt

we have

limr→∞

∫ π−δ2

δ1

e−r sin(t) dt <ε

3.

Hence,limr→∞

I (r) = 0.

Exercise 13. Find∫

γz−

12 dz where: (a)γ is the upper half of the unit circle from+1 to −1: (b) γ is the

lower half of the unit circle from+1 to −1.


Exercise 14.Prove that ifϕ : [a,b] → [c,d] is continuous andϕ(a) = c,ϕ(b) = d thenϕ is one-one iff ϕ isstrictly increasing.


Exercise 15.Show that the relation in Definition 1.16 is an equivalence relation.


Exercise 16.Show that ifγ andσ are equivalent rectifiable paths then V(γ) = V(σ).


Exercise 17.Show that ifγ : [a,b] → C is a path then there is an equivalent pathσ : [0,1]→ C.


Exercise 18.Prove Proposition 1.17.


Exercise 19.Letγ = 1+ eit for 0 ≤ t ≤ 2π and find∫

γ(z2 − 1)−1 dz.


Exercise 20.Letγ = 2eit for −π ≤ t ≤ π and find∫

γ(z2 − 1)−1 dz.


Exercise 21.Show that if F1 and F2 are primitives for f : G→ C and G is open end connected then thereis a constant c such that F1(z) = c+ F2(z) for each z in G.

Solution. Let F1 and F2 be primitives for f: G→ C. Recall then that F′1 = f = F′2. Let h= F1 − F2 forall z ∈ G. Then for all z∈ G,

h′(z) = F′1(z) − F′2(z) = f (z) − f (z) = 0

Since G is open and connected with h′(z) = 0 for all z ∈ G, then h is a constant, say c: h= c. Soh = F1 − F2 = c and F1(z) = F2(z) + c for all z ∈ G.

47

Exercise 22. Let γ be a closed rectifiable curve in an open set G and a< G. Show that for n≥ 2,∫

γ(z− a)−n dz= 0.

Solution. Let f(z) = (z− a)−n for all z ∈ G. Then f has the antiderivative F(z) =(z− a)1−n

1− non G. Since

F′(z) = f (z) where n≥ 2 (note F is undefined for n= 1). Now, letγ be a closed rectifiable curve in G with

endpoints z1, z2 ∈ C. Since z1 = z2, F(z1) = F(z2). So,∫

γ

(z− a)−ndz= F(z2) − F(z1) = 0.

Exercise 23. Prove the following integration by parts formula. Let f and gbe analytic in G and letγ be arectifiable curve from a to b in G. Then

∫

γf g′ = f (b)g(b) f (a)g(a) −

∫

γf ′g.

Solution. Let f,g be analytic in G and letγ be a rectifiable curve from a to b in G. Then

∫

γ

f g′ +∫

γ

f ′g =(def)

∫ b

af (γ(t))g′(γ(t)) dγ(t) +

∫ b

af ′(γ(t))g(γ(t)) dγ(t)

=Prop. 1.7 (a)

∫ b

af (γ(t))g′(γ(t)) + f ′(γ(t))g(γ(t)) dγ(t)

=(def)

∫

γ

f g′ + f ′g =∫

γ

( f g)′

Since( f g)′ is continuous ( f and g are analytic) by Theorem 1.18 we obtain∫

γ

( f g)′ = ( f g)(b) − ( f g)(a) = f (b)g(b) − f (a)g(a).

Hence, ∫

γ

f g′ = f (b)g(b) − f (a)g(a) −∫

γ

f ′g.

4.2 Power series representation of analytic functions

Exercise 1. Show that the function defined by (2.2) is continuous.


Exercise 2. Prove the following analogue of Leibniz.s rule (this exercise will be frequently used in the latersections.) Let G be an open set and letγ be a rectifiable curve inC. Suppose thatϕ : y × G → C is acontinuous function and define g: G→ C by

g(z) =∫

γ

ϕ(w, z) dw

then g is continuous. If∂ϕ∂z exists for each(w, z) in γ ×G and is continuous then g is analytic and

g′(z) =∫

γ

∂ϕ

∂z(w, z) dw.


48

Exercise 3. Suppose thatγ is a rectifiable curve inC andϕ is defined and continuous onγ. Use Exercise2 to show that

g(z) =∫

γ

ϕ(w)w− z

dw

is analytic onC − γ and

g(n)(z) = n!∫

γ

ϕ(w)(w− z)n+1

dw.


Exercise 4. (a) Prove Abel’s Theorem: Let∑

an(z− a)n have radius of convergence1 and suppose that∑

an converges to A. Prove thatlimr→1−

∑

anrn= A.

(Hint: Find a summation formula which is the analogue of integration by parts.)(b) Use Abel’s Theorem to prove thatlog 2= 1− 1

2 +13 − . . ..


Exercise 5. Give the power series expansion oflogz about z= i and find its radius of convergence.

Solution. Assume a∈ C is not zero, then

1z=

1a+ z− a

=1a

1a+z−a

a

=1a

11+ z−a

a

=1a

∞∑

n=0

(−1)n

an(z− a)n

=

∞∑

n=0

(−1)n

an+1(z− a)n,

where the radius of convergence is|a|. Let a= i, then integrate term by term yields

log(z) =π

2i +

∞∑

n=0

(

i2)n

(n+ 1)in+1(z− i)n+1

+π

2i −

∞∑

n=0

in+1

n+ z(z− i)n+1

=π

2i −

∞∑

n=1

(1+ iz)n

n,

since Log(i) = π2 i. The convergence radius is|i| = 1.

Exercise 6. Give the power series expansion of√

z about z= 1 and find its radius of convergence.

Solution. From the definition of numbers with rational exponents za= exp(a logz) we see that the square

root cannot be analytic on all ofC. In fact, taking the first derivatives we see that f(n)(z) = 2−n ∏n−1k=0(1 −

2k)z−n+ 12 and therefore

an =1n!

f (n)(1) =1

n!2n

n−1∏

k=0

,

and with these an,

f (z) =√

z=∞∑

n=0

an(z− 1)z =∞∑

k=0

1n!2n

n−1∏

k=0

(1− 2k)(z− 1)n.

Note that the logarithm is undefined at zero. Thus, of all branch cuts of the logarithm, the best radius ofconvergence we can achieve is 1.

49

Exercise 7. Use the results of this section to evaluate the following integrals:(a)

∫

γ

eiz

z2dz, γ(t) = eit , 0 ≤ t ≤ 2π;

(b)∫

γ

dzz− a

, γ(t) = a+ reit , 0 ≤ t ≤ 2π;

(c)∫

γ

sin(z)z3

dz, γ(t) = eit , 0 ≤ t ≤ 2π;

(d)∫

γ

logzzn

dz, γ(t) = 1+12

eit , 0 ≤ t ≤ 2π and n≥ 0.

Solution. a) Let a = 0, r = 1, n = 1, f (z) = eiz and f′(z) = ieiz. Clearly f(z) is analytic onC andB(0; 1)⊂ C. Now use Corollary 2.13.

f ′(0) =1!2πi

∫

γ

eiz

(z− 0)2dz ⇐⇒ i =

12πi

∫

γ

eiz

z2dz ⇐⇒

∫

γ

eiz

z2dz= −2π.

b) Let a= a, r = r, n = 0, and f(z) = 1. Clearly f(z) is analytic onC andB(a; r) ⊂ C. Now use Corollary2.13.

f (0)(a) =0!2πi

∫

γ

1(z− a)1

dz ⇐⇒ 1 =1

2πi

∫

γ

1z− a

dz ⇐⇒∫

γ

1z− a

dz= 2πi.

c) Let a= 0, r = 1, n = 2 and f(z) = sin(z). Then f′(z) = cos(z) and f′′(z) = − sin(z). Clearly f(z) isanalytic onC (p. 38) andB(0; 1)⊂ C. Now use Corollary 2.13.

f ′′(0) =2!2πi

∫

γ

sin(z)(z− 0)3

dz ⇐⇒ − sin(0)=1πi

∫

γ

sin(z)z3

dz ⇐⇒ 0 =1πi

∫

γ

sin(z)z3

dz

⇐⇒∫

γ

sin(z)z3

dz= 0.

d) We have that

f (z) =log(z)

zn

is analytic in the disk B(1; 12) where n≥ 0. Furthermore,γ is a closed rectifiable curve (γ(t) is a circle with

radius r = 1/2 and center(1,0) in C). Obviouslyγ(t) = 1+ 12eit ⊆ B(1; 1

2). Hence, by Proposition 2.15, fhas a primitive and therefore

∫

γf = 0. Therefore,

∫

γ

log(z)zn

dz= 0.

Exercise 8.Use a Möbuis transformation to show that Proposition 2.15 holds if the disk B(a; R) is replacedby a half plane.


50

Exercise 9. Evaluate the following integrals:(a)

∫

γ

ez − e−z

zndz where n is a positive integer andγ(t) = eit ,0 ≤ t ≤ 2π;

(b)∫

γ

dz(

z− 12

)n where n is a positive integer andγ(t) =12+ eit ,0 ≤ t ≤ 2π;

(c)∫

γ

dzz2 + 1

whereγ(t) = 2eit ,0 ≤ t ≤ 2π. (Hint: expand(z2+ 1)−1 by means of partial fractions);

(d)∫

γ

sinzz

dz whereγ(t) = eit ,0 ≤ t ≤ 2π;

(e)∫

γ

z1/m

(z− 1)mdz whereγ(t) = 1+

12

eit ,0 ≤ t ≤ 2π;

Solution. a) Let a= 0, r = 1, n = m− 1 and

f (z) = ez − e−z

f ′(z) = ez+ e−z

f ′′(z) = ez − e−z

...

f (m−1)(z) = ez − (−1)m+1e−z.

Clearly f(z) is analytic onC andB(0; 1)⊂ C. Now use Corollary 2.13.

f (m−1)(0) =(m− 1)!

2πi

∫

γ

ez − e−z

(z− 0)mdz

⇐⇒ (m− 1)!2πi

∫

γ

ez − e−z

zmdz=

0 ,m odd

2 ,m even

⇐⇒∫

γ

ez − e−z

zmdz=

0 ,m odd4πi

(m−1)! ,m even(m> 0).

b) Let a= 1/2, r = 1, n = m− 1 and

f (z) = 1

f ′(z) = 0

f ′′(z) = 0...

f (m−1)(z) = 0.

51

Clearly f(z) is analytic onC andB( 12; 1) ⊂ C. Now use Corollary 2.13.

f (m−1)

(

12

)

=(m− 1)!

2πi

∫

γ

1(

z− 12

)m dz

⇐⇒ (m− 1)!2πi

∫

γ

1(

z− 12

)m dz=

1 ,m= 1

0 ,m> 1

⇐⇒∫

γ

1(

z− 12

)m dz=

2πi ,m= 1

0 ,m> 1.

c) Using partial fraction decomposition, we obtain

1z2 + 1

!=

Az− i

+B

z+ i=

Az+ Ai + Bz− Biz2 + 1

which implies

A+ B = 0 (4.1)

Ai − Bi = 1. (4.2)

From (4.1), we get A= −B inserted into (4.2) yields−2Bi = 1 and therefore B= − 12i and A= 1

2i . Thus∫

γ

dzz2 + 1

=12i

∫

γ

dzz− i

− 12i

∫

γ

dzz+ i

.

For the first integral on the right side, use Proposition 2.6.Let a = 0, r = 2 and f(w) = 1, γ(t) = 2eit ,0 ≤ t ≤ 2π. Clearly f(z) is analytic onC. We also haveB(0; 2)⊂ C. Then

f (z) =1

2πi

∫

γ

f (w)w− i

dw ⇐⇒ 1 =1

2πi

∫

γ

1w− i

dw ⇐⇒∫

γ

1z− i

dz= 2πi,

where| − i − 0| < 2 ⇐⇒ 1 < 2.For the second integral on the right side, use Proposition 2.6. Let a= 0, r = 2 and f(w) = 1, γ(t) = 2eit ,0 ≤ t ≤ 2π. Clearly f(z) is analytic onC. We also haveB(0; 2)⊂ C. Then

f (z) =1

2πi

∫

γ

f (w)w+ i

dw ⇐⇒ 1 =1

2πi

∫

γ

1w+ i

dw ⇐⇒∫

γ

1z+ i

dz= 2πi,

where|i| < 2 ⇐⇒ 1 < 2. Hence∫

γ

dzz2 + 1

=12i

∫

γ

dzz− i

− 12i

∫

γ

dzz+ i

=12i

2πi − 12i

2πi = π − π = 0.

So ∫

γ

dzz2 + 1

= 0.

d) Let a = 0, r = 1, n = 0 and f(z) = sin(z). Clearly f(z) is analytic onC and B(0; 1) ⊂ C. Now useCorollary 2.13.

f (0)(0) =0!2πi

∫

γ

sin(z)(z− 0)1

dz ⇐⇒ 0 =1

2πi

∫

γ

sin(z)z

dz ⇐⇒∫

γ

sin(z)z

dz= 0.

52

e) Let a= 1, r = 1/2, n = m− 1 and

f (z) = z1/m

f ′(z) =1m

z1/m−1

f ′′(z) =1m

(

1m− 1

)

z1/m−2

...

f (m−1)(z) =1m

(

1m− 1

)

· . . . ·(

1m− (m− 2)

)

z1/m−(m−1).

Clearly f(z) is analytic onC andB(1; 12) ⊂ C. Now use Corollary 2.13.

f (m−1)(1) =(m− 1)!

2πi

∫

γ

z1/m

(z− 1)mdz

⇐⇒ 1m

(

1m− 1

)

· . . . ·(

1m− (m− 2)

)

·1 = (m− 1)!2πi

∫

γ

z1/m

(z− 1)mdz

⇐⇒∫

γ

z1/m

(z− 1)mdz=

2πi(m− 1)!

1m

(

1−mm

) (

1− 2mm

)

· . . . ·(

1− (m− 1)mm

)

⇐⇒∫

γ

z1/m

(z− 1)mdz=

2πi(m− 1)!mm−1

m−2∏

i=0

(1− im).

Exercise 10.Evaluate∫

γ

z2+1

z(z2+4) dz whereγ(t) = reit , 0 ≤ t ≤ 2π, for all possible values of r,0 < r < 2 and2 < r < ∞.

Solution. Let f(z) =z2+ 1

z(z2 + 4). Then f(z) can be decomposed using partial fractions as

f (z) =z2+ 1

z(z2 + 4)=

1/4z+

3/8z− 2i

+3/8

z+ 2i.

It follows,∫

γ

z2+ 1

z(z2 + 4)dz =

14

∫

γ

1z

dz+38

∫

γ

1z− 2i

dz+38

∫

γ

1z+ 2i

dz. By Proposition 2.6,∫

γ

1z

dz =

2πi f (0) = 2πi where f ≡ 1 as f ≡ 1 is analytic onB(0,2) andγ = 0+ reit with 0 < r < 2. Now, note 1z−2i is

analytic on B(0,2) andγ for 0 < r < 2 is in B(0,2). Thus, by Proposition 2.15,∫

γ

1z− 2i

dz= 0. Similarly,∫

γ

1z+ 2i

dz= 0. It follows for0 < r < 2,∫

γ

z2+ 1

z(z2 + 4)dz=

14

2πi =π

2i. Now, for2 < r < ∞, noteγ(t) = reit

soγ′(t) = ireit . Calculating the integrals in turn,∫

γ

1z

dz=∫ 2π

0

ireit

reitdt = i

∫ 2π

0dt = 2πi.

For the second integral, we use Proposition 2.6 with a= 0, 2 < r < ∞, f (w) = 1, andγ(t) = 0 + reit for

53

0 ≤ t ≤ 2π. Clearly f(z) is analytic onC andB(0; 2< r < ∞) ⊂ C. Then,

f (−2i) =1

2πi

∫

γ

f (w)w+ 2i

dw ⇐⇒ 1 =1

2πi

∫

γ

f (w)w+ 2i

dw

⇐⇒∫

γ

1z+ 2i

dz= 2πi

if 2 < r < ∞. Similarly,

f (2i) =1

2πi

∫

γ

f (w)w− 2i

dw ⇐⇒∫

γ

1z− 2i

dz= 2πi.

It follows that,

∫

γ

z2+ 1

z(z2 + 4)dz=

14

∫

γ

1z

dz+38

∫

γ

1z− 2i

dz+38

∫

γ

1z+ 2i

dz

=14

(2πi) +38

(2πi) +38

(2πi)

= 2πi

Exercise 11.Find the domain of analyticity of

f (z) =12i

log

(

1+ iz1− iz

)

;

also, show thattan f (z) = z (i.e., f is a branch of arctan z). Show that

f (z) =∞∑

k=0

(−1)kz2k+1

2k+ 1, for |z| < 1

(Hint: see Exercise III. 3.19.)


Exercise 12.Show that

secz= 1+∞∑

k=1

E2k

(2k)!z2k

for some constants E2,E4, · · · . These numbers are called Euler’s constants. What is the radius of conver-gence of this series? Use the fact that1 = coszsecz to show that

E2n −(

2n2n− 2

)

E2n−2 +

(

2n2n− 4

)

E2n−4 + · · · + (−1)n−1

(

2n2

)

E2 + (−1)n = 0.

Evaluate E2,E4,E6,E8. (E10 = 50521and E12 = 2702765).

Solution. It is easily seen that

secz= 1+∞∑

k=1

E2k

(2k)!z2k, (for some E2,E4, . . .),

54

sincesec(0)= 1cos(0) =

11 = 1 andsecz = 1

cosz is an even function. The radius of convergence is the closestdistance from 0 to the nearest non-analytic point which is preciselyπ2 .Since1 = coszsecz, we can multiply the series forsecz andcosz together to obtain

1 =∞∑

i=0

n∑

k=0

(−1)i−k E2k

(2k)!(2i − 2k)!

z2i .

Finally, we can compare the coefficients of z2i and multiply by(2i)! to obtain the formula

n∑

i=0

(−1)n−kE2k

(

2n2k

)

= 0

which is equivalent to

E2n −(

2n2n− 2

)

E2n−2 +

(

2n2n− 4

)

E2n−4 + · · · + (−1)n−1

(

2n2

)

E2 + (−1)n = 0.

In addition, we have E2 = 1, E4 = 5, E6 = 61and E8 = 1285.

Exercise 13.Find the series expansion ofez−1z about zero and determine its radius of convergence. Consider

f (z) = zez−1 and let

f (z) =∞∑

k=0

ak

k!zk

be its power series expansion about zero. What is the radius of convergence? Show that

0 = a0 +

(

n+ 11

)

a1 + · · · +(

n+ 1n

)

an.

Using the fact that f(z) + 12z is an even function show that ak = 0 for k odd and k> 1. The numbers

B2n = (−1)n−1a2n are called the Bernoulli numbers for n≥ 1. Calculate B2, B4, · · · , Bl0.

Solution. The Bernoulli numbers Bn are defined by

Bn = limz→∞

dn zez−1

dzn, n = 0,1,2, . . .

We determine the largest R such that

zez − 1

=

∞∑

k=0

Bk

k!zk for 0 < |z| < R.

The function f(z) = zez−1 has a removable singularity at z= 0, because it can be extended to0 by an

analytic function g on a small neighborhood B(0, r) of 0, setting

g(z) :=

zez−1 if 0 < |z| < r

1 if x = 0.

Furthermore the function f has essential singularities exactly at z= 2πk, k ∈ Z − 0. Therefore f(z) hasa Laurent series expansion on the punctured disk ann(0; 0,2π) and the negative indices disappear becausethe singularity at the origin is a removable one. The uniqueness of the Laurent expression together with the

55

formula for the coefficients in the expansion (Theorem 2.8, p. 72) give the desiredresult.First we compute B0 = limz→0

zez−1 = 1 and note that the result in question does not hold for B0. Also

compute B1. Taking the limits, B1 = limz→0(1−z)ez−1(ez−1)2 = −

12. Consider the equation z

ez−1 =∑∞

k=0Bk

k! zk forz ∈ ann(0; 0,2π). This is equivalent to

z=

∞∑

l=1

zl

l!

∞∑

k=0

Bk

k!zk

.

Divide by z, 0, then

1 =

∞∑

l=1

zl−1

l!

∞∑

k=0

Bk

k!

=

∞∑

l=0

zl

(l + 1)!

∞∑

k=0

Bk

k!

=

∞∑

n=0

n∑

m=0

Bm

m!zm zn−m

(n+ 1−m)!

=

∞∑

n=0

znn∑

m=0

Bm

m!(n+ 1−m)!

=

∞∑

n=0

zn

(n+ 1)!

n∑

m=0

(

n+ 1m

)

Bm

= B0 +

∞∑

n=1

zn

(n+ 1)!

n∑

m=0

(

n+ 1m

)

Bm.

Since B0 = 0 conclude that

0 =∞∑

n=1

zn

(n+ 1)!

n∑

m=0

(

n+ 1m

)

Bm

for all nonzero z∈ ann(0; 0,2π). This can only happen if the inner summation equals zero for all n ∈ N.This prove the first part of the statement. For the second claim, one only needs to show that f(z) + 1

2z is aneven function. It is true that

f (z) +12

z=2z+ (ez − 1)z

2(ez − 1)=

z(ez+ 1)

2(ez − 1).

as well as t’

f (−z) − 12

z=−z(ez

+ 1)e−z − 1

=−z(1+ ez)2(1− ez)

=z(1+ ez)2(ez − 1)

= f (z) +12

z.

With the power series expansion of f and recalling that B1 = − 12 this is the same as saying

1+∞∑

k=2

Bk

k!(−z)k

= 1+∞∑

k=2

Bk

k!(z)k

or, yet in other terms,Bk = −Bk = 0 for k odd, k > 1.

56

In part b) the values of B0 = 1 and B1 = − 12 have been computed already. Also B3 = B5 = B7 = 0 by

the second statement of part b). It remains to compute B2, B4, B6 and B8. The formula involving binomialcoefficients in the part above can be solved for Bn and

Bn = −1

n+ 1

n−1∑

k=0

(

n+ 1k

)

Bk

.

This gives

B2 = −13

(B0 + 3B1) = −13

(

1− 32

)

=16,

B4 = −15

(B0 + 5B1 + 10B2) = − 130,

B6 = −17

(B0 + 7B1 + 21B2 + 35B4) =142,

and

B8 = −19

(B0 + 9B1 + 36B2 + 126B4 + 84B6) = − 130.

Exercise 14. Find the power series expansion oftanz about z= 0, expressing the coefficients in terms ofBernoulli numbers. (Hint: use Exercise 13 and the formulacot 2z= 1

2 cotz− 12 tanz.)

Solution. To find the power series expansion oftanz about z= 0, we replace z by2iz in the function

f (z) =z(ez+ 1)

2(ez − 1)

which we obtained in the previous problem. We obtain

f (2iz) =2iz(e2iz

+ 1)2(e2iz − 1)

=iz(eiz

+ e−iz)eiz − e−iz

= zcos(z)sin(z)

= zcot(z).

Next, we replace z by2iz of the power series expansion for f(z) to obtain

zcot(z) = 1−∞∑

n=1

4nB2n

(2n)!z2n.

Multiplying the formulacot 2z= 12 cotz− 1

2 tanz with2z gives

ztanz= zcotz− 2zcot 2z

and therefore

ztanz=

1−∞∑

n=1

4nB2n

(2n)!z2n

−

1−∞∑

n=1

42nB2n

(2n)!z2n

=

∞∑

n=1

4n(4n − 1)B2n

(2n)!z2n.

Hence,

tan(z) =∞∑

n=1

4n(4n − 1)B2n

(2n)!z2n−1.

57

4.3 Zeros of an analytic function

Exercise 1. Let f be an entire function and suppose there is a constant M, an R> 0, and an integer n≥ 1such that| f (z)| ≤ M|z|n for |z| > R. Show that f is a polynomial of degree≤ n.

Solution. Since f is an entire function, that is f∈ A(C), we have by Proposition 3.3 p. 77, that f has apower series expansion

f (z) =∞∑

k=0

akzk, whereak =

f (k)(0)k!

with infinite radius of convergence. We have to show that f(k)(0) = 0 for k ≥ n+ 1, then we are done. Letr > R and letγ(t) = reit , 0 ≤ t ≤ 2π. Note thatγ is a closed and rectifiable curve. Then

| f (k)(0)| =Corollary 2.13 p.73

∣∣∣∣∣∣

k!2πi

∫

γ

f (z)zk+1

dz

∣∣∣∣∣∣≤ k!

2π

∫

γ

| f (z)||z|k+1

|dz|.

Since r> R, |γ(t)| = |reit | = r > R,0 ≤ t ≤ 2π. So| f (z)| ≤ M|z|n, z∈ γ, we get

k!2π

∫

γ

| f (z)||z|k+1

|dz| ≤ k!2π

∫

γ

M|z|n|z|k+1

|dz| = k!M2π

∫

γ

|z|n−k−1 |dz|

=|z|=r

k!M2π

rn−k−1∫

γ

|dz| = k!M2π

rn−k−12πr

= k!Mrn−k= k!M

1rk−n→ 0

as r→ ∞ since k≥ n+ 1 ⇐⇒ k− n ≥ 1. Hence,| f (k)(0)| = 0 ∀k ≥ n+ 1 and thus

f (z) =n∑

k=0

f (k)(0)k!

zk;

a polynomial of degree less or equal n.

Exercise 2. Give an example to show that G must be assumed to be connected in Theorem 3.7.


Exercise 3. Find all entire functions f such that f(x) = ex for x inR.

Solution. Let f(z) = ez. Clearly f is analytic onC since it is entire. Then for all x∈ R, f (x) = ex. Now letg be an entire function such that g(x) = ex for all x ∈ R. Now, set H= z ∈ G : f (z) = g(z). So,R ⊆ H.Clearly0 is a limit point of H and0 ∈ G. Hence, by Corollary3.8, f (z) ≡ g(z). So f(z) = ez is the set of allentire functions f such that f(x) = ex for all x ∈ R.

Exercise 4. Prove that ez+a= ezea by applying Corollary 3.8.


Exercise 5. Prove thatcos(a+ b) = cosacosb− sinasinb by applying Corollary 3.8.


Exercise 6. Let G be a region and suppose that f: G → C is analytic and a∈ G such that| f (a)| ≤ | f (z)|for all z in G. Show that either f(a) = 0 or f is constant.

58

Solution. Let f : G→ C be analytic, where G is a region. Let a∈ G such that

| f (a)| ≤ | f (z)| ∀z ∈ G. (4.3)

Obviously, f(a) satisfies (4.3) since0 ≤ | f (z)| ∀z ∈ G is true.Now, suppose f(a) , 0 ∀a ∈ G. Define

g(z) =1

f (z).

Clearly, g∈ A(G) (since f(a) , 0 ∀a ∈ G) and

|g(z)| =(def)

1| f (z)| ≤(4.3)

1| f (a)| =(def)

|g(a)|, ∀z ∈ G.

According to the Maximum Modulus Theorem 3.11 p. 79, g is constant and thus f is constant. Thus, weconclude: Either f(a) = 0 or f is constant.

Exercise 7. Give an elementary proof of the Maximum Modulus Theorem for polynomials.


Exercise 8. Let G be a region and let f and g be analytic functions on G such that f(z)g(z) = 0 for all z inG. Show that either f≡ 0 or g ≡ 0.

Solution. It is easy to verify the following claim:If f (z) =constant in B(a; R) ⊂ G, then f(z) =same constant for all z∈ G.Proof: Suppose f(z) = c in B(a; R). Define g(z) = f (z) − c, then g(z) ∈ A(G). g(n)(a) = 0 for alln = 0,1,2, . . .. So g(z) = f (z) − c ≡ 0 on G.

Let G be a region and let f,g ∈ A(G) such that

f (z)g(z) = 0, ∀z ∈ G.

Assume f(z) , 0, ∀z ∈ G. Let a∈ G such that f(a) , 0, then (since f is continuous, because f∈ A(G))there is an R> 0 such that f(z) , 0 whenever z∈ B(a; R). Hence g(z) = 0∀z ∈ B(a; R) to fulfill f (z)g(z) = 0∀z ∈ B(a; R). By the claim above we obtain that

g(z) = 0, ∀z ∈ G.

So g≡ 0.Similar, assume g(z) , 0 ∀z ∈ G, then by the same reasoning, we obtain f≡ 0. Thus, we conclude: Eitherf ≡ 0 or g ≡ 0.

Exercise 9. Let U : C → R be a harmonic function such that U(z) ≥ 0 for all z in C; prove that U isconstant.

Solution. There exists an analytic function f: C→ C such that

U(z) = Re( f (z)),

since U : C → R is assumed to be harmonic. Hence, Re( f (z)) ≥ 0 ∀z ∈ C by assumption. For all z∈ C,define g(z) = e− f (z) ∈ A(C). Then

|g(z)| =(def)|e− f (z)| = eRe(− f (z))

= e−Re( f (z)) ≤ 1, ∀z ∈ C,

where the last step follows since Re( f (z)) ≥ 0 by assumption. Thus, g(z) is bounded and an entire function,and therefore g is constant according to Liouville’s Theorem 3.4 p. 77 and hence f(z) is constant, too. Itfollows that U is constant.

59

Exercise 10. Show that if f and g are analytic functions on a region G such that f g is analytic then eitherf is constant or g≡ 0.


4.4 The index of a closed curve



Exercise 2. Give an example of a closed rectifiable curveγ in C such that for any integer k there is a pointa < γ with n(γ; a) = k.


Exercise 3. Let p(z) be a polynomial of degree n and let R> 0 be sufficiently large so that p never vanishesin z : |z| ≥ R. If γ(t) = Reit ,0 ≤ t ≤ 2π, show that

∫

γ

p′(z)p(z) dz= 2πin.

Solution. Since p(z) is a polynomial of degree n, we can write

p(z) = cn∏

k=1

(z− ak)

for some constant c (see Corollary 3.6 p. 77) where a1, . . . ,an are the zeros of p(z). By the product rule weobtain

p′(z) = cn∑

k=1

(z− ak)′

n∏

l=1l,k

(z− al) = cn∑

k=1

n∏

l=1l,k

(z− al).

Thus

∫

γ

p′(z)p(z)

dz =

∫

γ

c∑n

k=1∏n

l=1l,k

(z− al)

c∏n

k=1(z− ak)dz

=

∫

γ

(

1z− a1

+1

z− a2+ . . . +

1z− an

)

dz

=

∫

γ

(z− a1)−1 dz+∫

γ

(z− a2)−1 dz+ . . . +∫

γ

(z− an)−1 dz

= n(γ,a1)2πi + n(γ,a2)2πi + . . . + n(γ,an)2πi =n∑

k=1

2πi = 2πin,

since by Definition 4.2 p. 81γ(t) = Reit , 0 ≤ t ≤ 2π or γ(t) = Re2πit , 0 ≤ t ≤ 1 is a closed and rectifiablecurve and a1, . . . ,an < γ since R> 0 sufficiently large so that p never vanishes inz : |z| ≥ R. In addition,we have n(γ; ai) = 1 ∀1 ≤ i ≤ n. Hence,

∫

γ

p′(z)p(z)

dz= 2πin.

Exercise 4. Fix w = reiθ, 0 and letγ be a rectifiable path inC − 0 from 1 to w. Show that there is an

integer k such that∫

γz−1 dz= log r + iθ + 2πik.


60

4.5 Cauchy’s Theorem and Integral Formula

Exercise 1. Suppose f: G→ C is analytic and defineϕ : G ×G→ C byϕ(z,w) = [ f (z) − f (w)](z− w)−1

if z , w andϕ(z, z) = f ′(z). Prove thatϕ is continuous and for each fixed w, z→ ϕ(z,w) is analytic.


Exercise 2. Give the details of the proof of Theorem 5.6.


Exercise 3.Let B± = B(±1; 12), G = B(0; 3)−(B+∪B−). Letγ1, γ2, γ3 be curves whose traces are|z−1| = 1,

|z+1| = 1, and|z| = 2, respectively. Giveγ1, γ2, andγ3 orientations such that n(γ1; w)+n(γ2; w)+n(γ3; w) =0 for all w in C −G.


Exercise 4. Show that the Integral Formula follows from Cauchy’s Theorem.

Solution. Assume the conditions of Cauchy’s Theorem, that is: Let G be an open subset of the plane andf : G → C an analytic function. Ifγ is a closed rectifiable curve in G such that n(γ; w) = 0 ∀w ∈ G − C,then

∫

γf = 0.

Now, let a∈ G, where a< γ. Define

g(z) =

f (z)− f (a)z−a , z, a

f ′(a), z= a.

Clearly, g ∈ A(G). (Check the power series of f(z) like in the proof of Theorem 3.7 c)⇒ b) p. 78) ByCauchy’s Theorem, we obtain

∫

γ

g = 0 ⇒a<γ

∫

γ

f (z) − f (a)z− a

dz= 0

⇒∫

γ

f (z)z− a

dz−∫

γ

f (a)z− a

dz= 0

⇒∫

γ

f (z)z− a

dz= f (a)∫

γ

1z− a

dz

⇒Def. 4.2 p.81

∫

γ

f (z)z− a

dz= f (a)n(γ; a)2πi

⇒ 12πi

∫

γ

f (z)z− a

dz= n(γ; a) f (a).

This is exactly Cauchy’s Integral Formula.

Exercise 5. Letγ be a closed rectifiable curve inC and a< γ. Show that for n≥ 2∫

γ(z− a)−n dz= 0.

Solution. Sinceγ is a closed rectifiable curve inC, γ is bounded. So there exists R> 0 such thatγ ⊂ B(a; R). HenceC\B(a; R) belongs to the unbounded component of B(a; R). Therefore n(γ; w) = 0∀w ∈ C\B(a; R) by Theorem 4.4 p.82. We can apply Corollary 5.9 p.86 if we let G= B(a; R) the open setand f : G→ C given by f(z) = 1 is analytic. Furtherγ is a closed and rectifiable curve in G such that

n(γ; w) = 0, ∀w ∈ C\G,

61

then for a in B(a; R)\γ

f (n−1)(a)n(γ; a) =(n− 1)!

2πi

∫

γ

1(z− a)n

dz

⇒ 0·n(γ; a) =(n− 1)!

2πi

∫

γ

1(z− a)n

dz

since f(n−1)(a) = 0 if n ≥ 2. Hence∫

γ

(z− a)−n dz= 0, for n ≥ 2.

Exercise 6. Let f be analytic on D= B(0; 1)and suppose| f (z)| ≤ 1 for |z| < 1. Show| f ′(0)| ≤ 1.

Solution. We can directly apply Cauchy’s estimate 2.14 p. 73 with a= 0, R= 1, M = 1 and N= 1, that isby assumption f is analytic on B(0; 1)and | f (z)| ≤ 1 for all z in B(0; 1) (since|z| < 1). Then

| f ′(0)| ≤ 1!·111= 1.

So| f ′(0)| ≤ 1.

Exercise 7. Letγ(t) = 1+ eit for 0 ≤ t ≤ 2π. Find∫

γ

(z

z−1

)ndz for all positive integers n.

Solution. Let G = B(1;R) (open) where1 < R < ∞ and let f : G → C be given by f(z) = zn whichis analytic on G. By Theorem 4.4 p. 82, n(γ; w) = 0 ∀w ∈ C\G sinceC\G belongs to the unboundedcomponent of G andγ is a closed and rectifiable curve inC.Now, we can apply Corollary 5.9 with G and f defined above such that n(γ; w) = 0 ∀w ∈ C\G andγ is aclosed rectifiable curve in G, then for a= 1 in G\γ

f (n−1)(1)n(γ; a) =(n− 1)!

2πi

∫

γ

zn

(z− 1)ndz=

(n− 1)!2πi

∫

γ

( zz− 1

)ndz

⇒ n! =(n− 1)!

2πi

∫

γ

( zz− 1

)ndz

⇒∫

γ

( zz− 1

)ndz= 2πin,

since f(n−1)(1) = n! and n(γ; a) = 1.

Exercise 8. Let G be a region and suppose fn : G → C is analytic for each n≥ 1. Suppose that fnconverges uniformly to a function f: G→ C Show that f is analytic.

Solution. Let G be a region and suppose fn : G→ C is analytic for each n≥ 1, that us fn is continuous foreach n≥ 1.Let T be an arbitrary triangular path in G. Then T is closed andrectifiable. We also have, that n(T; w) = 0∀w ∈ C\G, since w is an element of the unbounded component of G (Theorem 4.4 p. 82).By Cauchy’s Theorem (First Version), p. 85 with m= 1, we have

∫

Tfn = 0, ∀n ∈ N, (4.4)

62

since we suppose fn : G→ C is analytic for each n≥ 1 and T is closed and rectifiable. We also have∫

Tf = lim

n→∞

∫

Tfn =

(4.4)limn→∞

0 = 0

The first equality follows by Lemma 2.7 p. 71 since T is closed and rectifiable and since each fn ∈ A(G),we have fn is continuous for each fn and by the uniform convergence to a function f , we get f is continuous(Theorem 6.1 p. 29).Finally, we can apply Morera’s Theorem 5.10 p. 86 to obtain that f is analytic in G, since f is continuousand

∫

γf = 0 for every triangular path T in G.

Exercise 9. Show that if f: C→ C is a continuous function such that f is analytic off [−1,1] then f is anentire function.

Solution. By Morera’s Theorem, it suffice to show that for every triangular path T inC, we have∫

Tf = 0,

since we already assume f is continuous. There are several cases to consider:Case 1: A triangle does not intersect I. In this case, we obtain by Cauchy’s Theorem, that

∫

Tf = 0,

since we can find an open neighborhood G containing T (is closed and rectifiable) such that f∈ A(G) (byassumption) and n(T; w) = 0 ∀w ∈ C\G (by Theorem 4.4 p. 82).Case 2: A triangle touches I exactly at one point P. This single point of intersection P is a removablesingularity, since f is continuous. Again apply Cauchy’s Theorem to obtain

∫

Tf = 0. (Procedure: Translate

the triangle by±εi depending if it lies above or below the x-axis. By Cauchy’s Theorem∫

Tf = 0 over the

translated triangle and letε → 0 since f is continuous).Case 3: One edge of the triangle touches I. Like in case 2 we cantranslate the triangle by±ε We canalso argue this way: Let G be an open neighborhood containingthe triangle T. LetTn be a sequence oftriangles that are not intersecting I, but whose limit is thegiven T. Then by the continuity of f , we get

∫

Tf = lim

n→∞

∫

Tn

f = limn→∞

0 = 0,

where the first equality follows by Lemma 2.7 and the second one by case 1.Case 4: A triangle T is cut into 2 parts by I. Then, we can alwaysdecompose the triangle T in three parts.Two triangles are of the kind explained in case 3 and one is of the kind explained in case 2 (a sketch mighthelp). Thus,

∫

Tf = 0 again.

Case 5: A triangle T contains parts of I (also here a sketch might help). In this instance, we can decomposethe triangle into 5 parts. Two triangles are of the kind explained in case 3 and the other three triangles areof the kind explained in case 2. Hence,

∫

Tf = 0.

Summary: Since∫

Tf = 0 for every triangular path inC and f is continuous, we get by Morera’s Theorem:

f ∈ A(C), that is f is an entire function.

Exercise 10. Use Cauchy’s Integral Formula to prove the Cayley–HamiltonTheorem: If A is an n× nmatrix overC and f(z) = det(z− A) is the characteristic polynomial of A then f(A) = 0. (This exercise wastaken from a paper by C. A. McCarthy, Amer. Math. Monthly, 82 (1975), 390–391).


4.6 The homotopic version of Cauchy’s Theorem and simple connec-tivity

Exercise 1. Let G be a region and letσ1, σ2 : [0,1] → G be the constant curvesσ1(t) ≡ a, σ2(t) ≡ b.Show that ifγ is closed rectifiable curve in G andγ ∼ σ1, thenγ ∼ σ2. (Hint: connect a and b by a curve.)

63

Solution. Assumeγ is a closed and rectifiable curve in G andγ ∼ σ1. Sinceσ1(t) ≡ a andσ2(t) ≡ b, wesurely have thatσ1 andσ2 are closed and rectifiable curves in G. If we can show thatσ1 ∼ σ2, then we getγ ∼ σ2, since∼ is an equivalence relation, that is:

γ ∼ σ1 and σ1 ∼ σ2⇒ γ ∼ σ2.

So, now we showσ1 ∼ σ2, that is, we have to find a continuous functionΓ : [0,1] × [0,1]→ G such that

Γ(s,0) = σ1(s) = a and Γ(s,1) = σ2(s) = b, for 0 ≤ s≤ 1

andΓ(0, t) = Γ(1, t) for 0 ≤ t ≤ 1. Let Γ(s, t) = tb + (1 − t)a. ClearlyΓ is a continuous function. (it isconstant with respect to s and a line with respect to t). In addition, it satisfiesΓ(s,0) = a ∀0 ≤ s ≤ 1 andΓ(s,1) = b ∀0 ≤ s≤ 1 andΓ(0, t) = tb+ (1− t)a = Γ(1, t) ∀0 ≤ t ≤ 1. Soσ1 ∼ σ2.

Exercise 2. Show that if we remove the requirement “Γ(0, t) = Γ(1, t) for all t” from Definition6.1 then thecurveγ0(t) = e2πit , 0 ≤ t ≤ 1, is homotopic to the constant curveγ1(t) ≡ 1 in the region G= C − 0.Solution. Not available.

Exercise 3. 3. LetC = all rectifiable curves in G joining a to b and show that Definition 6.11 gives anequivalence relation onC.


Exercise 4. Let G= C − 0 and show that every closed curve in G is homotopic to a closed curve whosetrace is contained inz : |z| = 1.Solution. Not available.

Exercise 5. Evaluate the integral∫

γ

dzz2+1 whereγ(θ) = 2| cos 2θ|eiθ for 0 ≤ θ ≤ 2π.

Solution. A sketch shows that the two zeros of z2+ 1 (they are±i) are inside of the closed and rectifiable

curveγ (The region looks like a clover with four leaves ). Using partial fraction decompositions gives∫

γ

dzz2 + 1

=12i

∫

γ

1z− i

dz− 12i

∫

γ

1z+ i

dz

= π1

2πi

∫

γ

1z− i

dz− π 12i

∫

γ

1z+ i

dz

=Def 4.2 p. 81

π (n(γ; i) − n(γ;−i))

= π·0 = 0,

since n(γ; i) = n(γ;−i) since i and−i are contained in the region generated byγ. Hence∫

γ

dzz2 + 1

= 0.

Exercise 6. Letγ(θ) = θeiθ for 0 ≤ θ ≤ 2π andγ(θ) = 4π − θ for 2π ≤ θ ≤ 4π. Evaluate∫

γ

dzz2+π2 .

Solution. A sketch reveals that the zero−πi is inside the region and the zeroπi is outside the region. Usingpartial fraction decomposition yields

∫

γ

dzz2 + π2

=1

2πi

∫

γ

1z− iπ

dz− 12πi

∫

γ

1z+ iπ

dz

=Def 4.2 p. 81

n(γ; iπ) − n(γ;−iπ) = 0− 1 = −1,

64

since iπ is not contained in the region generated byγ, so n(γ; iπ) = 0 and−iπ is contained in the region, son(γ;−iπ) = 1. Hence,

∫

γ

dzz2 + π2

= −1.

Exercise 7.Let f(z) = [(z− 12−i)· (z−1− 3

2 i)· (z−1− i2)· (z− 3

2−i)]−1 and letγ be the polygon[0,2,2+2i,2i,0].Find

∫

γf .


Exercise 8. Let G= C − a,b, a , b, and letγ be the curve in the figure below.(a) Show that n(γ; a) = n(γ; b) = 0.(b) Convince yourself thatγ is not homotopic to zero. (Notice that the word is “convince”and not “prove”.Can you prove it?) Notice that this example shows that it is possible to have a closed curveγ in a regionsuch that n(γ; z) = 0 for all z not in G withoutγ being homotopic to zero. That is, the converse to Corollary6.10 is false.

Solution. Let γ be the path depicted on p. 96. We can write it as a sum of 6 paths.Two of them willbe closed and have a and b in their unbounded component. Therefore, two integrals will be zero and wewill have another 2 pair of non-closed paths. The first pair begins at the leftmost crossing pair and goesaround a in opposite direction. The second pair begins at themiddle crossing pair and goes around b inopposite direction. They also meet at the rightmost crossing point. If we integrate over the path around ais equivalent to evaluate

∫

γ1−γ2

1z−a dz whereγ1(t) = a+ reit , 0 ≤ t ≤ π andγ2(t) = a+ reit , π ≤ t ≤ 2π for

some r> 0. It is easily seen that we have∫

γ1

1z− a

dz=∫

γ2

1z− a

dz= πi

and therefore n(γ; a) = 0 and similarly we obtain n(γ; b) = 0.

Exercise 9. Let G be a region and letγ0 andγ1 be two closed smooth curves in G. Supposeγ0 ∼ γ1, andΓsatisfies (6.2). Also suppose thatγt(s) = Γ(s, t) is smooth for each t. If w∈ C −G define h(t) = n(γt; w) andshow that h: [0,1]→ Z is continuous.


Exercise 10. Find all possible values of∫

γ

dz1+z2 , whereγ is any closed rectifiable curve inC not passing

through±i.

Solution. Letγ be a closed rectifiable curve inC not passing through±i. Using partial fraction decompo-sition and the definition of the winding number we obtain

∫

γ

dz1+ z2

=12i

(∫

γ

1z− 1

dz−∫

γ

1z+ i

dz

)

=12i

(2πin(γ; i) − 2πin(γ;−i)) = π (n(γ; i) − n(γ;−i)) .

Exercise 11.Evaluate∫

γ

ez−e−z

z4 dz whereγ is one of the curves depicted below. (Justify your answer.)

Solution. Using Corollary 5.9 p. 86 we have∀a ∈ G − γ∫

γ

f (z)(z− a)k+1

dz= 2πi f (k)(a)n(γ; a)1k!,

65

whereγ is a closed rectifiable curve in G (open) such that n(γ; w) = 0 for all w ∈ C −G and f : G→ C isanalytic.In our case a= 0 andγ satisfies the above assumptions for exercise a), b) and c). Let f (z) = ez− e−z whichis clearly analytic on any region G. We have f(3)(0) = e0

+ e−0= 2 and using the formula above for k= 3

we obtain ∫

γ

ez − e−z

z4dz= 2πi·2n(γ; 0)

16=

2πi3

n(γ; 0).

Hence, n(γ; 0) = 1 for part a) and thus the result is2πi3 . We have n(γ; 0) = 1 for part b) and thus the result

is 4πi3 . We obtain the same result for part c).

4.7 Counting zeros; the Open Mapping Theorem

Exercise 1. Show that if f : G → C is analytic andγ is a rectifiable curve in G then f γ is also arectifiable curve.


Exercise 2. Let G be open and suppose thatγ is a closed rectifiable curve in G such thatγ ≈ 0. Setr = d(γ, ∂G) and H = z ∈ C : n(γ; z) = 0. (a) Show thatz : d(z, ∂G) < 1

2r ⊂ H. (b) Use part (a)to show that if f : G → C is analytic then f(z) = α has at most a finite number of solutions z such thatn(γ; z) , 0.


Exercise 3. Let f be analytic in B(a; R) and suppose that f(a) = 0. Show that a is a zero of multiplicity miff f (m−1)(a) = . . . = f (a) = 0 and f(a)(a) , 0.

Solution. ⇐: Let f (n)(a) = 0 for all n = 1,2, . . . ,m− 1 (we also have f(a) = 0) and f(m)(a) , 0. Sincef ∈ A(B(a; R)), we can write

f (z) =

∞∑

k=0

ak(z− a)k where ak =f (k)(a)

k!and |z− a| < R

=

∞∑

k=0

f (k)(a)k!

(z− a)k=

∞∑

k=m

f (k)(a)k!

(z− a)k

= (z− a)m∞∑

k=0

f (k)(a)k!

(z− a)k−m

︸︷︷︸

g(z)

=g(a),0, since f (m),0

(z− a)mg(z)

where g∈ A(B(a; R)). Hence, a is a zero of multiplicity m.⇒: Suppose a is a zero of multiplicity m≥ 1 of f . By definition, there exists g∈ A(B(0;R)) such that

f (z) = (z− a)mg(z), g(a) , 0 (p. 97).

Then after a lengthy calculation, we obtain

f (n)(z) =n∑

k=0

(

nk

)

m!(m− (n− k))!

(z− a)m−(n−k)g(k)(z), n = 1,2, . . . ,m.

66

If z = a, then clearly f(n)(a) = 0 for m− (n− k) > 0 and f(n)(a) , 0 for m− (n− k) = 0. If k = 0 and m= n,then f(m)(a) = f (n)(a) = m!g(a) , 0 since g(a) , 0 by assumption. Hence, f(n)(a) = 0 for n = 1,2, . . . ,m−1and f(n)(a) , 0 for n = m.

Exercise 4. Suppose that f: G→ C is analytic and one-one; show that f′(z) , 0 for any z in G.

Solution. Assume f′(a) = 0 for some a∈ G. Then f(a) = α whereα is a constant. Define

F(z) = f (z) − α.

Then F(a) = f (a) − α = α − α = 0 and F′(a) = f ′(a) = 0 by assumption. Hence, F has a as a root ofmultiplicity m≥ 2.Then, we can use Theorem 7.4 p. 98 to argue that there is anε > 0 andδ > 0 such that for0 < |ξ − a| < δ,the equation f(z) = ξ has exactly m simple roots in B(a; ε), since f is analytic in B(a; R) andα = f (a). Alsof (z) − α has a zero of order m≥ 2 at z= a.Since f(z) = ξ has exactly m (our case m≥ 2) simple roots in B(a; ε), we can find at least two distinctpoints z1, z2 ∈ B(a; ε) ⊂ G such that

f (z1) = ξ = f (z2)

contradicting that f is1− 1. So f′(z) , 0 for any z in G.

Exercise 5. Let X andΩ be metric spaces and suppose that f: X→ Ω, is one-one and onto. Show that fis an open map iff f is a closed map. (A function f is a closed map if it takes closed sets onto closed sets.)


Exercise 6. Let P : C→ R be defined by P(z) = Re z; show that P is an open map but is not a closed map.(Hint: Consider the set F= z : Im z= (Re z)−1 and Re z, 0.)


Exercise 7. Use Theorem 7.2 to give another proof of the Fundamental Theorem of Algebra.


4.8 Goursat’s Theorem


67

Chapter 5

Singularities

5.1 Classification of singularities

Exercise 1. Each of the following functions f has an isolated singularity at z= 0. Determine its nature; ifit is a removable singularity define f(0) so that f is analytic at z= 0; if it is a pole find the singular part; ifit is an essential singularity determine f(z : 0 < |z| < δ) for arbitrarily small values ofδ.a)

f (z) =sinz

z;

b)

f (z) =cosz

z;

c)

f (z) =cosz− 1

z;

d)f (z) = exp(z−1);

e)

f (z) =log(z+ 1)

z2;

f)

f (z) =cos(z−1)

z−1;

g)

f (z) =z2+ 1

z(z− 1);

h)f (z) = (1− ez)−1;

i)

f (z) = zsin1z

;

68

j)

f (z) = zn sin1z.

Solution. a) According to Theorem 1.2 p.103, z= 0 is a removable singularity, since

limz→∞

(z− 0)sin(z)

z= lim

z→∞zsin(z)

z= lim

z→∞sin(z) = 0.

If we define,

f (z) =

sin(z)z , z, 0

1, z= 0,

then f is analytic at z= 0. A simple computation using L’Hospital’s rule yieldslimz→0sin(z)

z = 1.b) We have (see p. 38)

cos(z) =∞∑

k=0

(−1)kz2k

(2k)!

and thereforecos(z)

z=

∞∑

k=0

(−1)kz2k−1

(2k)!=

1z+ analytic part.

Thuscoszz has a simple pole at z= 0.

c) According to Theorem 1.2 p.103, z= 0 is a removable singularity, since

limz→∞

(z− 0)cos(z) − 1

z= lim

z→∞zcos(z) − 1

z= lim

z→∞cos(z) − 1 = 1− 1 = 0.

If we define,

f (z) =

cos(z)−1z , z, 0

1, z= 0,

then f is analytic at z= 0. A simple computation using L’Hospital’s rule yieldslimz→0cos(z)−1

z = 0.e) We know

log(1+ z) =∞∑

k=1

(−1)k+1 zk

k.

So we can write

log(1+ z)z2

=1z2

∞∑

k=1

(−1)k+1 zk

k=

∞∑

k=1

(−1)k+1 zk−2

k=

1z− 1

2+

∞∑

k=3

(−1)k+1 zk−2

k

and hence f(z) = log(1+z)z2 has a pole of order 1 at z= 0 and the singular part is1

z (by Equation 1.7 andDefinition 1.8 p. 105).g) First, we simplifyz2

+1z2−z = 1+ z+1

z(z−1). Now partial fraction decomposition ofz+1z(z−1) yields

z+ 1z(z− 1)

= −1z+

2z− 1

so

f (z) = 1− 1z+

2z− 1

69

and hence f(z) = z2+1

z2−z has poles of order 1 at z= 0 and z= 1 and the singular part is2z−1 − 1

z (by Equation1.7 and Definition 1.8 p. 105).i) We know (see p. 38)

sin(z) =∞∑

k=0

(−1)kz2k+1

(2k+ 1)!

so

f (z) = zsin1z= z

∞∑

k=0

(−1)k

(1z

)2k+1

(2k+ 1)!=

∞∑

k=0

(−1)k

(1z

)2k

(2k+ 1)!= 1− 1

3!z2+

15!z4− 1

7!z6+ . . .

Therefore f(z) = zsin 1z has an essential singularity at z= 0 (it is neither a pole nor a removable singularity

by Corollary 1.18c). We have

f (z : 0 < |z| < δ) = f [ann(0; 0, δ)] = C

by the Casorati-Weierstrass Theorem p. 109. But f(z : 0 < |z| < δ) = f [ann(0; 0, δ)] is eitherC or C\afor one a∈ C by the Great Picard Theorem (p. 300). In this case,

f (z : 0 < |z| < δ) = f [ann(0; 0, δ)] = C.

Exercise 2. Give the partial fraction expansion of r(z) = z2+1

(z2+z+1)(z−1)2 .


Exercise 3. Give the details of the derivation of (1.17) from (1.16).


Exercise 4. Let f(z) = 1z(z−1)(z−2); give the Laurent Expansion of f(z) in each of the following annuli: (a)

ann(0; 0,1); (b) ann(0; 1,2); (c) ann(0; 2,∞).

Solution. a) We have0 < |z| < 1 and using partial fraction decomposition yields

f (z) =1

z(z− 1)(z− 2)=

1z

1(z− 1)(z− 2)

=1z

( −1z− 1

+1

z− 2

)

=1z

(

11− z

− 12

11− z

2

)

=|z|<1and|z/2|<1

1z

∞∑

n=0

zn − 12

∞∑

n=0

( z2

)n

=

∞∑

n=0

zn−1 −∞∑

n=0

(

12

)n+1

zn−1=

∞∑

n=−1

1−(

12

)n+2 zn

=

∞∑

n=−1

(

1− 2−n−2)

zn, (0 < |z| < 1).

b) We have1 < |z| < 2 and thus

f (z) =1

z(z− 1)(z− 2)

=1z

−1z

1

1− 1z

− 12

11− z

2

=|1/z|<1and|z/2|<1

1z

−1z

∞∑

n=0

(

1z

)n

− 12

∞∑

n=0

( z2

)n

=

∞∑

n=0

−(

1z

)n+2

−∞∑

n=0

(

12

)n+1

zn−1=

∞∑

n=0

−z−n−2 −∞∑

n=0

(

12

)n+1

zn−1

=

−2∑

n=−∞−zn −

∞∑

n=−1

(

12

)n+2

zn, (1 < |z| < 2).

70

c) We have2 < |z| < ∞ and thus

f (z) =1

z(z− 1)(z− 2)

=1z

−1z

1

1− 1z

+1z

1

1− 2z

=|1/z|<1and|2/z|<1

1z

−1z

∞∑

n=0

(

1z

)n

+1z

∞∑

n=0

(

2z

)n

=

∞∑

n=0

−(

1z

)n+2

+

∞∑

n=0

2n

(

1z

)n+2

=

∞∑

n=0

(2n − 1)z−n−2

=

−2∑

n=−∞(2−n−2 − 1)zn, (2 < |z| < ∞).

Exercise 5. Show that f(z) = tanz is analytic inC except for simple poles at z= π2 + nπ, for each integer

n. Determine the singular part of f at each of these poles.

Solution. Define zn = π2 + nπ. We knowtan(z) = sin(z)

cos(z) and(cos(z))′ = − sin(z). Hence,tan(z) has a simple

pole at each zn with residue−1 and with singular part− 1z−zn

at each zn.

Exercise 6. If f : G→ C is analytic except for poles show that the poles of f cannot have a limit point inG.


Exercise 7. Let f have an isolated singularity at z= a and suppose f(z) . 0. Show that if either (1.19) or(1.20) holds for some s inR then there is an integer m such that (1.19) holds if s> m and (1.20) holds ifs< m.

Solution. Let limz→a |z− a|s| f (z)| = 0 hold, that islimz→a(z− a)s f (z) = 0 for some s inR. Then it holdsobviously for any greater s’s. Hence, there exist an integern > 0 such that

limz→a

(z− a)n f (z) = 0 (5.1)

and

limz→a

(z− a)n+1 f (z) = 0 (5.2)

(In fact we are free to choose any positive integer greater than s). Now,(z− a)n f (z) thus has a removablesingularity at z= a (by Theorem 1.2 p. 103) applied to (5.2)) and its extended value is also zero by (5.2).Thus(z− a)n f (z) has a zero of finite order k at z− a. So

(z− a)n f (z) = (z− a)kh(z)

where h(z) is analytic at a and h(a) , 0. Hence

limz→a

(z− a)s f (z) = limz→a

(z− a)s−n(z− a)n f (z) = limz→a

(z− a)s−n+kh(z) =

0, s> n− k

±∞, s< n− k

h(a) , 0, s= n− k.

Thus,

limz→a|z− a|s| f (z)| =

0, s> m

∞, s< m

71

where m= n− k.Now letlimz→a |z− a|s| f (z)| = ∞ for some s inR. Then there exists an integer n< 0 so that

limz→a|z− a|n| f (z)| = ∞ (5.3)

(In fact we are free to choose any negative integer less than s). Now(z− a)n f (z) thus has a pole at z= a(by Definition 1.3 p.105 and (5.1)) and(z− a)n f (z) has a pole of finite order l at z= a and so

(z− a)n f (z) = (z− a)−lk(z)

where k(z) is analytic at a and k(a) , 0 (Proposition 1.4 p.105). Hence

limz→a

(z− a)s f (z) = limz→a

(z− a)s−n+n f (z) = limz→a

(z− a)s−n−lk(z) =

0, s> n+ l

±∞, s< n+ l

k(a) , 0, s= n+ l.

Thus,


0, s> m

∞, s< m

where m= n+ l.

Exercise 8. Let f , a, and m be as in Exercise 7. Show: (a) m= 0 iff z = a is a removable singularity andf (a) , 0; (b) m< 0 iff z= a is a removable singularity and f has a zero at z= a of order−m; (c) m> 0 iffr = a is a pole of f of order m.

Solution. a)⇒: Let m= 0. Then by Exercise 7, we have

limz→a

(z− a)2 f (z) =

0, s> 0 = m

, 0, s= 0 = m. (5.4)

Thus using s= 1, f (z) has a removable singularity (by Theorem 1.2 p. 103) with the extended value equalto f(a) = limz→a f (z) , 0.⇐: Let z = a be a removable singularity and f(a) = limz→a f (z) , 0. Thenlimz→a(z− a) f (z) = 0 byTheorem 1.2 p.103. By Exercise 7, we obtain that there is an integer m such that


0, s> m

∞, s< m.

But m= 0, since

limz→a

(z− a)s f (z) =

0, s> 0

∞, s< 0,

which follows by (5.4).b)⇒: Let the algebraic order be defined at z= a and be equal to−m (m> 0). Then by Exercise 7, we have

limz→a

f (z) = 0 and limz→a

(z− a) f (z) = 0.

Recall limz→a(z− a)s f (z) = 0 if s > −m so it is true of s= 0 and s= 1. Thus, f(z) has a removablesingularity at z= a and also the extended function has the valuelimz→a f (z) = 0 at z= a (p. 103 Theorem

72

1.2). Now h(z) := (z− a)−m f (z) has a removable singularity at z= a (sincelimz→a(z− a)−m+1 f (z) = 0) buth(z) has extended value

h(a) = limz→a

(z− a)−m f (z) , 0

(by Exercise 7). Therefore f(z) = (z− a)mh(z) has a zero of order m at z= a.⇐: Assume f(z) has a removable singularity at z= a and let z= a be a zero of order m> 0. Thenf (z) = (z− a)mh(z) where h(z) is analytic in|z− a| < δ, δ > 0 and h(a) , 0. Then

limz→a|z− a|s| f (z)| = lim

z→a|z− a|s+m|h(z)| =

0, s> −m

∞, s< −m

by Exercise 7. Hence, the algebraic order of f at z= a is equal to−m (m> 0).c)⇒: Let the algebraic order be m> 0. Then

limz→a

(z− a)s f (z) = 0

for s> m by Exercise 7, that is

limz→a

(z− a)m+1 f (z) = 0 and limz→a

(z− a)m f (z) , 0

by Exercise 7. Then for z, a, we have h(z) = (z− a)m f (z) has a removable singularity at z= a withh(a) , 0. Now f(z) = (z− a)−mh(z) where h∈ A(B(a; δ)) and h(a) , 0. Thus f(z) has a pole of order m atz= a.⇐: Let f(z) have a pole of order m at z= a. Then f(z) = (z−a)−mg(z) where g(z) ∈ A(B(a; δ)) and g(a) , 0.Thuslimz→a(z− a)m f (z) = g(a) , 0 and

limz→a

(z− a)m+1 f (z) = limz→a

(z− a)g(z) = 0.

So

limz→a|z− a|s| f (z)| = lim

z→a|z− a|s−m|z− a|m| f (z)| =

0, s> m

∞, s< m

by Exercise 7. Therefore, m> 0 is the algebraic order of f at z= a.

Exercise 9. A function f has an essential singularity at z= a iff neither (1.19) nor (1.20) holds for any realnumber s.

Solution. ⇒: We prove this direction by proving the contrapositive (P→ Q ⇐⇒ ¬Q ⇒ ¬P). Assume(1.19) or (1.20) hold for some real number s. Then by Exercise7, there exists an integer m such that (1.19)holds if s> m and (1.20) holds if s< m. Then by Exercise 8, z= a is either a removable singularity or apole. Hence z= a is not an essential singularity (by Definition 1.3 p.105).⇐: Also this direction by is being proved by showing the contrapositive. Assume that f has an isolatedsingularity at z= a which is not an essential singularity. Then either z= a is a removable singularity or apole (by Definition 1.3 p.105). Again by Problem 7 and 8, in either case there exists an m such that (1.19)or (1.20) holds. (If z= a is removable, then either m= 0 or m < 0; if z = a is a pole, then m> 0 byExercise 8).

Exercise 10.Suppose that f has an essential singularity at z= a. Prove the following strengthened versionof the Casorati–Weierstrass Theorem. If c∈ C andε > 0 are given then for eachδ > 0 there is a numberα,|c− α| < ε, such that f(z) = α has infinitely many solutions in B(a; δ).

73


Exercise 11.Give the Laurent series development of f(z) = exp(

1z

)

. Can you generalize this result?


Exercise 12. (a) Letλ ∈ C and show that

exp

12λ

(

z+1z

)

= a0 +

∞∑

n=1

an

(

zn+

1zn

)

for 0 < |z| < ∞, where for n≥ 0

an =1π

∫ π

0eλ cost cosnt dt.

(b) Similarly, show

exp

12λ

(

z− 1z

)

= b0 +

∞∑

n=1

bn

(

zn+

(−1)n

zn

)

for 0 < |z| < ∞, where

bn =1π

∫ π

0cos(nt− λ sint) dt.


Exercise 13.Let R> 0 and G= z : |z| > R; a function f : G→ C has a removable singularity, a pole, oran essential singularity at infinity if f(z−1) has, respectively, a removable singularity, a pole, or an essentialsingularity at z= 0. If f has a pole at∞ then the order of the pole is the order of the pole of f(z−1) at z= 0.(a) Prove that an entire function has a removable singularity at infinity iff it is a constant.(b) Prove that an entire function has a pole at infinity of order m iff it is a polynomial of degree m.(c) Characterize those rational functions which have a removable singularity at infinity.(d) Characterize those rational functions which have a poleof order m at infinity.


Exercise 14.Let G= z : 0 < |z| < 1 and let f : G→ C be analytic. Suppose thatγ is a closed rectifiablecurve in G such that n(γ; a) = 0 for all a in C −G. What is

∫

γf ? Why?


Exercise 15.Let f be analytic in G= z : 0 < |z− a| < r except that there is a sequence of polesan in Gwith an→ a. Show that for anyω in C there is a sequencezn in G with a= lim zn andω = lim f (zn).


Exercise 16. Determine the regions in which the functions f(z) =(

sin 1z

)−1and g(z) =

∫ 1

0(t − z)−1 dt are

analytic. Do they have any isolated singularities? Do they have any singularities that are not isolated?


Exercise 17.Let f be analytic in the region G=ann(a; 0,R). Show that if!

G| f (x+ iy)|2 dx dy< ∞ then f

has a removable singularity at z= a. Suppose that p> 0 and!

G| f (x+ iy)|p dx dy< ∞; what can be said

about the nature of the singularity at z= a?


74

5.2 Residues

Exercise 1. Calculate the following integrals:a)

∫ ∞

0

x2 dxx4 + x2 + 1

b)∫ ∞

0

cosx− 1x2

dx

c)∫ π

0

cos 2θ dθ1− 2acosθ + a2

where a2 < 1

d)∫ π

0

dθ(a+ cosθ)2

where a> 1.

Solution. Note that ∫ ∞

0

x2 dxx4 + x2 + 1

=12

∫ ∞

−∞

x2 dxx4 + x2 + 1

since x2

x4+x2+1 is even. LetγR = [−R,R] ∪ CR where CR = Reit , 0 ≤ t ≤ π and R 0 (draw a sketch).

f (z) = z2

z4+z2+1 has two simple poles at a1 = ei π3 , a2 = ei 2π3 enclosed byγR. By the residual Theorem, we have

2πi(Res( f ,a1) + Res( f ,a2)) =∫

γR

f =∫ R

−R

x2

x4 + x2 + 1dx+

∫

CR

z2

z4 + z2 + 1dz.

Thus,∫ ∞

−∞

x2 dxx4 + x2 + 1

= − limR→∞

∫

CR

z2

z4 + z2 + 1dz+ 2πi

2∑

k=1

Res( f ,ak).

We have∫

CR

f (z) dz =

∫

CR

z2

z4 + z2 + 1dz≤

∣∣∣∣∣∣

∫

CR

z2

z4 + z2 + 1dz

∣∣∣∣∣∣≤

∫

CR

|z|2|z4 + z2 + 1| |dz| ≤

∫

CR

|z|2|z|4 − |z|2 − 1

|dz|

=R2

R4 − R2 − 1

∫

CR

|dz| = πR3

R4 − R2 − 1→ 0 as R→ ∞.

Next, we calculate Res( f ,a1).

Res( f ,ei π3 ) = limz→ei π3

(z− ei π3 )z2

z4 + z2 + 1= ei 2π

3 limz→ei π3

(z− ei π3 )1

z4 + z2 + 1

= ei 2π3 lim

z→ei π3

14z3 + 2z

=ei 2π

3

4ei 3π3 + 2ei π3

=ei π3

4ei 2π3 + 2

=

12 +

12

√3i

4(− 12 +

12

√3i) + 2

=

12 +

12

√3i

2√

3i=

1

4√

3i+

14

=14− 1

4√

3i.

75

Finally, we calculate Res( f ,a2).

Res( f ,ei 2π3 ) = lim

z→ei 2π3

(z− ei 2π3 )

z2

z4 + z2 + 1= ei 4π

3 limz→ei 2π

3

(z− ei 2π3 )

1z4 + z2 + 1

= ei 4π3 lim

z→ei 2π3

14z3 + 2z

=ei 4π

3

4ei 6π3 + 2ei 2π

3

=ei 2π

3

4ei 4π3 + 2

=− 1

2 +12

√3i

4(− 12 − 1

2

√3i) + 2

=− 1

2 +12

√3i

−2√

3i=

1

4√

3i− 1

4

= −14− 1

4√

3i.

So

2πi2∑

k=1

Res( f ,ak) = 2πi

(

14− 1

4√

3i +

1

4√

3i− 1

4

)

= 2πi

(

− 1

2√

3i

)

= − π√3

i2 =π√3.

Hence,∫ ∞

0

x2 dxx4 + x2 + 1

=12π√3.

b) Let f(z) = eiz−1z2 andγ = [−R,−r] + γr + [r,R] + γR whereγR = Reit , 0 ≤ t ≤ π andγr = re−it , 0 ≤ t ≤ π

and0 < r < R (A sketch might help to see thatγ is a closed and rectifiable curve). Clearly∫

γf (z) dz= 0

since f is analytic on the region enclosed byγ. Also∫

γ

f (z) dz=∫ −r

−R

eix − 1x2

dx+∫ R

r

eix − 1x2

dx+∫

γr

f +∫

γR

f .

We have∫ −r

−R

eix − 1x2

dx=∫ R

r

e−ix − 1x2

dx.

So∫ −r

−R

eix − 1x2

dx+∫ R

r

eix − 1x2

dx =

∫ R

r

e−ix+ e−ix − 2

x2dx= 2

∫ R

r

e−ix+e−ix

2 − 1

x2dx

= 2∫ R

r

cos(x) − 1x2

dx

Therefore,∫ R

r

cos(x) − 1x2

dx= −12

∫

γR

eiz − 1z2

dz− 12

∫

γr

eiz − 1z2

dz

which implies that∫ ∞

0

cos(x) − 1x2

dx= limR→∞

(

−12

∫

γR

eiz − 1z2

dz

)

− limr→0

(

12

∫

γr

eiz − 1z2

dz

)

The first term on the right hand side is zero, since∣∣∣∣∣∣

∫

γR

eiz − 1z2

dz

∣∣∣∣∣∣≤

∫

γR

|eiz − 1||z|2 |dz| ≤ 1

R2

(∫

γR

|dz| +∫

γR

|eiz| |dz|)

≤ 1R2

(πR+ πR) =2πR→ 0 as R→ ∞.

76

The second term on the right hand side isπ, since

limr→0

(∫

γr

eiz − 1z2

dz

)

= −12

2πi Res( f (z),0) = −πi2 = π

( f (z) can be written asiz+analytic part and therefore Res( f (z),0) = i). Hence,∫ ∞

0

cos(x) − 1x2

dx= −12·0− 1

2· π = −π

2.

c) If a = 0, then∫ π

0


=

∫ π

0cos 2θ dθ =

12

[sin 2θ]π0 =12

[0 − 0] = 0.

Assume a, 0. We can write∫ π

0


=12

∫ 2π

0


=12

∫ 2π

0

e2θi+e2θi

2

1− 2aeθi+eθi2 + a2

dθ.

Let z= eiθ. Then dz= iz dθ =⇒ dθ = dziz . Hence,

12

∫ 2π

0

e2θi+e2θi

2

1− 2aeθi+eθi2 + a2

dθ =12

∫

|z|=1

z2+

1z2

2

1− a(

z+ 1z

)

+ a2

dziz=

14i

∫

|z|=1

z4+ 1

z2(

z− az2 − a+ a2z) dz

= − 14ia

∫

|z|=1

z4+ 1

z2(

z2 − 1+a2

a z+ 1) dz= − 1

4ia

∫

|z|=1

z4+ 1

z2 (z− a)(

z− 1a

) dz

By the Residue Theorem and the fact that|a| < 1, since a2 < 1∫

|z|=1

z4+ 1

z2 (z− a)(

z− 1a

) dz= 2πi[

Res f( f ,0)+ Res( f ,a)]

where f(z) = z4+1

z2(z−a)(z− 1a)

. It is easily obtained

Res( f ,0) = limz→0

ddz

z2 z4

+ 1

z2 (z− a)(

z− 1a

)

= lim

z→0

ddz

z4+ 1

(z− a)(

z− 1a

)

= limz→0

−4z3(

z2 − az− 1az+ 1

)

− (z4 − 1)(

2z− a− 1a

)

(

z2 − az− 1az+ 1

)2= a+

1a

and

Res( f ,a) = limz→a

(z− a)z4+ 1

z2 (z− a)(

z− 1a

) = limz→a

z4+ 1

z2(

z− 1a

) =a4+ 1

a2(

a− 1a

) =a4+ 1

a(a2 − 1).

So ∫

|z|=1

z4+ 1

z2 (z− a)(

z− 1a

) dz= 2πi

(

a+1a+

a4+ 1

a(a2 − 1)

)

= 2πi2a4

a(a2 = 1)= − 4πia3

1− a2.

Finally,∫ π

0


= − 14ia·(

− 4πia3

1− a2

)

= πa2

1− a2.

77

Exercise 2. Verify the following equations:a)

∫ ∞

0

dx(x2 + a2)2

=π

4a3, a > 0;

b)∫ ∞

0

(log x)3

1+ x2dx= 0;

c)∫ ∞

0

cosax(1+ x2)2

dx=π(a+ 1)e−a

4, if a > 0;

d)∫ π/2

0

dθ

a+ sin2 θ=

π

2[a(a+ 1)]12

, if a > 0;

e)∫ ∞

0

log x(1+ x2)2

dx= −π4

;

f)∫ ∞

0

dx1+ x2

=π

2;

g)∫ ∞

−∞

eax

1+ exdx=

π

sinaπ, if 0 < a < 1;

h)∫ 2π

0log sin2 2θ dθ = 4

∫ 4π

0log sinθ dθ = −4π log 2.

Solution. a) Let f(z) = 1(z2+a2)2 andγR = [−R,R] +CR where CR = Reit , 0 ≤ t ≤ π. So f has z1 = ai as a

pole of order2 enclosed byγR. So we have by the Residue Theorem∫

γR

f (z) dz= 2πi Res( f , z1)

where

Res( f , z1) = limz→ai

ddz

[

(z− ai)2 1(z− ai)2(z+ ai)2

]

= limz→ai

ddz

[

1(z+ ai)2

]

= limz→ai

−2(z+ ai)3

= − 2(2ai)3

=1

4a3i.

Also

2πi1

4a3i=

π

2a3=

∫ R

−R

dx(x2 + a2)2

+

∫

CR

f (z) dz

where the latter integral on the right hand side is zero since∣∣∣∣∣∣

∫

CR

1(z2 + a2)2

dz

∣∣∣∣∣∣≤

∫

CR

1|z2 + a2|2 |dz| ≤

∫

CR

1(|z|2 − |a|2)2

|dz| ≤ 1(R2 − a2)2

∫

CR

|dz|

=πR

(R2 − a2)2→ 0 as R→ ∞.

78

Hence,∫ ∞

0

dx(x2 + a2)2

=12

∫ ∞

−∞

dx(x2 + a2)2

=12

limR→∞

∫ R

−R

dx(x2 + a2)2

=12π

2a3=

π

4a3.

c) Let f(z) = eiaz

(1+z2)2 andγR = [−R,R] +CR where CR = Reit , 0 ≤ t ≤ π. So f has z1 = i as a pole of order 2enclosed byγR. By the Residue Theorem, we have

∫

γR

f (z) dz= 2π Res( f , z1)

where

Res( f , z1) = limz→i

ddz

[

(z− i)2 eiaz

(z− i)2(z+ i)2

]

= limz→i

ddz

[

eiaz

(z+ i)2

]

= limz→i

[

(z+ i)2iaeiaz − 2(z+ i)eiaz

(z+ i)4

]

= limz→i

[

(z+ i)iaeiaz − 2eiaz

(z+ i)3

]

=2i2aei2a − 2ei2a

(2i)3=−2ae−a − 2e−a

−8i=

(a+ 1)e−a

4i.

Also

2πi(a+ 1)e−a

4i=π(a+ 1)e−a

2=

∫ R

−Rf (x) dx+

∫

CR

f (z) dz

where the latter integral on the right hand side is zero since∣∣∣∣∣∣

∫

CR

f (z) dz

∣∣∣∣∣∣=

∣∣∣∣∣∣

∫

CR

eiaz

(1+ z2)2dz

∣∣∣∣∣∣≤

∫

CR

|eiaz||1+ z2|2 |dz| ≤

∫

CR

1(|z|2 − 1)2

|dz|

=1

(R2 − 1)2Rπ→ 0 as R→ ∞,

since|eiaz| = e−Im(az) ≤ 1 since Im(az) ≥ 0 (a > 0 by assumption). Hence

∫ ∞

0

cosax(1+ x2)2

dx=12

∫ ∞

−∞

cosax(1+ x2)2

dx=12

limR→∞

∫ R

−R

cosax(1+ x2)2

dx=12π(a+ 1)e−a

2=π(a+ 1)e−a

4

provided a> 0.g) First of all substitute u= ex, so

∫ ∞

−∞

eax

1+ exdx=

∫ ∞

0

ua

1+ uduu=

∫ ∞

0

ua−1

1+ udu.

Now, let u= t2, then∫ ∞

0

ua−1

1+ udu= 2

∫ ∞

0

t2a−2t1+ t2

dt = 2∫ ∞

0

t2a−1

1+ t2dt.

Hence,∫ ∞

−∞

eax

1+ exdx= 2

∫ ∞

0

x2a−1

1+ x2dx.

Let f(z) = z2a−1

1+z2 andγ = [−R,−r] + γr + [r,R] + γR (0 < r < R). Definelogz on G= z ∈ C : z , 0,− π2 <arg(z) < 3π

2 with log(z) = log |z| + iθ, θ = arg(z). f (z) has a simple pole i enclosed byγ. By the Residue

79

Theorem∫

γ

f (z) dz = 2πRes( f , i) = 2πi limz→i

(z− i)z2a−1

(z− i)(z+ i)= 2πi

i2a−1

2i

= πi2a−1= πe(2a−1) logi

= πe(2a−1)(0+i π2 )= πeaπie−i π2 = −πieaπi .

Also∫

γ

f (z) dz=∫ −r

−Rf (x) dx

︸︷︷︸

I

+

∫

γr

f (z) dz

︸︷︷︸

II

+

∫ R

rf (x) dx

︸︷︷︸

III

+

∫

γR

f (z) dz

︸︷︷︸

IV

.

I)

∫ −r

−Rf (x) dx =

∫ R

rf (−x) dx=

∫ R

r

(−x)2a−1

1+ x2dx=

∫ R

r

e(2a−1)(log|−x|+iπ)

1+ x2dx

= e(2a−1)iπ∫ R

r

e(2a−1) logx

1+ x2dx= −e2aπi

∫ R

r

e(2a−1) logx

1+ x2dx

= −e2aπi∫ R

r

x2a−1

1+ x2dx.

IV) ∣∣∣∣∣∣

∫

γR

f (z) dz

∣∣∣∣∣∣=

∣∣∣∣∣∣

∫

γR

z2a−1

1+ z2dz

∣∣∣∣∣∣≤

∫

γR

|z|2a−1

|1+ z2| |dz| ≤ R2a−1

R2 − 1πR=

πR2a

R2 − 1→ 0 as R→ ∞

since a< 1.II) Similar to IV) ∣

∣∣∣∣∣

∫

γr

f (z) dz

∣∣∣∣∣∣≤ πr2a

r2 − 1→ 0 as r→ 0

since a> 0.Hence, if r→ 0 and R→ ∞, we get

−πieaπi=

∫

γ

f (z) dz= −e2aπi∫ ∞

0f (x) dx+ 0+

∫ ∞

0f (x) dx+ 0 = (1− e2aπi)

∫ ∞

0f (x) dx.

Thus,

−πieaπi= (1− e2aπi)

∫ ∞

0f (x) dx

and so ∫ ∞

0f (x) dx=

−πieaπi

1− e2aπi=

−iπe−aπi − eaπi

=iπ

eaπi − e−aπi=

iπ2i sin(aπ)

=π

2 sin(aπ).

Therefore,∫ ∞

−∞

eax

1+ exdx= 2

∫ ∞

0

x2a−1

1+ x2dx= 2

π

2 sin(aπ)=

π

sin(aπ)

provided0 < a < 1.

Exercise 3. Find all possible values of∫

γexpz−1 dz whereγ is any closed curve not passing through z= 0.


80

Exercise 4. Suppose that f has a simple pole at z= a and let g be analytic in an open set containing a.Show that Res( f g; a) = g(a)Res( f ; a).


Exercise 5. Use Exercise 4 to show that if G is a region and f is analytic in Gexcept for simple poles ata1, . . . ,an; and if g is analytic in G then

12πi

∫

γ

f g =n∑

k=1

n(γ; ak)g(ak)Res( f ; ak)

for any closed rectifiable curveγ not passing through a1, . . . ,an such thatγ ≈ 0 in G.


Exercise 6. Let γ be the rectangular path[n+ 12 + ni,−n− 1

2 + ni,−n− 12 − ni,n+ 1

2 − ni,n+ i + ni] andevaluate the integral

∫

γπ(z+ a)−2 cotπz dz for a, an integer. Show that limn→∞

∫

γπ(z+ a)−2 cotπz dz= 0

and, by using the first part, deduce that

π2

sin2 πa=

∞∑

n=−∞

1(a+ n)2

(Hint: Use the fact that for z= x+ iy, | cosz|2 = cos2 x+ sinh2 y and| sinz|2 = sin2 x+ sinh2 y to show that| cotπz| ≤ 2 for z onγ if n is sufficiently large.)


Exercise 7. Use Exercise 6 to deduce that

π2

8=

∞∑

n=0

1(2n+ 1)2


Exercise 8. Letγ be the polygonal path defined in Exercise 6 and evaluate∫

γπ(z2 − a2)−1 cotπz dz for a,

an integer. Show thatlimn→∞ π(z2 − a2)−1 cotπz dz= 0, and consequently

π cotπa =1a+

∞∑

n=1

2aa2 − n2

for a , an integer.


Exercise 9. Use methods similar to those of Exercises 6 and 8 to show that

π

sinπa=

1a+

∞∑

n=1

2(−1)naa2 − n2

for a , an integer.


81

Exercise 10.Letγ be the circle|z| = 1 and let m and n be non-negative integers. Show that

12πi

∫

γ

(z2 ± 1)m dzzm+n+1

=

(±1)p(n+2p)!p!(n+p)! , if m = 2p+ n, p ≥ 0

0, otherwise


Exercise 11. In Exercise 1.12, consider an and bn as functions of the parameterλ and use Exercise 10 tocompute power series expansions for an(λ) and bn(λ). (bn(λ) is called a Bessel function.)


Exercise 12.Let f be analytic in the plane except for isolated singularities at a1,a2, . . . ,am. Show that

Res( f ;∞) = −∞∑

k=1

Res( f ,ak).

(Res( f ;∞) is defined as the residue of−z−2 f (z−1) at z = 0. Equivalently, Res( f ;∞) = − 12πi

∫

f whenγ(t) = Reit , 0 ≤ t ≤ 2π, for sufficiently large R.) What can you say if f has infinitely many isolatedsingularities?


Exercise 13. Let f be an entire function and let a,b ∈ C such that|a| < R and|b| < R. If γ(t) = Reit ,0 ≤ t ≤ 2pi, evaluate

∫

γ[(z− a)(z− b)]−1 f (z) dz. Use this result to give another proof of Liouville’s

Theorem.


5.3 The Argument Principle

Exercise 1. Prove Theorem 3.6.


Exercise 2. Suppose f is analytic onB(0; 1)and satisfies| f (z)| < 1 for |z| = l. Find the number of solutions(counting multiplicities) of the equation f(z) = zn where n is an integer larger than or equal to 1.


Exercise 3. Let f be analytic inB(0;R) with f(0) = 0, f ′(0) , 0 and f(z) , 0 for 0 < |z| ≤ R. Putρ = min| f (z)| : |z| = R > 0. Define g: B(0;ρ)→ C by

g(ω) =1

2πi

∫

γ

z f′(z)f (z) − ω dz

whereγ is the circle|z| = R. Show that g is analytic and discuss the properties of g.


Exercise 4. If f is meromorphic on G andf : G → C∞ is defined byf (z) = ∞ when z is a pole of f andf (z) = f (z) otherwise, show thatf is continuous.

82


Exercise 5. Let f be meromorphic on the region G and not constant; show that neither the poles nor thezeros of f have a limit point in G.


Exercise 6. Let G be a region and let H(G) denote the set of all analytic functions on G. (The letter“H” stands for holomorphic. Some authors call a differentiable function holomorphic and call functionsanalytic if they have a power series expansion about each point of their domain. Others reserve the term“analytic” for what many call the complete analytic function, which we will not describe here.) Show thatH(G) is an integral domain; that is, H(G) is a commutative ring with no zero divisors. Show that M(G), themeromorphic functions on G, is a field.We have said that analytic functions are like polynomials; similarly, meromorphic functions are analoguesof rational functions. The question arises, is every meromorphic function on G the quotient of two analyticfunctions on G? Alternately, is M(G) the quotient field of H(G)? The answer is yes but some additionaltheory will be required before this answer can be proved.


Exercise 7. State and prove a more general version of Rouche’s Theorem for curves other than circles inG.


Exercise 8. Is a non-constant meromorphic function on a region G an open mapping of G intoC? Is it anopen mapping of G intoC∞?


Exercise 9. Letλ > 1 and show that the equationλ − z− e−z= 0 has exactly one solution in the half plane

z : Re z> 0. Show that this solution must be real. What happens to the solution asλ→ 1?


Exercise 10.Let f be analytic in a neighborhood of D= B(0; 1). If | f (z)| < 1 for |z| = 1, show that there isa unique z with|z| < 1 and f(z) = z. If | f (z)| ≤ 1 for |z| = 1, what can you say?

Solution. Let g(z) = f (z) − z and h(z) = z. Clearly g,h are analytic in a neighborhood ofB(0,1) since f isanalytic. Note that g,h have no poles. Letγ = z : |z| = 1. We have

|g(z) + h(z)| = | f (z) − z+ z| = | f (z)| < 1 = |z| = |h(z)| ≤ |g(z)| + |h(z)|onγ since| f (z)| < 1 for |z| = 1. By Roché’s Theorem, we get

Zg − Pg = Zh − Ph.

But Pg = Ph = 0, soZg = Zh (inside the unit circle).

Since h(z) = z has only one zero (z= 0) inside the unit circle, we get that

g(z) = f (z) − z

has exactly one zero inside the unit circle. Hence, there is aunique z such that f(z) − z= 0 ⇐⇒ f (z) = zwith |z| < 1.

If we choose f(z) = z, then| f (z)| ≤ 1 for |z| = 1 and therefore we have infinitely many fixed points.If we choose f(z) = 1, then| f (z)| ≤ 1 for |z| = 1 and therefore we do not have any fixed points.

83

Chapter 6

The Maximum Modulus Theorem

6.1 The Maximum Principle

Exercise 1. Prove the following Minimum Principle. If f is a non-constant analytic function on a boundedopen set G and is continuous on G−, then either f has a zero in G or| f | assumes its minimum value on∂G.(See Exercise IV. 3.6.)

Solution. Since f∈ C(G) we have| f | ∈ C(G). Hence∃ a ∈ G such that| f (a)| ≤ | f (z)| ∀z ∈ G.If a ∈ ∂G, then| f | assumes its minimum value on∂G and we are done. Otherwise, if a< ∂G, then a∈ Gand we can write G=

⋃

Ai where Ai are the components of G, that is a∈ Ai for some i. But each Ai is aregion, so we can use Exercise IV 3.6 which yields either f(a) = 0 or f is constant. But f was assumedto be non-constant, so f has to have a zero in G. Therefore, either f has a zero in G or| f | assumes itsminimum value on∂G.

Exercise 2. Let G be a bounded region and suppose f is continuous on G− and analytic on G. Show thatif there is a constant c≥ 0 such that| f (z)| = c for all z on the boundary of G then either f is a constantfunction or f has a zero in G.

Solution. Assume there is a constant c≥ 0 such that| f (z)| = c for all z ∈ ∂G. According to the MaximumModulus Theorem (Version 2), we get

max| f (z)| : z ∈ G = max| f (z)| : z ∈ ∂G = c.

So

| f (z)| ≤ c, ∀z ∈ G. (6.1)

Since f∈ C(G) which implies| f | ∈ C(G) and hence there exists an a∈ G such that

| f (a)| ≤ | f (z)| ≤(6.1)

c,

then by Exercise IV 3.6 either f is constant of f has a zero in G.

Exercise 3. (a) Let f be entire and non-constant. For any positive real number c show that the closure ofz : | f (z)| < c is the setz : | f (z)| ≤ c.(b) Let p be a polynomial and show that each component ofz : |p(z)| < c contains a zero of p. (Hint: Use

84

Exercise 2.)(c) If p is a polynomial and c> 0 show thatz : |p(z)| = c is the union of a finite number of closed paths.Discuss the behavior of these paths as c→ ∞.


Exercise 4. Let 0 < r < R and put A= z : r ≤ |z| ≤ R. Show that there is a positive numberε > 0 suchthat for each polynomial p,

sup|p(z) − z−1| : z ∈ A ≥ εThis says that z−1 is not the uniform limit of polynomials on A.


Exercise 5. Let f be analytic onB(0;R) with | f (z)| ≤ M for |z| ≤ R and| f (0)| = a > 0. Show that thenumber of zeros of f in B(0; 1

3R) is less than or equal to1log 2 log

(Ma

)

. Hint: If z1, . . . , zn are the zeros of f

in B(0; 13R); consider the function

g(z) = f (z)

n∏

k=1

(

1− zzk

)

−1

,

and note that g(0) = f (0). (Notation:∏n

k=1 ak = a1a2 . . . an.)


Exercise 6. Suppose that both f and g are analytic onB(0;R) with | f (z)| = |g(z)| for |z| = R. Show that ifneither f nor g vanishes in B(0;R) then there is a constantλ, |λ| = 1, such that f= λg.


Exercise 7. Let f be analytic in the disk B(0;R) and for0 ≤ r < R define A(r) = maxRe f(z) : |z| = r.Show that unless f is a constant, A(r) is a strictly increasing function of r.


Exercise 8. Suppose G is a region, f: G→ C is analytic, and M is a constant such that whenever z is on∂∞G andzn is a sequence in G with z= lim zn we havelim sup| f (zn)| ≤ M. Show that| f (z)| ≤ M, foreach z in G.

Solution. We need to showlim sup

z→a| f (z)| ≤ M ∀u ∈ ∂∞G.

Then we can use the Maximum Modulus Theorem (Version 3). Instead of showing that

lim sup| f (zn)| ≤ M ⇒ lim supz→a

| f (z)| ≤ M,

we show the contrapositive, that is

lim supz→a

| f (z)| > M ⇒ lim sup| f (zn)| > M.

So assumelim supz→a | f (z)| > M. But this implieslim supz→a | f (zn)| > M since zn → z as z→ ∞, that is zngets arbitrarily close to z and since f is analytic, that is continuous, we have f(zn) gets arbitrarily close tof (z).

85

6.2 Schwarz’s Lemma

Exercise 1. Suppose| f (z)| ≤ 1 for |z| < 1 and f is a non-constant analytic function. By considering thefunction g: D→ D defined by

g(z) =f (z) − a

1− a f(z)

where a= f (0), prove that| f (0)| − |z|

1+ | f (0)| |z| ≤ | f (z)| ≤ | f (0)| + |z|1− | f (0)| |z|

for |z| < 1.

Solution. We claim that|g(z)| ≤ 1 and g(0) = 0. It is easy to check that g(0) = 0 since

g(0) =f (0)− a

1− a f(0)=

a− a1− aa

= 0.

We also have

|g(z)| = | f (z) − a||1− a f(z)| ≤ 1

since| f (z)| ≤ 1. (Can you see it?) Now, apply Schwarz Lemma.

|g(z)| ≤ |z| for |z| < 1

⇐⇒ | f (z) − a| ≤ |z|· |1− a f(z)|⇒ || f (z)| − |a|| ≤ | f (z) − a| ≤ |z|· |1− a f(z)| ≤ |z| + |z|· |a|· | f (z)|⇒ || f (z)| − |a|| ≤ |z| + |z|· |a|· | f (z)|⇒ −|z| − |z|· |a|· | f (z)| ≤ | f (z)| − |a| ≤ |z| + |z|· |a|· | f (z)|

⇒ |a| − |z|1+ |a| |z| ≤ | f (z)| ≤ |a| + |z|

1− |a| |z| .

Set a= f (0) to obtain the desired result.

Exercise 2. Does there exist an analytic function f: D→ D with f( 12) = 3

4 and f′( 12) = 2

3?

Solution. The answer is no. Assume there exist an analytic function f: D→ D with f( 12) = 3

4. Accordingto the definition, we have

Eaα = f ∈ A(D) : | f (z)| ≤ 1, f (a) = α.

In our case, we have a= 12 andα = 3

4. We know,

∣∣∣ f ′(a)

∣∣∣ ≤ 1− |α|2

1− |a|2

and therefore we must have

∣∣∣∣∣∣f ′

(

12

)∣∣∣∣∣∣≤

1−(

34

)2

1−(

12

)2=

71634

=716

34=

712= 0.583.

But f′( 12) = 2

3 = 0.6 which is not possible.

86

Exercise 3. Suppose f: D→ C satisfies Re f(z) ≥ 0 for all z in D and suppose that f is analytic.(a) Show that Re f(z) > 0 for all z in D.(b) By using an appropriate Möbius transformation, apply Schwarz’s Lemma to prove that if f(0) = 1 then

| f (z)| ≤ 1+ |z|1− |z|

for |z| < 1. What can be said if f(0) , 1?(c) Show that f(0) = 1, f also satisfies

| f (z)| ≥ 1− |z|1+ |z|

(Hint: Use part (a)).


Exercise 4. Prove Caratheodory’s Inequality whose statement is as follows: Let f be analytic onB(0;R)and let M(r) = max| f (z)| : |z| = r, A(r) = maxRe f(z) : |z| = r; then for0 < r < R, if A(r) ≥ 0,

M(r) ≤ R+ rR− r

[A(R) + | f (0)|]

(Hint: First consider the case where f(0) = 0 and examine the function g(z) = f (Rz)[2A(R) + f (Rz)]−1 for|z| < 1.)


Exercise 5. Let f be analytic in D= z : |z| < 1 and suppose that| f (z)| ≤ M for all z in D. (a) If f(zk) = 0for 1 ≤ k ≤ n show that

| f (z)| ≤ Mn∏

k=1

|z− zk||1− zkz|

for |z| < 1. (b) If f (zk) = 0 for 1 ≤ k ≤ n, each zk , 0, and f(0) = Meia(z1z2, . . . , zn), find a formula for f .


Exercise 6. Suppose f is analytic in some region containingB(0; 1) and | f (z)| = 1 where|z| = 1. Find aformula for f . (Hint: First consider the case where f has no zeros in B(0; 1).)


Exercise 7. Suppose f is analytic in a region containingB(0; 1) and | f (z)| = 1 when|z| = 1. Suppose thatf has a zero at z= 1

4(1+ i) and a double zero at z= 12. Can f(0) = 1

2?


Exercise 8. Is there an analytic function f on B(0; 1) such that| f (z)| < 1 for |z| < 1, f (0) = 12, and

f ′(0) = 34? If so, find such an f . Is it unique?


87

6.3 Convex functions and Hadamard’s Three Circles Theorem

Exercise 1. Let f : [a,b] → R and suppose that f(x) > 0 for all x and that f has a continuous secondderivative. Show that f is logarithmically convex iff f ′′(x) f (x) − [ f ′(x)]2 ≥ 0 for all x.


Exercise 2. Show that if f: (a,b)→ R is convex then f is continuous. Does this remain true if f is definedon the closed interval[a,b]?

Solution. Since f is convex on(a,b) and if a< s< t < u < b, we have

f (t) − f (s)t − s

≤ f (u) − f (s)u− s

≤ f (u) − f (t)u− t

(6.2)

which we show now. Let a< s < t < u < b, t = λs+ (1− λ)u with λ = u−tu−s ∈ (0,1). Since f is convex on

(a,b), we get

f (t) ≤ u− tu− s

f (s) +t − su− s

f (u)

which implies(t − s) f (u) + (u− t) f (s) − (u− s) f (t) ≥ 0


t f (u) − s f(u) + u f(s) − t f (s) − u f(t) + s f(t) ≥ 0. (6.3)

From (6.3), we get (rearrange terms and add/subtract a term)

t f (u) − t f (s) − s f(u) + s f(s) − u f(t) − u f(s) + s f(t) − s f(s) ≥ 0

⇐⇒ (t − s)( f (u) − f (s)) − (u− s)( f (t) − f (s)) ≥ 0

⇐⇒ f (t) − f (s)t − s

≤ f (u) − f (s)u− s

.

Similar, from (6.3), we get

u f(u) − s f(u) − u f(t) + s f(t) − u f(u) + u f(s) + t f (u) − t f (s) ≥ 0

⇐⇒ (u− s)( f (u) − f (t)) − (u− t)( f (u) − f (s)) ≥ 0

⇐⇒ f (u) − f (s)u− s

≤ f (u) − f (t)u− t

.

Thus, we have proved (6.2).Given x∈ (a,b) chooseδ > 0 such that[x− δ, x+ δ] ⊂ (a,b).Claim:

f (x) − f (x− δ)δ

≤ f (z) − f (x)z− x

≤ f (x+ δ) − f (x)δ

∀z ∈ (x− δ, x+ δ). (6.4)

If the claim is true, (6.4) is equivalent to

f (x) − f (x− δ)δ

(z− x) ≤ f (z) − f (x) ≤ f (x+ δ) − f (x)δ

(z− x).

Taking the limit z→ x, we have thatf (x)− f (x−δ)δ

(z− x)→ 0 and f (x+δ)− f (x)δ

(z− x)→ 0. Thus f(z)− f (x)→ 0,that is |z− x| < δ⇒ | f (z) − f (x)| < ε which shows that f is continuous.

88

Thus it remains to show the claim. The proof uses (6.2).First, consider a point z∈ (x− δ, x) and apply the second inequality in (6.2) gives

f (x) − f (x− δ)δ

≤ f (z) − f (x)z− x

which gives the first inequality in (6.4).Applying the outer inequality in (6.2) to the three points z< x < x+ δ gives

f (x) − f (z)x− z

≤ f (x+ δ) − f (x)x+ δ − x

⇐⇒ f (z) − f (x)z− x

≤ f (x+ δ) − f (x)δ

which gives the second inequality in (6.4).Now, consider the case x< z< x+ δ. Then, the first inequality in (6.4) follows from the outer inequality in(6.2) applied to the three points x− δ, x, z.

f (x) − f (x− δ)x− x+ δ

≤ f (z) − f (x)z− x

⇐⇒ f (x) − f (x− δ)δ

≤ f (z) − f (x)z− x

.

The second inequality in (6.4) follows from the first inequality in (6.2) applied to x, z, x+ δ.

f (z) − f (x)z− x

≤ f (x+ δ) − f (x)x+ δ − x

⇐⇒ f (z) − f (x)z− x

≤ f (x+ δ) − f (x)δ

.

Thus, we have proved the claim.The statement is not true if f is defined on the closed interval[a,b]! Here is a counterexample: Definef : [a,b] → R by

f (x) =

0, x ∈ [a,b)

1, x = b.

Clearly, f is convex on[a,b], but f is not continuous at x= b (but the function is continuous on(a,b)).

Exercise 3. Show that a function f: [a,b] → R is convex iff any of the following equivalent conditions issatisfied:

(a) a≤ x < u < y ≤ b givesdet

f (u) u 1f (x) x 1f (y) y 1

≥ 0;

(b) a≤ x < u < y ≤ b gives f (u)− f (x)u−x ≤ f (y)− f (x)

y−x ;

(c) a≤ x < u < y ≤ b gives f (u)− f (x)u−x ≤ f (y)− f (u)

y−u ;Interpret these conditions geometrically.


Exercise 4. Supply the details in the proof of Hadamard’s Three Circle Theorem.


Exercise 5. Give necessary and sufficient conditions on the function f such that equality occursin theconclusion of Hadamard’s Three Circle Theorem.


89

Exercise 6. Prove Hardy’s Theorem: If f is analytic onB(0;R) and not constant then

I (r) =12π

∫ 2π

0| f (reiθ)| dθ

is strictly increasing andlog I (r) is a convex function oflog r. Hint: If 0 < r1 < r < r2 find a con-tinuous functionϕ : [0,2π] → C such thatϕ(θ) f (reiθ) = | f (reiθ)| and consider the function F(z) =∫ 2π

0f (zeiθ)ϕ(θ) dθ. (Note that r is fixed, soϕ may depend on r.)


Exercise 7. Let f be analytic in ann(0;R1,R2) and not identically zero; define

I2(r) =12π

∫ 2π

0| f (reiθ)|2 dθ.

Show thatlog I2(r) is a convex function oflog r, R1 < r < R2.


6.4 The Phragmén-Lindelöf Theorem

Exercise 1. In the statement of the Phragmén-Lindelöf Theorem, the requirement that G be simply con-nected is not necessary. Extend Theorem 4.1 to regions G withthe property that for each z in∂∞G there isa sphere V inC∞ centered at z such that V∩G is simply connected. Give some examples of regions thatare not simply connected but have this property and some which don’t.


Exercise 2. In Theorem 4.1 suppose there are bounded analytic functionsϕ1, ϕ2, . . . , ϕn on G that nevervanish and∂∞G = A∪ B1 ∪ . . . ∪ Bn such that condition (a) is satisfied and condition (b) is alsosatisfiedfor eachϕk and Bk. Prove that| f (z)| ≤ M for all z in G.

Solution. We haveϕ1, . . . , ϕn ∈ A(G), bounded andϕk . 0 by assumption. Hence,|ϕk(z)| ≤ κk ∀ z in G foreach k= 1, . . . ,n. Further a) for every a in A,lim supz→a | f (z)| ≤ M and b) for every b in Bk andηk > 0,lim supz→b | f (z)||ϕk(z)|ηk ≤ M ∀ k = 1, . . . ,n.Also because G is simply connected, there is an analytic branch of logϕk(z) on G for each k= 1, . . . ,n(Corollary IV 6.17). Hence, gk(z) = expηk logϕk(z) is an analytic branch ofϕk(z)ηk for ηk > 0 and|gk(z)| = |ϕk(z)|ηk ∀ k = 1, . . . ,n. Define F: G→ C by

F(z) = f (z)n∏

k=1

gk(z)κ−ηk

k ;

then F is analytic on G and

|F(z)| ≤ | f (z)|n∏

k=1

|ϕk(z)|ηkκ−ηk

k ≤|ϕk(z)|≤κk ∀k

| f (z)|n∏

k=1

κηk

k κ−ηk

k |ϕk(z)| = | f (z)| ∀z ∈ G.

Hence,

lim supz→a

| f (z)| =(a),(b)

M, a ∈ A

Mκ−ηk

k , a ∈ Bk ∀k.

90

By the Maximum Modulus Theorem III, we have

|F(z)| ≤ maxM,Mκ−η1

1 , . . . ,Mκ−ηnn ∀z ∈ G.

Thus,

| f (z)| = |F(z)|n∏

k=1

|gk(z)|−1κηk

k ≤n∏

k=1

∣∣∣∣∣

κk

ϕk(z)

∣∣∣∣∣

ηk

maxM,Mκ−η1

1 , . . . ,Mκ−ηnn

for all z in G and for allηk > 0. Lettingηk → 0+ ∀k = 1, . . . ,n gives that| f (z)| ≤ M for all z in G.

Exercise 3. Let G= z : |Im z| < 12π and suppose f: G→ C is analytic andlim supz→w | f (z)| ≤ M for w

in ∂G. Also, suppose A< ∞ and a< 1 can be found such that

| f (z)| < exp[Aexp(a|Re z|)]

for all z in G. Show that| f (z)| ≤ M for all z in G. Examineexp(expz) to see that this is the best possiblegrowth condition. Can we take a= 1 above?


Exercise 4. Let f : G→ C be analytic and suppose M is a constant such thatlim sup| f (zn)| ≤ M for eachsequencezn in G which converges to a point in∂∞G. Show that| f (z)| ≤ M. (See Exercise 1.8).

Solution. Claim: Let M be a constant such thatlim supn→∞ | f (zn)| ≤ M for each sequencezn in G with

z= limn→∞

zn, ∀z ∈ ∂∞G (6.5)

implies

lim supz→a

| f (z)| ≤ M for all a ∈ ∂∞G. (6.6)

If G would be a region together with the claim would give| f (z)| ≤ M ∀z ∈ G by the Maximum ModulusTheorem III.However, G is assumed to be an open set. But every open set can be written as a union of components(components are open and connected). So each component is a region.Therefore, we can apply the Maximum Modulus Theorem III to each component once we have shown theclaim and additionally we have to show that the boundary of G is the same as taking the boundaries of thecomponents:Clearly, since the components of G add up to G, we have that theboundary of the components must includethe boundary of G.So it remains to show that the boundary of the component does not include more than the boundary of G.Assume it would be the case. Then there needs to be a boundary of a component lying inside G (since everycomponent is lying inside G). But this would imply that it must be connected to another component of Gcontradicting the fact a component is a maximally connectedsubset.Hence, the boundary of G equals the boundary of all its components.It remains to show the claim which we will do by showing the contrapositive.Proof of the claim: Assume not (6.6), that is

∀M > 0 ∃a ∈ ∂∞G : lim supz→a

| f (z)| > M


∀M > 0 ∃a ∈ ∂∞G : limr→0+

sup| f (z)| : z ∈ G∩ B(a; r) > M

91

by definition. Then, clearly∃δ > 0 such thatlimr→0+ sup| f (z)| : z ∈ G ∩ B(a; r) > M + δ. Let zn be asequence withlimn→∞ = a and let rn = 2|zn − a|. Obviously, we havelimn→∞ rn = 0. Next, consider thesequenceαn = sup| f (z)| : z ∈ F ∩ B(a; rn). Thenlimn→∞ αn > M + δ. In addition,αn is a nonincreasingsequence, thereforelimn→∞ αn > M + δ impliesαn > M + δ ∀n. Thus,∃ yn ∈ G ∩ B(a; rn) such that| f (yn)| > M + δ. Takinglim sup in this inequality yields

lim supn→∞

| f (yn)| ≥ M + δ,

that islim sup

n→∞| f (yn)| ≥ M

wherelimn→∞ yn = a, which gives “not (6.5)”. Hence (6.5) implies (6.6).

Exercise 5. Let f : G → C be analytic and suppose that G is bounded. Fix z0 in ∂G and suppose thatlim supz→w | f (z)| ≤ M for w in ∂G, w , z0. Show that iflimz→z0 |z− z0|ε | f (z)| = 0 for everyε > 0 then| f (z)| ≤ M for every z in∂G. (Hint: If a < G, considerϕ(z) = (z− z0)(z− a)−1.)


Exercise 6. Let G= z : Re z> 0 and let f : G → C be an analytic function withlim supz→w | f (z)| ≤ Mfor w in ∂G, and also suppose that for everyε > 0,

limr→∞

supexp(−ε/r)| f (reiθ)| : |θ| < 12π = 0.

Show that| f (z)| ≤ M for all z in G.


Exercise 7. Let G = z : Re z> 0 and let f : G → C be analytic such that f(1) = 0 and such thatlim supz→w | f (z)| ≤ M for w in ∂G. Also, suppose that for everyδ, 0 < δ < 1, there is a constant P such that

| f (z)| ≤ Pexp(|z|1−δ).

Prove that

| f (z)| ≤ M

[

(1− x)2+ y2

(1+ x)2 + y2

] 12

.

Hint: Consider f(z)(

1+z1−z

)

.


92

Chapter 7

Compactness and Convergence in theSpace of Analytic Functions

7.1 The space of continuous functionsC(G,Ω)

Exercise 1. Prove Lemma 1.5 (Hint: Study the function f(t) = t1+t for t > −1.)

Solution. To show thatµ : S × S→ R, µ(s, t) = d(s,t)1+d(s,t) is a metric we proof thatµ satisfies the conditions

of a metric function.For s, t ∈ S the valueµ(s, t) = d(s,t)

1+d(s,t) is well defined in the nonnegative real numbers because d is a metric.Also µ(s, t) = 0 if and only if d(s, t) = 0 and since d is a metric it follows that s= t. The metric d issymmetric and thereforeµ is also. It remains to show the triangle inequality. Let therefore s, t,u ∈ S . Themetric d satisfies the triangle inequality

d(s,u) ≤ d(s, t) + d(t,u). (7.1)

The function f(t) = t1+t is a strictly increasing, concave function on the interval[0,∞). Thus if d(s,u) ≤

maxd(s, t),d(t,u), then alsoµ(s,u) ≤ maxµ(s, t), µ(t,u) and by nonnegativity ofµ also

µ(s,u) ≤ µ(s, t) + µ(t,u).

If insteadmaxd(s, t),d(t,u) < d(s,u) then

11+ d(s,u)

< min

11+ d(s, t)

,1

1+ d(t,u)

.

Together with equation (7.1) we have

µ(s,u) =d(s,u)

1+ d(s,u)≤ d(s, t) + d(t,u)

1+ d(s,u)

=d(s, t)

1+ d(s,u)+

d(t,u)1+ d(s,u)

≤ d(s, t)1+ d(s, t)

+d(t,u)

1+ d(t,u)

= µ(s, t) + µ(t,u),

which gives the triangle inequality.To show that the metrics d andµ are equivalent on S , let O be on open set in(S,d) and let x∈ O. There is

93

ε ∈ (0,1) such that Bd(x, ε) ⊂ O. Chooseδ positive such thatδ < ε1+ε , then Bµ(x, δ) ⊂ Bd(x, ε). Similarly, if

O is an open set in(S, µ) andδ is such that x∈ Bµ(x, δ) ⊂ O then chooseε > 0 with ε ≤ δ1−δ which implies

Bd(x, ε) ⊂ Bµ(x, δ). Since this can be done for any element x∈ O, open sets in(S,d) are open in(S, µ) andvice-versa.The fact that exactly the open sets in(S,d) are open in(S, µ) leads to the statement about Cauchy sequences.Assume thatxnn is a Cauchy sequence in(S,d). To see that it is Cauchy in(S, µ), let ε > 0 be arbitrarybut fixed. By the above there is aδ small enough that Bd(x, δ) ⊂ Bµ(x, ε) andδ depends only onε, but noton x ∈ S . Sincexnn is Cauchy in(S,d) there is N∈ N such that d(xm, xn) < δ whenever m,n ≥ N andtherefore alsoµ(xm, xn) < ε for m,n ≥ N. The opposite statement holds with a similar argument.

Exercise 2. Find the sets Kn obtained in Proposition 1.2 for each of the following choices of G: (a) G is anopen disk; (b) G is an open annulus; (c) G is the plane with n pairwise disjoint closed disks removed; (d) Gis an infinite strip; (e) G= C − Z.

Solution. a) G is an open disk,

Let a∈ C and r> 0 such that G= B(a, r). Set

Kn = B

(

a, r − 1n

)

where a ball of negative radius is to be understood as the empty set. Depending on the size of r, afinite number of Kn may be empty, not violating the condition Kn ⊂ int Kn+1, and obviously G=

⋃∞n=1.

b) G is an open annulus,

Let a∈ C be the center of the annulus with radii r,R,0 ≤ r < R< ∞ so that G= ann(a, r,R). Define

Kn = ann

(

a, r +1n,R− 1

n

)

,

again with the interpretation that an annulus with inner radius larger than the outer radius is theempty set.

d) G= z ∈ C : |Imz| < 1,For this open set G define

Kn = z= x+ iy ∈ C : x, y ∈ R, |x| ≤ n ∩

z= x+ iy ∈ C : x, y ∈ R|y| ≤ 1− 1n

.

Although G is unbounded each Kn is bounded and closed and hence compact and G=⋃∞

n=1 Kn.

e) G= C − Z.

Here we can define

Kn = z ∈ C : |z| ≤ n ∩

n⋃

j=−n

B

(

( j,1n

)

C

,

an intersection of closed set, one of which is bounded, henceKn is compact. Also Kn ⊂ int Kn+1 bydefinition and G=

⋃∞n=1.

Exercise 3. Supply the omitted details in the proof of Proposition 1.18.

94


Exercise 4. Let F be a subset of a metric space(X,d) such that F− is compact. Show that F is totallybounded.


Exercise 5. Suppose fn is a sequence in C(G,Ω) which converges to f andzn is a sequence in G whichconverges to a point z in G. Showlim fn(zn) = f (z).


Exercise 6. (Dini’s Theorem) Consider C(G,R) and suppose that fn is a sequence in C(G,R) which ismonotonically increasing (i.e., fn(z) ≤ fn+1(z) for all z in G) and lim fn(z) = f (z) for all z in G wheref ∈ C(G,R). Show that fn→ f .

Solution. By Proposition 1.2 we can find a sequenceKn of compact subsets of G such that⋃∞

n=1 Kn. Thus,WLOG we can assume that G is compact. Define gn = f − fn ∀n. Clearly gn is continuous∀n, since fand fn are continuous by assumption. Since fn is monotonically increasing, gn is monotonically decreasing.Since fn converges to f , gn converges pointwise to zero, that is

∀ε > 0 ∃N ∈ N ∀x ∈ G : 0 ≤ gN(x) <ε

2.

Since gN is continuous, there is an open neighborhood B(x; r) around x (with r< ∞). For any y∈ B(x; r)we have gN(y) < ε, that is

∀ε > 0 ∀x ∈ G ∃N ∈ N ∃B(x; r) : ∀y ∈ B(x; r) : gN(y) < ε.

Since gN is monotonically decreasing, we surely get gn(y) < ε for every n≥ N and for every y∈ B(x; r).These open neighborhoods will cover G. Since G is compact, wecan find finitely many x1, . . . , xk such thatB(xi ; r) cover G for every value ofε. Note that the values of N can change the chosen center x of eachthose neighborhoods, say Ni . Therefore, let M= maxi Ni . Because of the monotonicity of gn, we have thatgn(y) < ε ∀ n ≥ M in every open neighborhood B(xi ; r), that is, for every value in G. Hence,

∀ε > 0 ∃M ∈ N,∀n ≥ M ∀x ∈ G : | fn(x) − f (x)| < ε,

which shows fn→ f .

Exercise 7. Let fn ⊂ C(G,Ω) and suppose that fn is equicontinuous at each point of G. If f∈ C(G,Ω)and f(z) = lim fn(z) for each z then show that fn→ f .

Solution. By Proposition 1.2 we can find a sequenceKn of compact subsets of G such that⋃∞

n=1 Kn. Thus,WLOG we can assume that G is compact. Since fn is equicontiuous at each point of G, we have that

∀ε > 0 ∃δ > 0 : d( fn(x), fn(y)) <ε

3whenever|x− y| < δ. (7.2)

Taking the limit as n→ ∞, we also get

d( f (x), f (y)) <ε

3whenever|x− y| < δ. (7.3)

by the pointwise convergence ( f(z) = lim fn(z) ∀z ∈ G by assumption). As in the previous exercise, wecan find finitely many points y1, . . . , yk ∈ G such that the open neighborhoods B(yi ; δ) cover G, since G is

95

compact. To be more precise: For every point x in G, we can find an open neighborhood B(yi ; δ) for somei = 1, . . . , k. Since fn→ f pointwise, we have

d( fn(yi), f (yi)) <ε

3∀n ≥ Ni .

Set N= maxi Ni , then we have

d( fn(yi), f (yi)) <ε

3∀n ≥ N ∀i = 1, . . . , k. (7.4)

Finally, for an arbitrary x∈ G and n≥ N we have

d( fn(x), f (x)) ≤ d( fn(x), fn(yi)) + d( fn(yi), f (yi)) + d( f (yi), f (x)) <(7.2),(7.3),(7.4)

ε

3+ε

3+ε

3= ε

where we choose yi such that|x− yi | < δ. This shows fn→ f .

Exercise 8. (a) Let f be analytic on B(0;R) and let f(z) =∑∞

n=0 anzn for |z| < R. If fn(z) =∑n

k=0 akzk, showthat fn→ f in C(G;C).(b) Let G= ann(0; 0,R) and let f be analytic on G with Laurent series development f(z) =

∑∞n=−∞ anzn. Put

fn(z) =∑n

k=−∞ akzk and show that fn→ f in C(G;C).

Solution. a) Let G= B(0;R). This result is proved with the Arzela-Ascoli Theorem. Withthe given functionf define an analytic function

g : G→ C, g(z) :=∞∑

k=0

|ak|zk. (7.5)

Let z∈ G and define r:= |z|+R2 (< R) then z∈ B(0; r). LetF = fnn. Then for this fixed z and for any n∈ N

we have

| fn(z)| = |n∑

k=0

akzk| ≤

n∑

k=0

|ak||z|k = g(|z|) < ∞.

Thus for a given z∈ G, fn(z) : fn ∈ F is bounded and therefore has compact closure.Next, we want to show equicontinuity at each point of G. Fix z0 ∈ G and letε be positive. Let r> 0 be suchthat B(z0; r) ⊂ G. With the function g(z) defined in equation (7.5) and the computations from above we havefor all w ∈ ∂B(z0; r) and for all n∈ N

| fn(w)| ≤ g(|w|).The boundary of the disk with radius r is compact and so is g(∂B(z0; r)). Thus there is a positive number Msuch that|g(∂B(z0, r))| ≤ M.Chooseδ ∈

(

0,min r2 , εr2M

)

and let |z− z0| < δ then as in the proof of Montel’s theorem the followingestimate holds

| fn(z) − fn(z0)| = | 12πi

∮

|w−z0|=r

fn(w)(z− z0)(w− z)(w− z0)

dw|

≤ 2Mrδ ≤ ε.

By Arzela-Ascoli the sequence fn is normal in C(G) so there is a subsequence fnk that converges in C(G).Since f is analytic and fn is the sequence of partial sums of f , fn itself is an analytic Cauchy sequencein C(G). Hence every subsequence converges to the same limit, whichshows that fn→ f in C(G). As a laststep recall that H(G) is complete as a subspace of C(G) (Corollary 2.3) and fn ⊂ H(G) thus also the limitfunction f ∈ H(G).

96

7.2 Spaces of analytic functions

Exercise 1. Let f, f1, f2, . . . be elements of H(G) and show that fn→ f iff for each closed rectifiable curveγ in G, fn(z)→ f (z) uniformly for z inγ.


Exercise 2. Let G be a region, let a∈ R, and suppose that f: [a,∞] × G → C is a continuous func-tion. Define the integral F(z) =

∫ ∞a

f (t, z) dt to be uniformly convergent on compact subsets of G if

limb→∞∫ b

af (t, z) dt exists uniformly for z in any compact subset of G. Suppose that this integral does con-

verge uniformly on compact subsets of G and that for each t in(a,∞), f (t, · ) is analytic on G. Prove that Fis analytic and

F(k)(z) =∫ ∞

a

∂k f (t, z)∂zk

dt


Exercise 3. The proof of Montel’s Theorem can be broken up into the following sequence of definitions andpropositions: (a) Definition. A setF ⊂ C(G,C) is locally Lipschitz if for each a in G there are constantsM and r > 0 such that| f (z) − f (a)| ≤ M|z− a| for all f in F and |z− a| < r. (b) If F ∈ C(G,C) is locallyLipschitz thenF is equicontinuous at each point of G. (c) IfF ⊂ H(G) is locally bounded thenF is locallyLipschitz.


Exercise 4. Prove Vitali’s Theorem: If G is a region and fn ⊂ H(G) is locally bounded and f∈ H(G) thathas the property that A= z ∈ G : lim fn(z) = f (z) has a limit point in G then fn→ f .

Solution. The sequence fn is a locally bounded sequence in H(G) and by Montel’s Theorem then fn isnormal, so there is a subsequence fnk ⊂ fn that converges to f in H(G).Suppose that fn 9 f in H(G) then there must be a compact set K⊂ G the convergence is not uniform onK. In other words there must be anε > 0 so that for all n∈ N there is zn ∈ K with | fn(zn) − f (zn)| ≥ ε. Bycompactness of K extract a subsequence ofzn, sayznm that converges to a point z0 ∈ K.The sequence fn is locally bounded, so in particular the subsequence fnm with indices correspondingto znm is, and again by Montel’s Theorem now applied to the subsequence and again the completenessof H(G) there is an analytic function g such that fnm → g in H(G). On the set A of points of pointwiseconvergence fn(z)→ f and fn(z)→ g.Now G is a region and A has a limit point in G so by the Identity Theorem (Chapter 4, Corollary 3.8)already f = g on G which gives a contradiction on the set K and the point z0 because

| fnm(znm) − f (znm)| → |g(z0) − f (z0)| ≥ ε.

Hence we can conclude that fn→ f in H(G).

Exercise 5. Show that for a setF ⊂ H(G) the following are equivalent conditions:(a)F is normal;(b) For everyε > 0 there is a number c> 0 such thatc f : f ∈ F ⊂ B(0;ε) (here B(0;ε) is the ball inH(G) with center at0 and radiusε).


Exercise 6. Show that ifF ⊂ H(G) is normal thenF ′ = f ′ : f ∈ F is also normal. Is the converse true?Can you add something to the hypothesis thatF ′ is normal to insure thatF is normal?

97


Exercise 7. SupposeF is normal in H(G) andΩ is open inC such that f(G) ⊂ Ω for every f inF . Showthat if g is analytic onΩ and is bounded on bounded sets theng f : f ∈ F is normal.


Exercise 8. Let D= z : |z| < 1 and show thatF ⊂ H(D) is normal iff there is a sequenceMn of positiveconstants such thatlim sup n

√Mn ≤ 1 and if f(z) =

∑∞0 anzn is inF then|an| ≤ Mn for all n.


Exercise 9. Let D = B(0; 1) and for0 < r < 1 let γr (t) = re2πit , 0 ≤ t ≤ 1. Show that a sequence fn inH(D) converges to f iff

∫

γr| f (r) − fn(z)| |dz| → 0 as n→ ∞ for each r,0 < r < 1.

Solution. ⇒: Assumelimn→∞ fn(z) = f (z) ∀z ∈ D. Then∫

γr

| f (z) − fn(z)| |dz| ≤ supz∈γr

| f (z) − fn(z)|∫

γr

|dz|

=γr=re2πit ,0≤t≤1

supz∈γr

| f (z) − fn(z)|·2πr, ∀0 < r < 1.

Butsupz∈γr| f (z) − fn(z)| → as n→ ∞ sincelimn→∞ fn(z) = f (z) ∀ z ∈ D. Therefore,

∫

γr

| f (z) − fn(z)| |dz| → 0

as n→ ∞ for each r,0 < r < 1.⇐: Assume

∫

γr| f (z)− fn(z)| |dz| → 0 as n→ ∞ for each r,0 < r < 1. Let K be an arbitrary compact subset

of D = B(0; 1) and choose0 < r < 1 such that K⊂ B(0; r) and denote the shortest distance between anarbitrary point in K and an arbitrary point on∂B byδ > 0. By the assumption, we have

∫

∂B| f (z) − fn(z)| |dz| < 2πεδ. (7.6)

Let a∈ K, then by Cauchy’s Integral formula, we have

f (a) − fn(a) =1

2πi

∫

∂B

f (z) − fn(z)z− a

dz.

Thus,

| f (a) − fn(a)| ≤ 12π

∫

∂B

| f (z) − fn(z)||z− a| |dz| ≤ 1

2πδ

∫

∂B| f (z) − fn(z)| |dz| <

(7.6)

2πεδ2πδ

= ε.

Hence, fn(a)→ f (z) uniformly∀a ∈ K. Since it converges uniformly on all compact subsets of D, Proposi-tion 1.10 b) yields that fn converges to f∀ z ∈ D.

Exercise 10. Let fn ⊂ H(G) be a sequence of one-one functions which converge to f . If G isa region,show that either f is one-one or f is a constant function.

Solution. Assume fn ⊂ H(G) is a sequence of one-one functions which converge to f where Gis a region(open and connected). Suppose f is not the constant function. Then, we have to show that f is one-one.Choose an arbitrary point z0 ∈ G and define the sequence gn(z) = fn(z) − fn(z0). Clearly, the sequencegnconverges to g(z) = f (z) − f (z0) on the open connected set G\z0, which is again a region. In addition, we

98

have that gn never vanishes on G\z0, since fn are assumed to be one-one functions.Therefore, the sequencegn satisfies the conditions of the Corollary 2.6 resulting fromHurwitz’s Theorem,where the region is G\z0. Thus, either g≡ 0 or g never vanishes (g has no zeros). Since g(z) = f (z)− f (z0)is not identically zero on G\z0, we must have g(z) = f (z) − f (z0) has no zeros in G\z0. This implies

f (z) , f (z0) ∀z ∈ G\z0.

Since z0 was an arbitrary point in G, we have shown that f is one-one on G.

Exercise 11. Suppose that fn is a sequence in H(G), f is a non-constant function, and fn → f in H(G).Let a ∈ G andα = f (a); show that there is a sequencean in G such that: (i) a= lim an; (ii) f n(an) = αfor sufficiently large n.


Exercise 12.Show thatlim tannz= −i uniformly for z in any compact subset of G= z : Im z> 0.


Exercise 13. (a) Show that if f is analytic on an open set containing the disk B(a; R) then

| f (a)|2 ≤ 1πR2

∫ 2π

0

∫ R

0| f (a+ reiθ)|2r dr dθ.

(b) Let G be a region and let M be a fixed positive constant. LetF be the family of all functions f in H(G)such that

!

G| f (z)|2 dx dy≤ M. Show thatF is normal.

Solution. a) Let B(a; R) ⊂ G and f : G → C be analytic. Letγ(t) = a + reit for 0 ≤ t ≤ 2π, then byCauchy’s Integral Formula we have

g(a) =1

2πi

∫

γ

g(w)w− a

dw=12π

∫ 2π

0f (a+ reit ) dt.

Thus, letting g(z) = f (z)2, we obtain

f 2(a) =12π

∫ 2π

0

[

f (a+ reit )]2

dt.

Multiply by r, we get

f 2(a)r =12π

∫ 2π

0

[

f (a+ reiθ)]2

r dθ (7.7)

and then integrate with respect to r yields∫ R

0f 2(a)r dr = f 2(a)

R2

2=

(7.7)

∫ R

0

12π

∫ 2π

0

[

f (a+ reiθ)]2

r dθ dr =12π

∫ 2π

0

∫ R

0

[

f (a+ reiθ)]2

r dr dθ

which implies

f 2(a) =1πR2

∫ 2π

0

∫ R

0

[

f (a+ reiθ)]2

r dr dθ.

So

| f (a)|2 ≤ 1πR2

∫ 2π

0

∫ R

0

∣∣∣ f (a+ reiθ)

∣∣∣2

r dr dθ.

99

b) Let G be a region and M be a fixed constant. LetF be the family of all functions f in H(G) such that"

G| f (z)|2 dx dy≤ M. (7.8)

To show thatF is normal.According to Montel’s Theorem, it suffices to show thatF is locally bounded, that is∃ r > 0 such thatsup| f (z)| : |z− a| < r, f ∈ F < ∞ (p. 153). We will prove this fact by contradiction:AssumeF is not locally bounded. Then there is a compact set K⊂ G such thatsup| f (z)| : z ∈ K, f ∈ F =∞. That is, there is a sequence fn in F such thatsup| fn(z)| : z ∈ K ≥ n. Therefore,∃ zn ∈ K such that

| fn(zn)| ≥ n2. (7.9)

Every sequence has a convergence subsequence, sayznk, with znk → z0 ∈ K since K is compact. WLOG,we writeznk = zn. Clearly z0 ∈ G since z0 ∈ K (K ⊂ G), hence∃ R > 0 such that B(z0; R) ⊂ G (sinceG is open). If we pick n large enough, we can get zn ∈ B(z0; R

2 ) (a picture might help). In addition, we can

find anR > 0 such thatB(zn; R) ⊂ B(z0; R2 ) because B(z0; R

2 ) is open. Apply part a) to the ballB(zn; R) toobtain:

| fn(zn)|2 ≤ 1

πR2

∫ 2π

0

∫ R

0

∣∣∣ fn(zn + reiθ)

∣∣∣2

r dr dθ ≤B(zn;R)⊂G

1

πR2

"

G| fn(z)|2 dx dy ≤

(7.8)

1

πR3M < ∞.

So

| fn(zn)| ≤√

Mπ

1

R< ∞.

But by (7.9) we have

| fn(zn)| ≥ n2.

This gives a contradiction if we let n→ ∞. Thus,F is locally bounded and therefore normal.

7.3 Spaces of meromorphic functions


Solution. a) Let a∈ C and r > 0. To show: There is a numberδ > 0 such that B∞(a; δ) ⊂ B(a; r). LetR> 0 such that B(a; r) ⊂ B(0;R) and B∞(a; δ) ⊂ B(0;R).Claim: Pick

δ =2r√

1+ R2√

1+ |a|2to obtain

B∞(a; δ) ⊂ B(a; R).

Proof of the claim: Let z∈ B∞(a; δ). This is equivalent to

2|z− a|√

1+ |z|2√

1+ |a|2< δ

⇐⇒ 2|z− a|√

1+ |z|2√

1+ |a|2<

2r√1+ R2

√

1+ |a|2

⇐⇒ |z− a| < r√1+ R2

√

1+ |z|2

100

which implies (z∈ B∞(a; δ)⇒ z ∈ B(0;R), that is|z| < R)

|z− a| < r√1+ R2

√1+ R2

⇐⇒ |z− a| < r.

So z∈ B(a; r). ThereforeB∞(a; δ) ⊂ B(a; r).

b) Letδ > 0 and a∈ C. To show: There is a number r> 0 such that B(a; r) ⊂ B∞(a; δ).Claim: Pick r< δ

2 to obtainB(a; r) ⊂ B∞(a; δ).

Proof of the claim: Let z∈ B(a; r). This is equivalent to

|z− a| < r

⇒r< δ

2

|z− a| < δ

2

⇐⇒ |z− a| < δ

2·1

⇒1≤√

1+|z|2√

1+|a|2|z− a| < δ

2

√

1+ |z|2√

1+ |a|2

⇐⇒ 2|z− a|√

1+ |z|2√

1+ |a|2< δ

⇐⇒ d(z,a) < δ.

So z∈ B∞(a; δ). Therefore,B(a; r) ⊂ B∞(a; δ).

c) Letδ > 0. To show: There is a compact set K⊂ C such thatC∞ − K ⊂ B∞(∞; δ).

Claim: Choose K= B(0; r) where r>√

4δ2 − 1 to obtain

C∞ − K ⊂ B∞(∞; δ).

(Clearly K is compact and r> 0 sinceδ ≤ 2.)

101

Proof of the claim:

C∞ − K = C∞ − B(0; r)

= C − B(0; r) ∪ ∞= C − z ∈ C : |z| ≤ r ∪ ∞

⊂r>

√

4δ2−1

C −

z ∈ C : |z| ≤

√

4δ2− 1

∪ ∞

=

z ∈ C : |z| >

√

4δ2− 1

∪ ∞

=

z ∈ C :

√

4δ2− 1 < |z|

∪ ∞

=

z ∈ C :

2√

1+ |z|2< δ

∪ ∞

= z ∈ C : d(∞, z) < δ ∪ ∞= B∞(∞; δ).

Therefore,C∞ − K ⊂ B∞(∞; δ).

d) Let K ⊂ C where K is compact. To show: There is a numberδ > 0 such that B∞(∞; δ) ⊂ C∞ − K. LetB(0; r) such that B(0; r) ⊃ K where r> 0. Clearly

C∞ − B(0; r) ⊂ C∞ − K.

So it suffices to show: There is a numberδ > 0 such that

B∞(∞; δ) ⊂ C∞ − B(0; r)

for a given r> 0.Claim: Chooseδ ≤ 2√

r2+1to obtain

B∞(∞; δ) ⊂ C∞ − B(0; r)

and henceB∞(∞; δ) ⊂ C∞ − K.

Proof of the claim:

B∞(∞; δ) = z ∈ C : d(∞, z) < δ ∪ ∞

=

z ∈ C :

2√

1+ |z|2< δ

∪ ∞

⊂δ≤ 2√

r2+1

z ∈ C :

2√

1+ |z|2<

2√r2 + 1

∪ ∞

= z ∈ C : r ≤ |z| ∪ ∞= C ∪ z ∈ C : |z| < r ∪ ∞= C∞ − z ∈ C : |z| < r= C∞ − B(0; r).

102

Therefore,B∞(∞; δ) ⊂ C∞ − B(0; r).

Exercise 2. Show that ifF ⊂ M(G) is a normal family in C(G,C∞) thenµ(F ) is locally bounded.


7.4 The Riemann Mapping Theorem

Exercise 1. Let G andΩ be open sets in the plane and let f: G → Ω be a continuous function whichis one-one, onto, and such that f−1 : Ω → G is also continuous (a homeomorphism). Supposezn isa sequence in G which converges to a point z in∂G; also suppose that w= lim f (zn) exists. Prove thatw ∈ ∂Ω.


Exercise 2. (a) Let G be a region, let a∈ G and suppose that f: (G − a) → C is an analytic functionsuch that f(G − a) = Ω is bounded. Show that f has a removable singularity at z= a. If f is one-one,show that f(a) ∈ ∂Ω.(b) Show that there is no one-one analytic function which maps G = z : 0 < |z| < 1 onto an annulusΩ = z : r < |z| < R where r> 0.


Exercise 3. Let G be a simply connected region which is not the whole planeand suppose thatz ∈ Gwhenever z∈ G. Let a∈ G ∩ R and suppose that f: G→ D = z : |z| < 1 is a one-one analytic functionwith f(a) = 0, f ′(a) > 0 and f(G) = D. Let G+ = z ∈ G : Im z> 0. Show that f(G+) must lie entirelyabove or entirely below the real axis.


Exercise 4. Find an analytic function f which mapsz : |z| < 1,Re z> 0 onto B(0; 1) in a one-one fashion.


Exercise 5. Let f be analytic on G= z : Re z> 0, one-one, with Re f(z) > 0 for all z in G, and f(a) = afor some real number a. Show that| f ′(a)| ≤ 1.

Solution. Clearly G is a simply connected region (not the whole plane).Let a∈ G. Then, there is a uniqueanalytic function g: G→ D having the propertiesa) g(a) = 0 and g′(a) > 0b) g is 1-1c) g(G) = D

(by the Riemann Mapping Theorem). Since, we have f: G1−1−−−−−−→ G (since Re f(z) > 0) and g :

1−1−−−−−−→onto

D so g−1 : D1−1−−−−−−→onto

G we can define h(z) = g( f (g−1(z))). Clearly h(z) is analytic and one-

one, since f and g are analytic and one-one, and we have

h(D) = D

by construction. We have

h(0) = g( f (g−1(0))) =g−1(0)=a

g( f (a)) =f (a)=a

g(a) = 0.

103

We also have|h(z)| ≤ 1 ∀z ∈ D

since h: D→ D. Thus, the hypothesis of Schwarz’s Lemma are satisfied, andhence, we get

|h′(0)| ≤ 1.

We have

h′(z) =

[

g( f (g−1(z)))]′= g′( f (g−1(z))) f ′(g−1(z))

(

g−1(z))′

= g′( f (g−1(z))) f ′(g−1(z))1

g′(g−1(z))

where the last step follows from Proposition 2.20 provided g′(g−1(z)) , 0. So

h′(0) = g′( f (g−1(0))) f ′(g−1(0))1

g′(g−1(0))

= g′( f (a)) f ′(a)1

g′(a)= g′(a) f ′(a)

1g′(a)

= f ′(a)

(g′(a) > 0 by assumption). Therefore|h′(0)| = | f ′(a)| ≤ 1.

Thus, we have shown that| f ′(a)| ≤ 1.

Exercise 6. Let G1 and G2 be simply connected regions neither of which is the whole plane. Let f be aone-one analytic mapping of G1 onto G2. Let a∈ G1 and putα = f (a). Prove that for any one-one analyticmap h of G1 into G2 with h(a) = α it follows that|h′(a)| ≤ | f ′(a)|. Suppose h is not assumed to be one-one;what can be said?

Solution. Define the function F(z) = f −1(h(z)). This is well-defined since f: G11−1−−−−−−→onto

G2 and h :

G11−1−−−−−−→ G2. Clearly F(z) is analytic since f and h are analytic, F(z) is one-one and F: G1 → G1 by

construction. We haveF(a) = f −1(h(a)) = f −1(α) = a.

Thus by Exercise 5, we get|F′(a)| ≤ 1.

We have

F′(z) =(

f −1(h(z)))′= f ′(h(z))h′(z) =

1f ′( f −1(h(z)))

h′(z)

where the last step follows by Proposition 2.20 provided f′( f −1(h(z))) , 0. So

F′(a) =1

f ′( f −1(h(a)))h′(a) =

h′(a)f ′( f −1(α))

=h′(a)f ′(a)

which is well-defines since f′(a) , 0 by assumption (f is one-one). Since

|F′(a)| = |h′(a)|| f ′(a)| ≤ 1

104

gives|h′(a)| ≤ | f ′(a)|.

If h is not one-one, then we get the same result, since I do not need the assumption that F(z) has to beone-one for exercise 5.

Exercise 7. Let G be a simply connected region and suppose that G is not thewhole plane. Let∆ = ξ :|ξ| < 1 and suppose that f is an analytic, one-one map of G onto∆ with f(a) = 0 and f′(a) > 0 for somepoint a in G. Let g be any other analytic, one-one map of G onto∆ and express g in terms of f .

Solution. Let a ∈ G such that f(a) = 0 and f′(a) > 0. Since g is one-one mapping from G onto∆, weclearly have g(a) = α, whereα ∈ ∆ (α does not have to be zero). Next, define

g1(z) =z− α1− αz

.

Clearly, g1 is a Möbius transformation mapping∆ to ∆ and in addition g1 is one-one and analytic. Nowdefine

g2(z) = g1(g(z)).

Then, g2 is one-one and analytic and maps G onto∆. In addition, we have

g2(a) = g1(g(a)) = g1(α) = 0. (7.10)

But g′2(a) need not necessarily satisfy g′2(a) > 0. However, g′2(a) , 0 ∀z ∈ G since g2 is one-one. Therefore,define the rotation

g3(z) = eiθz

for θ ∈ [0,2π). Clearly g3 maps∆ onto∆. g3 is one-one and analytic. Finally, let

h(z) = g3(g2(z)).

Also h is one-one, analytic and maps G onto∆. We have

h(a) = g3(g2(a)) =(7.10)

g3(0) = 0.

And,

h′(z) = (g3(g2(z)))′ = (g3(g1(g(z))))′ = g′3(g1(g(z)))g′1(g(z))g′(z)

= eiθ (1− αg(z))·1− (g(z) − α)· (−α)(1− αg(z))2

g′(z).

So,

h′(a) = eiθ (1− αα) − 0(1− αα)2

g′(a) = eiθ 1− |α|2(1− |α|2)2

g′(a) = eiθ 11− |α|2 g′(a).

Clearly, we can choose aθ ∈ [0,2π) such that h′(a) > 0 (depending on the sign of g′(a). Recall g′(a) , 0since g is one-one andα ∈ ∆, so 1

1−|α|2 > 0). For example: If g′(a) < 0, then pickθ = π, so eiθ = −1 orθ = −Arg(g′(a)) will work, too.By uniqueness of the map f ( f satisfies the properties of the Riemann Mapping Theorem), we have

f (z) = h(z)

105

since h satisfies the properties of the Riemann Mapping Theorem, too. Thus,

f (z) = h(z) = g3(g2(z)) = eiθg2(z) = eiθg1(g(z)) = eiθ g(z) − α1− αg(z)

.

So

1− αg(z) = eiθg(z) − αeiθ

⇐⇒ f (z) − αg(z) f (z) = eiθg(z) − αeiθ

⇐⇒ (eiθ+ α f (z))g(z) = f (z) + αeiθ

⇐⇒ g(z) =f (z) + αeiθ

eiθ + α f (z).

Exercise 8. Let r1, r2,R1,R2, be positive numbers such that R1/r1 = R2/r2; show that ann(0; r1,R1) andann(0; r2,R2) are conformally equivalent.

Solution. The function f: C→ C; f (z) = r2r1

z is analytic and it maps the annulus ann(0; r1,R1) bijectivelyinto ann(0; r2,R2). The inner boundaryz : |z| = r1 of the domain is mapped to the inner boundary ofann(0; r2,R2). For the outer boundary note that by assumption R1 =

R2r1r2

, so if |z| = R1 then| f (z)| = r2r1R1=

R2. The inverse function is f−1(z) = r1r2

z, an analytic and non-constant function also. Therefore the result isproved.

Exercise 9.Show that there is an analytic function f defined on G=ann(0; 0,1) such that f′ never vanishesand f(G) = B(0; 1).


7.5 The Weierstrass Factorization Theorem

Exercise 1. Show that∏

(1+ zn) converges absolutely iff∏

(1+ |zn|) converges.

Solution. Assume Re(zn) > −1. Then the product∏

(1 + zn) converges absolutely iff∑∞

k=1 zn convergesabsolutely (by Corollary 5.6 p. 166 applied to wn = 1 + zn).

∑∞k=1 zn converges absolutely iff

∑∞n=1 |zn|

converges (by definition).Claim:

∑∞n=1 |zn| converges iff

∑∞n=1 log(1+ |zn|) converges.

Then∑∞

n=1 log(1+ |zn|) converges iff∏

(1+ |zn|) converges (by Proposition 5.2 p.165 applied to wn = 1+ |zn|).Thus, ones we have proved the claim, we obtain:

∏

(1+ zn) converges absolutely iff∏

(1+ |zn|) converges.Proof of the claim: Let xn = |zn| for convenience. Clearly xn ≥ 0 ∀n ∈ N.⇒: Assume

∑ |zn| converges, that is∑

xn converges. Therefore, we have xn→ 0. So givenε > 0, we have

0 ≤ log(1+ |xn|) ≤ (1+ ε)xn

for sufficiently large n where the last inequality follows sincelimz→0log(1+z)

z = 1 by Exercise 2. By thecomparison test,

∑

log(1+ xn) converges, that is∑

log(1+ |zn|) converges.⇐: Assume

∑

log(1+ |zn|) converges, that is∑

log(1+ xn) converges. Therefore, we havelog(1+ xn) → 0and therefore xn→ 0. So givenε > 0, we have

(1− ε)xn ≤ log(1+ xn)

for sufficiently large n where the last inequality follows sincelimz→0log(1+z)

z = 1 by Exercise 2. By thecomparison test,

∑

xn converges, that is∑ |zn| converges.

Therefore,∑∞

n=1 |zn| converges iff∑∞

n=1 log(1+ |zn|) converges.

106

Exercise 2. Prove that limz→0log(1+z)

z = 1.

Solution. First note that,

f (z) =log(1+ z)

zhas a removable singularity at z= 0, since

limz→0

z f(z) = limz→0

zlog(1+ z)

z= lim

z→0log(1+ z) = log(1)= 0

(by Theorem 1.2 Chapter 5 p. 103). We know

log(1+ z) =p.165

∞∑

k=1

(−1)k+1 zk

k

if |z| < 1. Therefore,

f (z) =log(1+ z)

z=

∑∞k=1(−1)k+1 zk

k

z=

∞∑

k=1

(−1)k+1 zk−1

k= 1+

∞∑

k=2

(−1)k+2 zk−1

k= 1+

∞∑

k=1

(−1)kzk

k+ 1

where the sum is an analytic function. Hence,

limz→0

f (z) = limz→0

1+∞∑

k=1

(−1)kzk

k+ 1

= 1+∞∑

k=1

(−1)k(limz→0 z)k

k+ 1= 1+

∞∑

k=1

0 = 1

and therefore

limz→0log(1+ z)

z= 1.

Exercise 3. Let f and g be analytic functions on a region G and show that there are analytic functionsf1,g1, and h on G such that f(z) = h(z) f1(z) and g(z) = h(z)g1(z) for all z in G; and f1 and g1 have nocommon zeros.

Solution. Let f and g be analytic functions on the region G. Leta j be the zeros of f with multiplicity nj .Let b j be the zeros of g with multiplicityn j .If f and g have no common zeros, let h= 1. Then f1 = f and g1 = g. Clearly f1, g1 and h are analytic onG such that f(z) = h(z) f1(z) and g(z) = h(z)g1(z) for all z in G and f1 and g1 have no common zeros.Otherwise letzj be the set of common zeros of f and g with multiplicity mj = min(n j , n j). Thenzj forma sequence of distinct points in G with no limit points on G (since the zeros are isolated). Then there is ananalytic function h defined on G whose only zeros are at the points zj with multiplicity mj (by Theorem 5.15p. 170). Let f1 =

fh and g1 =

gh, then clearly f1 and g1 are analytic functions on G such that f(z) = h(z) f1(z)

and g(z) = h(z)g1(z) for all z in G. (since f1 and g1 have removable singularities at the zj ’s by construction;the multiplicity of zj ’s in f and g is greater or equal than the multiplicity of zj ’s in h). In addition, we havethat f1 and g1 have no common zeros by construction.

Exercise 4. (a) Let0 < |a| < 1 and |z| ≤ r < 1; show that∣∣∣∣∣

a+ |a|z(1− az)a

∣∣∣∣∣≤ 1+ r

1− r

(b) Letan be a sequence of complex numbers with0 < |an| < 1 and∑

(1− |an|) < ∞. Show that the infiniteproduct

B(z) =∞∏

n=1

|an|an

(

an − z1− anz

)

107

converges in H(B(0; 1))and that|B(z)| ≤ 1. What are the zeros of B? (B(z) is called a Blaschke Product.)(c) Find a sequencean in B(0; 1)such that

∑

(1− |an|) < ∞ and every number eiθ is a limit point ofan.Solution. Not available.

Exercise 5. Discuss the convergence of the infinite product∏∞

n=11np for p > 0.


Exercise 6. Discuss the convergence of the infinite products∏[

1+ in

]

and∏∣

∣∣1+ i

n

∣∣∣.

Solution. First consider∏∞

n=1

(

1+ in

)

. We claim that this product does not converge.

Let zn := 1 + in = rneiθn with rn =

√

1+ 1n2 andθ = arctan

(1n

)

. If there is z= reiθ such that∏∞

n=1 zn = z,

then∏∞

n=1 rn = r and∑∞

n=1 θn = θ.Sincearctan(x)

x ≥ 12 for all values x∈ (0,1] the estimate

∞∑

n=1

θn =

∞∑

n=1

arctan

(

1n

)

≥ 12

∞∑

n1

1n

is valid. The right hand side does not converge, therefore the angles do not converge and there is no z∈ Cwith the desired properties (note that rn > 1 for all n, thus z, 0).

The second infinite product is a product of real numbers and zn = |1 + in | =

√

1+ 1n2 . The estimate

1+ 1n2 ≤

(

1+ 1n2

)2implies that1 ≤ zn ≤ 1+ 1

n2 . Thus for any integer N∈ NN∏

n=1

1 ≤N∏

n=1

zn ≤N∏

n=1

(

1+1n2

)

where the last term is absolutely convergent by an example inclass. Therefore the sequence of the partialproducts

∏Nn=1 zn is an increasing, bounded sequence that converges in the real numbers. Withlog(zn) > 0

for all n, the comparison with the sum of logarithms (Proposition 5.4) gives absolute convergence.

Exercise 7. Show that∏∞

n=2

(

1− 1n2

)

=12.

Solution. Consider

Pn =

n∏

k=2

(

1− 1k2

)

=

n∏

k=2

(

1− 1k

) (

1+1k

)

=

n∏

k=2

(

k− 1k

) (

k+ 1k

)

= exp

log

n∏

k=2

(

k− 1k

) (

k+ 1k

)

= exp

n∑

k=2

log

[(

k− 1k

) (

k+ 1k

)]

= exp

n∑

k=2

[

log(k− 1)− log(k)]

+ exp

n∑

k=2

[

log(k+ 1)− log(k)]

= exp[

log(1)− log(n)]

+ exp[− log(2)+ log(n+ 1)

]

= exp[− log(2n) + log(n+ 1)

]

= exp

log

(

n+ 12n

)

=n+ 12n

.

So,∞∏

n=2

(

1− 1n2

)

= limn→∞

Pn = limn→∞

n+ 12n=

12.

108

Exercise 8. For which values of z do the products∏

(1 − zn) and∏

(1 + z2n) converge? Is there an openset G such that the product converges uniformly on each compact subset of G? If so, give the largest suchopen.

Solution. Consider the infinite product∏∞

n=1(1− zn). By definition and Corollary 5.6 this converges abso-lutely if and only if

∑∞n=1−zn converges absolutely. From Calculus it is known that the convergence is also

uniform for |z| ≤ r < 1 with r ∈ [0,1).If z is an nth root of unity for a positive integer n, then one of the factorsin the infinite product becomes0and the product converges to 0, but not absolutely accordingto Definition 5.5. If there is no n such that zis nth root of unity then no factor of the infinite product equals zero and there is a subsequence(1 − znkksuch that|1− znk | > 1.5. We conclude that

∏∞n=1(1+ zn) does not converge.

With the denseness of⋃

n∈Nz ∈ C : z is nth root of unity in the unit sphere it follows that the maximal set Gsuch that the convergence is uniform on compact subsets is the unit disk.The behavior of

∏∞n=1(1+ z2n) is almost the same as the previous one. Substitute w:= z2, then the conver-

gence is uniform for|w| ≤ r < 1, so |z| ≤ sqrtr < 1. Also if |z| -and hence|w|- is greater than1, then theinfinite product diverges. Similar to the previous case one factor equals0 if z is a2nth root of−1.The maximal region G such that the convergence is uniform on compact sets is again the open unit disk.

Exercise 9. Use Theorem 5.15 to show there is an analytic function f on D= z : |z| < 1 which is notanalytic on any open set G which properly contains D.


Exercise 10. Suppose G is an open set and fn is a sequence in H(G) such that f(z) =∏

fn(z) convergesin H(G). (a) Show that

∞∑

k=1

f ′k(z)

∏

n,k

fn(z)

converges in H(G) and equals f′(z). (b) Assume that f is not the identically zero function and let K be acompact subset of G such that f(z) , 0 for all z in K. Show that

f ′(z)f (z)

=

∞∑

n=1

f ′n(z)fn(z)

and the convergence is uniform over K.


Exercise 11. A subsetJ of H(G), G a region, is an ideal iff: (i) f and g inJ implies a f+ bg is inJ forall complex numbers a and b; (ii) f inJ and g any function in H(G) implies f g is inJ . J is called aproper ideal ifJ , (0) andJ , H(G); J is a maximal ideal ifJ is a proper ideal and wheneverL is anideal withJ ⊂ L then eitherL = J or L = H(G); J is a prime ideal if whenever f and g∈ H(G) andf g ∈ J then either f∈ J or g ∈ J . If f ∈ H(G) letZ( f ) be the set of zeros of f counted according totheir multiplicity. SoZ((z− a)3) = a,a,a. If S ⊂ H(G) thenZ(S) = ∩Z( f ) : f ∈ S, where the zerosare again counted according to their multiplicity. So ifS = (z− a)3(z− b), (z− a)2 thenZ(S) = a,a.(a) If f and g∈ H(G) then f divides g (in symbols, f|g if there is an h in H(G) such that g= f h. Show thatf |g iffZ ⊂ Z(g).(b) If S ⊂ H(G) andS contains a non-zero function then f is a greatest common divisor ofS if: (i) f |g foreach g inS and (ii) whenever h|g for each g inS, h| f . In symbols, f= g.c.d.S. Prove that f= g.c.d.S. iffZ( f ) = Z(S) and show that each non-empty subset of H(G) has a g.c.d.(c) If A ⊂ G letJ(a) = f ∈ H(G) : Z( f ) ⊃ A. Show thatJ(A) is a closed ideal in H(G) andJ(A) = (0)

109

iff A has a limit point in G.(d) Let a∈ G andJ = J(a). Show thatJ is a maximal ideal.(e) Show that every maximal ideal in H(G) is a prime ideal.(f) Give an example of an ideal which is not a prime ideal.


Exercise 12. Find an entire function f such that f(n + in) = 0 for every integer n (positive, negative orzero). Give the most elementary example possible (i.e., choose the pn to be as small as possible).


Exercise 13.Find an entire function f such that f(m+ in) = 0 for all possible integers m,n. Find the mostelementary solution possible.


7.6 Factorization of the sine function

Exercise 1. Show thatcosπz=∏∞

n=1

[

1− 4z2

(2n−1)2

]

.

Solution. We know by the double-angle identity of sinesin(2z) = 2 sin(z) cos(z) (this is proved easily byusing the definition) orsin(2πz) = 2 sin(πz) cos(πz). Since, we knowsin(πz) = πz

∏∞n=1

(

1− z2

n2

)

, we obtain

sin(2πz) = 2 sin(πz) cos(πz)

⇐⇒ 2πz∞∏

n=1

(

1− 4z2

n2

)

= 2πz∞∏

n=1

(

1− z2

n2

)

cos(πz)

⇐⇒ 2πz∞∏

m=1

(

1− 4z2

(2m)2

) ∞∏

m=1

(

1− 4z2

(2m− 1)2

)

= 2πz∞∏

n=1

(

1− z2

n2

)

cos(πz)

where the last statement follows by splitting the product into a product of the even and odd terms (rear-rangement of the terms is allowed). Hence

2πz∞∏

n=1

(

1− 4z2

(2n)2

) ∞∏

n=1

(

1− 4z2

(2n− 1)2

)

= 2πz∞∏

n=1

(

1− z2

n2

)

cos(πz)

⇐⇒ 2πz∞∏

n=1

(

1− z2

n2

) ∞∏

n=1

(

1− 4z2

(2n− 1)2

)

= 2πz∞∏

n=1

(

1− z2

n2

)

cos(πz).

Thus,

cos(πz) =∞∏

n=1

(

1− 4z2

(2n− 1)2

)

.

Exercise 2. Find a factorization forsinhz andcoshz.

110

Solution. We know

sinh(z) =ez − e−z

2=

e−i(iz) − e−(−i)(iz)

2

=−ei(iz)

+ e−i(iz)

2= i−ei(iz)

+ e−i(iz)

2i

= −iei(iz) − e−i(iz)

2i= −i sin(iz)

= −i(iz)∞∏

n=1

(

1− (iz)2

n2π2

)

= z∞∏

n=1

(

1+z2

n2π2

)

since by p. 175 Equation 6.2

sin(πz) = πz∞∏

n=1

(

1− z2

n2

)

⇒ sin(z) = z∞∏

n=1

(

1− z2

n2π2

)

.

Therefore,

sinh(z) = z∞∏

n=1

(

1+z2

n2π2

)

.

We have

cosh(z) =ez+ e−z

2=

e−i(iz)+ e−(−i)(iz)

2

=ei(iz)+ e−i(iz)

2= cos(iz)

=

∞∏

n=1

(

1− 4(iz)2

(2n− 1)2π2

)

=

∞∏

n=1

(

1+4z2

(2n− 1)2π2

)

since by Exercise 1 p. 176

cos(πz) =∞∏

n=1

(

1− 4z2

(2n− 1)2

)

⇒ cos(z) =∞∏

n=1

(

1− 4z2

(2n− 1)2π2

)

.

Therefore,

cosh(z) =∞∏

n=1

(

1+4z2

(2n− 1)2π2

)

.

Exercise 3. Find a factorization of the functioncos(πz4

)

− sin(πz4

)

.


Exercise 4. Prove Wallis’s formula:π2 =∏∞

n=1(2n)2

(2n−1)(2n+1).


111

7.7 The gamma function

Exercise 1. Show that0 < γ < 1. (An approximation toγ is .57722. It is unknown whetherγ is rational orirrational.)


Exercise 2. Show thatΓ(z)Γ(1− z) = π cscπz for z not an integer. Deduce from this thatΓ( 12) =

√π.

Solution. Assume z is not an integer, then by Gauss’s Formula p. 178 we get

Γ(z)Γ(1− z) = limn→∞

n!nz

z(z+ 1) · · · (z+ n)lim

m→∞m!m1−z

(1− z)(2− z) · · · (m+ 1− z)

= limn→∞

n!n!nz+1−z

z(1+ z)(1− z)(2+ z)(2− z) · · · (n+ z)(n− z)(n+ 1− z)

= limn→∞

(n!)2nz(12 − z2)(22 − z2) · · · (n2 − z2)(n+ 1− z)

= limn→∞

(n!)2n

z(n!n!)(

1− z2

12

) (

1− z2

22

) (

1− z2

23

)

· · ·(

1− z2

n2

)

(n+ 1− z)

= limn→∞

1

z(

1− z2

12

) (

1− z2

22

) (

1− z2

32

)

· · ·(

1− z2

n2

)n+1−z

n

=1

zlimn→∞(

1− z2

12

) (

1− z2

22

) (

1− z2

32

)

· · ·(

1− z2

n2

)

limn→∞

n+ 1− zn

︸︷︷︸

=1

=1

z∏∞

n=1

(

1− z2

n2

) =π

πz∏∞

n=1

(

1− z2

n2

)

=π

sin(πz)= π csc(πz)

where the step before the last step follows by p. 175 Equation6.2. So

Γ(z)Γ(1− z) = π csc(πz)

for z not an integer. Now, let z= 12, then

Γ

(

12

)

Γ

(

1− 12

)

= π csc(π

2

)

which implies

Γ

(

12

)

=

√π·1 =

√π.

Exercise 3. Show:√πΓ(2z) = 22z−1

Γ(z)Γ(z+ 12). (Hint: Consider the functionΓ(z)Γ(z+ 1

2)Γ(2z)−1.)

Solution. Following the hint define a function f on its domain G= z ∈ C : z < (Z − N) by

f (z) =2z−1

√πΓ

( z2

)

Γ

(

z2+

12

)

.

112

Let x be a positive real number. It suffices to show that f(x) satisfies the Bohr-Mollerup Theorem. By thelogarithmic convexity ofΓ is follows that

log f (x) = (x− 1) log 2− log√π + logΓ

( x2

)

+ logΓ

(

x2+

12

)

is a convex function. Also f satisfies the functional equation of the Gamma-function:

f (x+ 1) =2x

√πΓ

(

x2+

12

)

Γ

( x2+ 1

)

=2x

√πΓ

(

x2+

12

)

x2Γ

( x2

)

=x2x−1

√πΓ

(

x2+

12

)

Γ

( x2

)

= x f(x).

Lastly, also f(1) =Γ( 1

2)√π= 1; thus by Theorem 7.13, f agrees withΓ on the positive real numbers. With the

Identity Theorem then f agrees withΓ on the whole domain G.

Exercise 4. Show thatlogΓ(z) is defined for z inC − (−∞,0] and that

logΓ(z) = − logz− γz−∞∑

n=1

[

log(

1+zn

)

− zn

]

.


Exercise 5. Let f be analytic on the right half plane Re z> 0 and satisfy: f(1) = 1, f (z+ 1) = z f(z), andlimn→∞

f (z+n)nz f (n) = 1 for all z. Show that f= Γ.


Exercise 6. Show that

Γ(z) =∞∑

n=0

(−1)n

n!(z+ n)+

∫ ∞

1e−ttz−1 dt

for z, 0,−1,−2, . . . (not for Re z> 0 alone).

Solution. WriteΓ(z) = Φ(z) + Ψ(z) where

Φ(z) =∫ 1

0e−ttz−1 dt

and

Ψ(z) =∫ ∞

1e−ttz−1 dt. (7.11)

We can write

Φ(z) =

∫ 1

0e−ttz−1 dt =

∫ 1

0

∞∑

n=0

(−t)n

n!tz−1 dt =

∞∑

n=0

(−1)n

n!

∫ 1

0tntz−1 dt

=

∞∑

n=0

(−1)n

n!

[

tn+z

n+ z

]1

0

=

∞∑

n=0

(−1)n

n!(z+ n)

113

Thus,

Γ(z) =∞∑

n=0

(−1)n

n!(z+ n)+

∫ ∞

1e−ttz−1 dt (7.12)

Claim 1: Γ(z) given by (7.12) is the analytic continuation of (7.11), that isΓ(z) given by (7.12) is definedfor all z ∈ C − 0,−1,−2, . . ..Proof of Claim 1: We know from the book thatΨ(z) is analytic for Re(z) > 0.Claim 2:Ψ(z) is analytic for Re(z) ≤ 0. ThusΨ(z) is analytic onC.Proof of Claim 2: Assume Re(z) ≤ 0. Then

|tz−1| = tRe(z)−1 ≤t∈[1,∞),Re(z)≤0

t−1.

But since e−12 ttRe(z)−1 → 0 as t→ ∞, there exists a constant C> 0 such that tRe(z)−1 ≤ Ce

12 when t≥ 1.

Hence, we have|e−ttz−1| ≤ |e−t |· |tz−1| = e−ttRe(z)−1 ≤ e−tCe

12 t= Ce−

12 t

and therefore Ce−12 t is integrable on(1,∞). By Fubini’s Theorem for anyγ ⊂ G = z : Re(z) ≤ 0,

∫

γ

∫ ∞

1e−ttz−1 dt dz=

∫ ∞

1

∫

γ

e−ttz−1 dz dt= 0

which implies∫ ∞

1e−ttz−1 dt ∈ H(G).

In summary,

Ψ(z) =∫ ∞

1e−ttz−1 dt ∈ H(C).

Thus, Claim 2 is proved.It remains to show that

Φ(z) =∞∑

n=0

(−1)n

n!(z+ n)

is analytic onC − 0,−1,−2, . . .. Note thatΦ(z) is uniformly and absolutely convergent as a series in anyclosed domain which contains none of the points0,−1,−2, . . . and thus provides the analytic continuationofΦ(z). Since

∞∑

n=0

(−1)n

n!(z+ n)+

∫ ∞

1e−ttz−1 dt

is analytic and we know (Theorem 7.15 p. 180)

Γ(z) =∫ ∞

0e−ttz−1 dt

for Re(z) > 0, we get

Γ(z) =∞∑

n=0

(−1)n

n!(z+ n)+

∫ ∞

1e−ttz−1 dt

is the analytic continuation of (7.11) for z ∈ C − 0,−1,−2, . . ..

114

Exercise 7. Show that∫ ∞

0sin(t2) dt =

∫ ∞

0cos(t2) =

12

√

12π.

Solution. We have shown thatΓ(

12

)

=√π (see Exercise 2 p. 185). Thus

√π = Γ

(

12

)

=

∫ ∞

0e−tt−

12 dt = 2a

∫ ∞

0e−a2t2 dt

(because of Theorem 7.15 p. 180). The last step involves integrating a complex integral. Hence, we have

∫ ∞

0e−a2t2 dt =

√π

2a. (7.13)

Let a = 1−i√2, thena2

=1−2i+i2

2 = −i and therefore

∫ ∞

0e−(

1−i√2

)2t2

dt =

∫ ∞

0e−(−i)t2 dt =

∫ ∞

0eit2

dt

=

∫ ∞

0cos(t2) + i sin(t2) dt =

∫ ∞

0cos(t2) dt+ i

∫ ∞

0sin(t2) dt

=(7.13)

√π

21−i√2

=

√π√2

1+ i(1− i)(1+ i)

=12

√

12π(1+ i) =

12

√

12π + i

12

√

12π.

Thus,∫ ∞

0cos(t2) dt+ i

∫ ∞

0sin(t2) dt =

12

√

12π + i

12

√

12π


0cos(t2) dt =

12

√

12π

and∫ ∞

0sin(t2) =

12

√

12π.

Exercise 8. Let u > 0 and v> 0 and expressΓ(u)Γ(v) as a double integral over the first quadrant of theplane. By changing to polar coordinates show that

Γ(u)Γ(v) = 2Γ(u+ v)∫ π

2

0(cosθ)2u−1(sinθ)2v−1 dθ.

The function

B(u, v) =Γ(u)Γ(v)Γ(u+ v)

is called the beta function. By changes of variables show that

B(u, v) =∫ 1

0tu−1(1− t)v−1 dt =

∫ ∞

0

tu−1

(1+ t)u+vdt

Can this be generalized to the case when u and v are complex numbers with positive real part?

115

Solution. Let u> 0 and v> 0, then by Theorem 7.15 p. 180 and three changes of variables (indicated by“CoV”) we obtain

Γ(u)Γ(v) =

∫ ∞

0e−ssu−1 ds

∫ ∞

0e−ttv−1 dt

=CoV:s=x2,t=y2

2∫ ∞

0e−x2

x2u−2x dx·2∫ ∞

0e−y2

y2v−2y dy

= 4∫ ∞

0

∫ ∞

0e−x2−y2

x2u−1y2v−1 dx dy

=CoV:x=r cos(θ),y=r sinθ

4∫ π/2

0

∫ ∞

0e−r2

r2u−1(cosθ)2u−1r2v−1(sinθ)2v−1r dr dθ

= 4∫ π/2

0

∫ ∞

0e−r2

r2u+2v−1(cosθ)2u−1(sinθ)2v−1r dr dθ

= 4∫ ∞

0e−r2

r2u+2v−1 dr∫ π/2

0(cosθ)2u−1(sinθ)2v−1 dθ

=CoV:t=r2

412

∫ ∞

0e−ttu+v−1 dt

∫ π/2


= 2Γ(u+ v)∫ π/2


This implies


= 2∫ π/2


Next, we show

B(u, v) =∫ 1

0tu−1(1− t)v−1 dt.

We have


= 2∫ π/2


=CoV:t=cos2 θ

2∫ 1

0

(√t)2u−2 (√

1− t)2v−2

dt

=

∫ 1

0tu−1(1− t)v−1 dt.

Hence,

B(u, v) =∫ 1

0tu−1(1− t)v−1 dt.

Finally, we show

B(u, v) =∫ ∞

0

tu−1

(1+ t)u+vdt.

116

We have

B(u, v) =

∫ 1

0tu−1(1− t)v−1 dt

=CoV:s= t

1−t

∫ ∞

0

( s1+ s

)u−1 (

1− s1+ s

)v−1(1+ s)−2 ds

=

∫ ∞

0su−1 1

(1+ s)u−1

1(1+ s)v−1

(1+ s)−2 ds

=

∫ ∞

0su−1 1

(1+ s)u+v−2(1+ s)−2 ds

=

∫ ∞

0

su−1

(1+ s)u+vds.

Hence,

B(u, v) =∫ ∞

0

tu−1

(1+ t)u+vdt.

Yes, the beta function can be generalized to the case when u and v are complex numbers with positive realpart. We have seen that

B(u, v) =∫ ∞

0

tu−1

(1+ t)u+vdt = 2

∫ π/2


for u > 0 and v> 0. Thus it remains to show that

B(u, v) =∫ ∞

0

tu−1

(1+ t)u+vdt ∈ H(G ×G)

where G×G ⊂ C × C with G = z : Re(z) > 0.Let Re(u) > 0 and Re(v) > 0.

∫ ∞

0

tu−1

(1+ t)u+vdt =

∫ 1

0

tu−1

(1+ t)u+vdt

︸︷︷︸

I

+

∫ ∞

1

tu−1

(1+ t)u+vdt

︸︷︷︸

II

.

I) We have

∫ 1

0|tu−1(1+ t)−u−v| dt ≤

∫ 1

0|tu−1|· |(1+ t)−u−v| dt

≤∫ 1

0tRe(u)−1(1+ t)−Re(u)−Re(v) dt

≤∫ 1

0tRe(u)−1 < ∞

where the last step follows since Re(u) > 0 and the previous one since Re(u) > 0 and Re(v) > 0 andt ∈ (0,1). This implies|tu−1(1+ t)−u−v| is integrable on(0,1). By Fubini’s Theorem for anyγ ⊂ G ×G

∫

γ

∫ 1

0

tu−1

(1+ t)u+vdt dz=

∫ 1

0

∫

γ

tu−1

(1+ t)u+vdz dt= 0

117

which implies∫ 1

0

tu−1

(1+ t)u+vdt ∈ H(G ×G).

II) We have∫ ∞

1|tu−1(1+ t)−u−v| dt ≤ lim

n→∞

∫ n

1|tu−1(1+ t)−u−v| dt

≤ limn→∞

∫ n

1|tu−1|· |(1+ t)−u−v| dt

= limn→∞

∫ n

1tRe(u)−1(1+ t)−Re(u)−Re(v) dt

= limn→∞

∫ n

1tRe(u)−1 1

tRe(u)+Re(v)(

1t + 1

)Re(u)+Re(v)dt

= limn→∞

∫ n

1t−1−Re(u) 1

(1t + 1

)Re(u)+Re(v)dt

≤ limn→∞

∫ n

1

1

t1+Re(v)dt < ∞

where the last step follows since Re(v) > 0 and the previous one since Re(u) + Re(v) > 0, t ≥ 1 andtherefore 1

( 1t +1)Re(u)+Re(v)

≤ 1. This implies|tu−1(1+ t)−u−v| is integrable on(1,∞). By Fubini’s Theorem for

anyγ ⊂ G ×G

∫

γ

∫ ∞

1

tu−1

(1+ t)u+vdt dz=

∫ ∞

1

∫

γ

tu−1

(1+ t)u+vdz dt= 0


1

tu−1

(1+ t)u+vdt ∈ H(G×G).

In summary∫ ∞

0

tu−1

(1+ t)u+vdt ∈ H(G×G).

Exercise 9. Letαn be the volume of the ball of radius one inRn (n ≥ 1). Prove by induction and iteratedintegrals that

αn = 2αn−1

∫ 1

0(1− t2)(n−1)/2 dt

Solution. Define the n-dimensional ball with radius r by

Bn(r) = x ∈ Rn : |x| ≤ r

and define the volume of Bn(r) by Vn(r). Clearly, V1(r) = 2r, V2(r) = πr2, V3(r) = 43πr3 and therefore

V1(1) = 2, V2(1) and V3 =43π.

Claim: For arbitrary n we haveVn(r) = rnVn(1)

118

which implies

Vn(r) = rnVn(1) =(def)

rnαn (7.14)

Assume the claim is true, then we write

Bn(1) = x ∈ Rn : |x| ≤ 1 = (x1, x2, . . . , xn−1, t) ∈ Rn : x21 + x2

2 + . . . + x2n−1 + t2 ≤ 1.

For a fixed t∈ [−1,1] we have the balls

Bn−1

(√1− t2

)

= (x1, x2, . . . , xn−1) ∈ Rn−1 : x21 + x2

2 + . . . + x2n−1 ≤ 1

and by (7.14) we obtain

Vn−1

(√1− t2

)

=

(√1− t2

)n−1Vn−1(1) =

(

1− t2) n−1

2αn−1.

Clearly, the ball Bn(1) is the union of all disjoint balls Bn−1

(√1− t2

)

as t varies over[−1,1]. Hence

αn = Vn(1) =∫ 1

−1

(

1− t2) n−1

2αn−1 dt = 2αn−1

∫ 1

0

(

1− t2) n−1

2 dt

which proves the formula above.Finally, we prove the claim Vn(r) = rnVn(1). We have by using a symmetry argument

Vn(r) =

∫ r

−r

∫√

r2−x21

−√

r2−x21

∫√

r2−x21−x2

2

−√

r2−x21−x2

2

· · ·∫√

r2−∑n−1i=1 x2

i

−√

r2−∑n−1i=1 x2

i

1 dxn dxn−1 · · · dx2 dx1

= 2n∫ r

0

∫√

r2−x21

0

∫√

r2−x21−x2

2

0· · ·

∫√

r2−∑n−1i=1 x2

i

01 dxn dxn−1 · · · dx2 dx1

=CoV:xi=ryi

2n∫ r

0

∫√

1−y21

0

∫√

1−y21−y2

2

0· · ·

∫√

1−∑n−1i=1 y2

i

01· rn

︸︷︷︸

Jacobian

dyn dyn−1 · · ·dy2 dy1

= rn∫ 1

−1

∫√

1−y21

−√

1−y21

∫√

1−y21−y2

2

−√

1−y21−y2

2

· · ·∫√

1−∑n−1i=1 y2

i

−√

1−∑n−1i=1 y2

i

1 dyn dyn−1 · · · dy2 dy1

= rnVn(1).

This implies Vn(r) = rnVn(1). (Note that this is kind of obvious, since we scale the unit ball in Rn by r ineach dimension to obtain the ball with radius r inRn). We could have proved the formula by mathematicalinduction. Using spherical coordinates simplifies the computation if preferred.

Exercise 10.Show that

αn =πn/2

(n/2)Γ(n/2)

whereαn is defined in problem 9. Show that if n= 2k, k≥ 1, thenαn = πk/k!


119

Exercise 11.The Gaussian psi function is defined by

Ψ(z) =Γ′(z)Γ(z)

(a) Show thatΨ is meromorphic inC with simple poles at z= 0,−1, . . . and Res(Ψ;−n) = −1 for n ≥ 0.(b) Show thatΨ(1) = −γ.(c) Show thatΨ(z+ 1)− Ψ(z) = 1

z.(d) Show thatΨ(z) − Ψ(1− z) = −π cotπz.(e) State and prove a characterization ofΨ analogous to the Bohr-Mollerup Theorem.


7.8 The Riemann zeta function

Exercise 1. Let ξ(z) = z(z − l)π−12 zζ(z)Γ( 1

2z) and show thatξ is an entire function which satisfies thefunctional equationξ(z) = ξ(1− z).

Solution. The functionξ is analytic onC − (−2k : k ∈ N0 ∪ 1) as a product of analytic functions. Thefactors z, z− 1 andπ−

12 z are entire. The zeta-function has a simple pole at z= 1, hencelimz→1(z− 1)ζ(z)

exists inC and the functionξ is well-defined at z= 1.The functionΓ

(12z

)

has simple poles at the non-positive even integers. For z= 0, limz→0 zΓ(

12z

)

∈ C. Fornegative even integers z= 2k,−k ∈ N the poles of Gamma coincide with the simple zeros of the zeta-function, hencelimz→2k ζ(z)Γ

(12z

)

exists inC also. Therefore the functionξ is analytic at each point.The functional equation ofξ is related to the functional equation ofζ (8.3, p. 192), evaluated at1− z, i.e.

ζ(1− z) = 2(2π)−zΓ(z)ζ(z) sin

(

12π(1− z)

)

. (7.15)

Furthermore the proof uses variants of problems 1, 2 of the homework set 5, in particular

Γ

(

12− z

2

)

Γ

(

12+

z2

)

= π csc

(

12π(1− z)

)

(7.16)

√πΓ(z) = 2z−1

Γ

( z2

)

Γ

(

12+

z2

)

. (7.17)

120

Then the following chain of equalities holds.

ξ(1− z) = (1− z)(−z)π−12+

z2 ζ(1− z)Γ

(

12− z

2

)

(7.15)= z(z− 1)π−

12+

z2Γ

(

12− z

2

)

21−zπ−zζ(z)Γ(z) sin

(

12π(1− z)

)

= ξ(z)π−12 21−z

Γ(z)Γ(

12 −

z2

)

Γ

(z2

) sin

(

12π(1− z)

)

(7.16)= ξ(z)π−

12 21−z Γ(z)

Γ

(z2

)

π csc(

12π(1− z)

)

Γ

(12 +

z2

) sin

(

12π(1− z)

)

= ξ(z)π12 21−z Γ(z)

Γ

(z2

)

Γ

(12 +

z2

)

(7.17)= ξ(z)π

12 21−z 1

Γ

(12 +

z2

)2z−1

π12

Γ

(

12+

z2

)

= ξ(z).

Exercise 2. Use Theorem 8.17 to prove that∑

p−1n = ∞. Notice that this implies that there are an infinite

number of primes.

Solution. Euler’s Theorem asserts that for real values x> 1, ζ(x) =∏∞

n=1

(1

1−p−xn

)

, where pn are the primenumbers. The left hand side is unbounded as x 1 and so is the right hand side.Now the argument goes as follows. Seeking contradiction suppose that

∑∞n=1

1pn

is finite. All summands

are positive, therefore the sum converges absolutely. By Proposition 5.4, p.165, also∑∞

n=1 log(

1− 1pn

)

converges absolutely. But this implies that

−∞∑

n=1

log

(

1− 1pn

)

= log∞∏

n=1

(

11− pn

)

= logζ(1) < ∞

which gives the desired contradiction because thelog(ζ(z)) has a simple pole at z= 1.

Exercise 3. Prove thatζ2(z) =∑∞

n=1d(n)nz for Re z> 1, where d(n) is the number of divisors of n.


Exercise 4. Prove thatζ(z)ζ(z− 1) =∑∞

n=1σ(n)nz , for Re z> 1, whereσ(n) is the sum of the divisors of n.


Exercise 5. Prove thatζ(z−1)ζ(z) =

∑∞n=1

ϕ(n)nz for Re z> 1, whereϕ(n) is the number of integers less than n and

which are relatively prime to n.


Exercise 6. Prove that 1ζ(z) =

∑∞n=1

µ(n)nz for Re z> 1, whereµ(n) is defined as follows. Let n= pk1

1 pk22 · · · p

kmm

be the factorization of n into a product of primes p1, . . . , pm and suppose that these primes are distinct. Letµ(1) = 1; if k1 = . . . = km = 1 then letµ(n) = (−1)m; otherwise letµ(n) = 0.


121

Exercise 7. Prove thatζ′(z)ζ(z) = −

∑∞n=1

Λ(n)nz for Re z> 1, whereΛ(n) = log p if n = pm for some prime p and

m≥ 1; andΛ(n) = 0 otherwise.


Exercise 8. (a) Letη(z) = ζ′(z)/ζ(z) for Re z> 1 and show thatlimz→z0(z− z0)η(z) is always an integer forRe z0 ≥ 1. Characterize the point z0 (in its relation toζ) in terms of the sign of this integer.(b) Show that forε > 0

Reη(1+ ε + it) = −∞∑

n=1

Λ(n)n−1(1+ε) cos(t logn)

whereΛ(n) is defined in Exercise 7.(c) Show that for allε > 0,

3Reη(1+ ε) + 4Reη(1+ ε + it) + Reη(1+ ε + 2it) ≤ 0.

(d) Show thatζ(z) , 0 if Re z= 1 (or 0).


122

Chapter 8

Runge’s Theorem

8.1 Runge’s Theorem

Exercise 1. Prove Corollary 1.14 if it is only assumed that E− meets each component ofC∞ −G.


Exercise 3. Let G be the open unit disk B(0; 1) and let K= z : 14 ≤ |z| ≤ 3

4. Show that there is a functionf analytic on some open subset G1 containing K which cannot be approximated on K by functions in H(G).

Remarks. The next two problems are concerned with the following question. Given a compact set Kcontained in an open set G1 ⊂ G, can functions in H(G1) be approximated on K by functions in H(G)?Exercise 2 says that for an arbitrary choice of K, G, and G1 this is not true. Exercise 4 below gives criteriafor a fixed K and G such that this can be done for any G1. Exercise 3 is a lemma which is useful in provingExercise 4.


Exercise 3. Let K be a compact subset of the open set G and suppose that any bounded component D ofG − K has D− ∩ ∂G , . Then every component ofC∞ − K contains a component ofC∞ −G.


Exercise 4. Let K be a compact subset of the open set G; then the following are equivalent:(a) If f is analytic in a neighborhood of K andε > 0 then there is a g in H(G) with | f (z) − g(z)| < ε for allz in K;(b) If D is a bounded component of G− K then D− ∩ ∂G , ;(c) If z is any point in G− K then there is a function f in H(G) with

| f (z)| > sup| f (w)| : w in K.


Exercise 5. Can you interpret part (c) of Exercise 4 in terms ofK?


123

Exercise 6. Let K be a compact subset of the region G and defineKG = z ∈ G : | f (z)| ≤ ‖ f ‖K for all f inH(G).(a) Show that ifC∞ −G is connected thenKG = K.(b) Show that d(K,C −G) = d(KG,C −G).(c) Show thatKG ⊂ the convex hull of K≡ the intersection of all convex subsets ofC which contain K.(d) If KG ⊂ G1 ⊂ G and G1 is open then for every g in H(G1) andε > 0 there is a function f in H(G) suchthat | f (z) − g(z)| < ε for all z in KG. (Hint: see Exercise 4.)(e) KG = the union of K and all bounded components of G− K whose closure does not intersect∂G.


8.2 Simple connectedness

Exercise 1. The set G= reit : −∞ < t < 0 and1+ et < r < 1+ 2et is called a cornucopia. Show that Gis simply connected. Let K= G−; is intK connected?

Solution. he result follows once it is established that every closed piecewise smooth curve in G is0-homotopic. This requires the following subclaim: let z∈ G then the argument of z is uniquely defined.Suppose otherwise then there are real r1, r2, t1, t2 such that z= r1eit1 = r2eit2. Suppose WLOG that t1 < t2 <0 then there is a positive integer k such that t1 = 2kπ = t2 and therefore r2 > 1+et2 = 1+et1ekπ > 1+2et1 > r1

contradicting the fact that|z| = r1 = r2.Define

R : (−∞,0)→ G; R(t) = 1+32

et.

Let γ : [0,1] → G be an arbitrary closed piecewise smooth curve in G with radius functionρ(s) andargument functionτ(s) such that

γ(s) = ρ(s)eiτ(s),

and define

Γ : [0,1] × [0,1]→ G,

Γ(s,0) = γ(s),

Γ(s,1) = R(τ(s)),

Γ(0,u) = Γ(1,u) ∀ u ∈ [0,1].

Also define a function

Θ : [0,1] × [0,1]→ G,

Θ(s,0) = R(τ(s)),

Θ(s,1) = R

(

τ

(

12

))

,

Θ(0,u) = Θ(1,u) ∀ u ∈ [0,1].

The functionΘ Γ satisfies Definition IV 6.1 (p.88), the curveγ is homotopic to the point R(

τ(

12

))

and byDefinition IV 6.14 (p.93) the set G is simply connected.

Exercise 2. If K is polynomially convex, show that the components of the interior of K are simply connected.


124

8.3 Mittag-Leffler’s Theorem

Exercise 1. Let G be a region and letan and bm be two sequences of distinct points in G without limitpoints in G such that an an , bm for all n,m. Let Sn(z) be a singular part at an and let pm be a positiveinteger. Show that there is a meromorphic function f on G whose only poles and zeros arean and bmrespectively, the singular part at z= an is Sn(z), and z= bm is a zero of multiplicity pm.

Solution. Let G be a region and letbm be a sequence of distinct points in G with no limit point in G;and letpm be a sequence of integers. By Theorem 5.15 p.170 there is an analytic function g defined on Gwhose only zeros are at the points bm; furthermore, bm is a zero of g of multiplicity pm.Since g∈ H(g) andan ∈ G, g has a Taylor series in a neighborhood B(an; Rn) of each an, that is

gn(z) =∞∑

k=0

αk(z− an)k ∈ B(an; Rn)

whereαk =1k! g

(k)(an). Goal: Try to use this series to create a singular part rn(z) at an such that rn(z)gn(z) =sn(z) or

rn(z)∞∑

k=0

αk(z− an)k=

mn∑

j=1

A jn

(z− an) j⇐⇒

mn∑

j=1

A jn

(z− an) j=

∞∑

k=0

αkrn(z)(z− an)k.

Claim:

rn(z) =mn∑

j=1

Bjk(z− an)− j−k (8.1)

works.Proof of the claim:

∞∑

k=0

αkrn(z)(z− an)k=

∞∑

k=0

αk

mn∑

j=1

Bjk(z− an)− j−k(z− an)k

=

∞∑

k=0

αk

mn∑

j=1

Bjk(z− an)− j

=

mn∑

j=1

(z− an)− j∞∑

k=0

αkBjk =

mn∑

j=1

A jn

(z− an) j

where the last step follows by choosing∞∑

k=0

αkBjk = A jn.

Since G is a region, G is open. Letan be a sequence of distinct points without a limit point in G andsuchthat an , bm for all n,m. Letrn(z) be the sequence of rational functions given by

rn(z) =mn∑

j=1

Bjk

(z− an) j+k

(see (8.1)). By Mittag-Leffler’s Theorem, there is a meromorphic function h on G whose poles are exactlythe pointsan and such that the singular part of h at an is rn(z).Set f= g·h. Then by construction f is the meromorphic function on G whose only poles and zeros areanandbm respectively, the singular part at z= an is Sn(z), and z= bm is a zero of multiplicity pm. (Note thatthe zeros do not cancel the poles since by assumption an , bm ∀ n,m).

125

Exercise 2. Let an be a sequence of points in the plane such that|an| → ∞, and letbn be an arbitrarysequence of complex numbers.(a) Show that if integerskn can be chosen such that

∞∑

k=n

(

ran

)kn bn

an(3.4)

converges absolutely for all r> 0 then

∞∑

k=n

(

ran

)kn bn

z− an(3.5)

converges in M(C) to a function f with poles at each point z= an.(b) Show that iflim sup|bn| < ∞ then (3.4) converges absolutely if kn = n for all n.(c) Show that if there is an integer k such that the series

∞∑

n=1

bn

ak+1n

(3.6)

converges absolutely, then (3.4) converges absolutely if kn = k for all n.(d) Suppose there is an r> 0 such that|an − am| ≥ r for all n , m. Show that

∑ |an|−3 < ∞. In particular,if the sequencebn is bounded then the series (3.6) with k= 2 converges absolutely. (This is somewhatinvolved and the reader may prefer to prove part (f) directlysince this is the only application.)(e) Show that if the series (3.5) converges in M(C) to a meromorphic function f then

f (z) =∞∑

n=1

bn

z− an+

bn

an

1+

(

zan

)

+ . . . +

(

zan

)kn−1

(f) Letω andω′ be two complex numbers such that Im(ω′/ω) , 0. Using the previous parts of this exerciseshow that the series

ζ(z) =1z+

∑′(

1z− w

+1w+

zw2

)

,

where the sum is over all w= 2nω+2n′ω for n, n= 0,±1,±2, . . . but not w= 0, is convergent in M(C) to ameromorphic functionζ with simple poles at the points2nω+2n′ω′. This function is called the Weierstrasszeta function.(g) LetP(z) = −ζ′(z); P is called the Weierstrass pe function. Show that

P(z) =1z2+

∑′(

1(z− w)2

− 1w2

)

where the sum is over the same w as in part (f). Also show that

P(z) = P(z+ 2nω + 2n′ω′)

for all integers n and n′. That is,P is doubly periodic with periods2ω and2ω′.


126

Exercise 3. This exercise shows how to deduce Weierstrass’s Theorem forthe plane (Theorem VII. 5.12)from Mittag-Leffler’s Theorem.(a) Deduce from Exercises 2(a) and 2(b) that for any sequencean in C with lim an = ∞ and an , 0 thereis a sequence of integerskn such that

h(z) =∞∑

n=1

1z− an

+1an+

1an

(

zan

)

+ . . . +1an

(

zan

)kn−1

is a meromorphic function onC with simple poles at a1,a2, . . .. The remainder of the proof consists ofshowing that there is a function f such that h= f ′/ f . This function f will then have the appropriate zeros.(b) Let z be an arbitrary but fixed point inC − a1,a2, . . .. Show that ifγ1 andγ2 are any rectifiable curvesin C − a1,a2, . . . from0 to z and h is the function obtained in part (a), then there is aninteger m such that

∫

γ1

h−∫

γ2

h = 2πim.

(c) Again let h be the meromorphic function from part (a). Prove that for z, a1,a2, . . . andγ any rectifiablecurve inC − a1,a2, . . .,

f (z) = exp

(∫

γ

h

)

defines an analytic function onC − a1,a2, . . . with f ′/ f = h. (That is, the value of f(z) is independent ofthe curveγ and the resulting function f is analytic.(d) Suppose that z∈ a1,a2, . . .; show that z is a removable singularity of the function f defined in part(c). Furthermore, show that f(z) = 0 and that the multiplicity of this zero equals the number of times that zappears in the sequencea1,a2, . . ..(e) Show that

f (z) =∞∏

n=1

(

1− zan

)

exp

zan+

12

(

zan

)2

+ . . . +12

(

zan

)kn

(3.7)

Remark. We could have skipped parts (b), (c), and (d) and gonedirectly from (a) to (e). However thiswould have meant that we must show that (3.7) converges in H(C) and it could hardly be classified as a newproof. The steps outlined in parts (a) through (d) give a proof of Weierstrass’s Theorem without introducinginfinite products.


Exercise 4. This exercise assumes a knowledge of the terminology and results of Exercise VII. 5.11.(a) Define two functions f and g in H(G) to be relatively prime (in symbols,( f ,g) = 1) if the only commondivisors of f and g are non-vanishing functions in H(G). Show that( f ,g) = 1 iffZ( f ) ∩Z(g) = .(b) If ( f ,g) = 1, show that there are functions f1, g1 in H(G) such that f f1+gg1 = 1 (Hint: Show that thereis a meromorphic functionϕ on G such that f1 = ϕg ∈ H(G) and g|(1− f f1).)(c) Let f1, . . . , fn ∈ H(G) and g= g.c.d f1, . . . , fn. Show that there are functionsϕ1, . . . , ϕn in H(G) suchthat g= ϕ1 f1 + . . . + ϕn fn. (Hint: Use (b) and induction.)(d) If Jα is a collection of ideals in H(G), show thatJ = ⋂

αJα is also an ideal. IfS ⊂ H(G) then letJ = ∩S : S is an ideal of H(G) andS ⊂ J. Prove thatJ is the smallest ideal in H(G) that containsSandJ = ϕ1 f1 + . . . + ϕn fn : ϕk ∈ H(G), fk ∈ S for 1 ≤ k ≤ n. J is called the ideal generated byS andis denoted byJ = (S). If S is finite then(S) is called a finitely generated ideal. IfS = f for a singlefunction f then( f ) is called a principal ideal.(e) Show that every finitely generated ideal in H(G) is a principal ideal.

127

(f) An idealJ is called a fixed ideal ifZ(J) , ; otherwise it is called a free ideal. Prove that ifJ = (S)thenZ(J) = Z(S) and that a proper principal ideal is fixed.(g) Let fn(z) = sin(2−nz) for all n ≥ 0 and letJ = ( f1, f2, . . .). Show thatJ is a fixed ideal in H(C) whichis not a principal ideal.(h) LetJ be a fixed ideal and prove that there is an f in H(G) withZ( f ) = Z(J) andJ ⊂ ( f ). Also showthatJ = ( f ) if J is finitely generated.(i) LetM be a maximal ideal that is fixed. Show that there is a point a in Gsuch thatM = ((z− a)).(j) Let an be a sequence of distinct points in G with no limit point in G. LetJ = f ∈ H(G) : f (an) = 0for all but a finite number of the an. Show thatJ is a proper free ideal in H(G).(k) If J is a free ideal show that for any finite subsetS of J , Z(S) , . Use this to show thatJ cancontain no polynomials.l) LetJ be a proper free ideal; thenJ is a maximal ideal iff whenever g∈ H(G) andZ(g) ∩Z( f ) , forall f in J then g∈ J .


Exercise 5. Let G be a region and letan be a sequence of distinct points in G with no limit point in G.For each integer n≥ 1 choose integers kn ≥ 0 and constants A(k)

n , 0 ≤ k ≤ kn. Show that there is an analyticfunction f on G such that f(k)(an) = k!A(k)

n . (Hint: Let g be an analytic function on G with a zero at an ofmultiplicity kn + 1. Let h be a meromorphic function on G with poles at each an of order kn + 1 and withsingular part Sn(z). Choose the Sn so that f= gh has the desired property.)


Exercise 6. Find a meromorphic function with poles of order 2 at1,√

2,√

3, . . . such that the residue ateach pole is 0 andlim(z− √n)2 f (z) = 1 for all n.

Solution. We claim that the function

f (z) =∞∑

k=1

3√

nz2 − 2z3

n3/2(z− √n)2

has the desired properties. f has poles of order2 at z=√

n,n ∈ N (and no other poles) and it is analyticfor every z∈ C that is not one of the poles. Moreover f converges in M(C). To see this let r> 0 and chooseN ∈ N so large that

√n > 2r for all n ≥ N. Then for|z| ≤ r and n∈ N

| 3√

nz2 − 2z3

n3/2(z− √n)2| ≤ r2(3

√n+√

n)

n3/2(√

n− r)2≤ 16r2

n2=: Mn.

This is the majorant independent of z for the Weierstrass’ Criterion.Next fix m∈ N and consider the limit in equationlimz→√n(z− √n)2 f (z) = 1,

limz→√m

(z−√

m)2 f (z) = limz→√m

(z−√

m)2∞∑

n=1

3√

nz2 − 2z3

n3/2(z− √n)2

= limz→√m

m−1∑

n=1

(z− √m)(3√

nz2 − 2z3)

n3/2(z− √n)2+

3m3/2 − 2m3/2

m3/2

+ limz→√m

∞∑

m+1

(z− √n)(3√

nz2 − 2z3)

n3/2(z− √n)2

= 0+ 1+ 0 = 1.

128

Lastly we have to check that the residuals are0 at the poles. By Proposition V 2.4, p.113, for a functiong(z) := (z− √m)2 f (z), it has to be verified that g′(z)|z=√m = 0.First find the derivative of f as

f ′(z) =∞∑

n=1

6√

nz2 − 6nz− 2z3

n3/2(z− √m)3

then compute the residuum at z=√

m

Res(f ,√

m) = g′(z)|z=√m = 2(z−√

m) f (z) + (z−√

m)2 f ′(z)∣∣∣z=√

m

=

∞∑

n=1

2(z− √m)(3√

nz2 − 2z3)

n3/2(z− √n)2

+

∞∑

n=1

(z− √m)2(6√

nz2 − 6nz− 2z3)

n3/2(z− √m)3

z=√

m

=

∞∑

n=1

(

2(z−√

(n))(3√

nz2 − 2z3) + (z−√

m)(6√

nz2) − 6nz− 2z3))

· z− √m

n3/2(z− √n)3

]

z=√

m

If n , m and z=√

m, the last factor equals zero, in the case n= m, (z− √m)2 cancels directly and we areleft with

12√

m− 2z3 − 6mz

n3/2(z− √m)

∣∣∣∣∣∣z=m

=−6z(z− √m)

n3/2

∣∣∣∣∣∣z=m

= 0.

The residuum vanishes for all√

m,m ∈ N and therefore the function considered in this exercise has all therequired properties.

129

Chapter 9

Analytic Continuation and RiemannSurfaces

9.1 Schwarz Reflection Principle

Exercise 1. Letγ be a simple closed rectifiable curve with the property that there is a point a such that forall z onγ the line segment[a, z] intersectsγ only at z; i.e.[a, z] ∩ γ = z. Define a point w to be insideγ if [a,w] ∩ γ = and let G be the collection of all points that are insideγ.(a) Show that G is a region and G− = G∪ γ.(b) Let f : G− → C be a continuous function such that f is analytic on G. Show that

∫

γf = 0.

(c) Show that n(γ; z) = ±1 if z is insideγ and n(γ; z) = 0 if z < G−.Remarks. It is not necessary to assume thatγ has such a point a as above; each part of this exercise remainstrue if γ is only assumed to be a simple closed rectifiable curve. Of course, we must define what is meantby the inside ofγ. This is difficult to obtain. The fact that a simple closed curve divides the plane into twopieces (an inside and an outside) is the content of the JordanCurve Theorem. This is a very deep result oftopology.


Exercise 2. Let G be a region in the plane that does not contain zero and letG∗ be the set of all points zsuch that there is a point w in G where z and w are symmetric withrespect to the circle|ξ| = 1. (See III.3.17.)(a) Show that G∗ = z : (1/z) ∈ G.(b) If f : G→ C is analytic, define f∗ : G∗ → C by f∗(z) = f (1/z). Show that f∗ is analytic.(c) Suppose that G= G∗ and f is an analytic function defined on G such that f(z) is real for z in G with|z| = 1. Show that f= f ∗.(d) Formulate and prove a version of the Schwarz Reflection Principle where the circle|ξ| = 1 replacesR.Do the same thing for an arbitrary circle.


Exercise 3. Let G, G+, G−, G0 be as in the statement of the Schwarz Reflection Principle andlet f :G+ ∪G0 → C∞ be a continuous function such that f is meromorphic on G+. Also suppose that for x in G0f (x) ∈ R. Show that there is a meromorphic function g: G→ C∞ such that g(z) = f (z) for z in G+ ∪G0. Isit possible to allow f to assume the value∞ on G0?

130


9.2 Analytic Continuation Along a Path

Exercise 1. The collectionD0,D1, . . . ,Dn of open disks is called a chain of disks if Dj−1 ∩ D j , for1 ≤ j ≤ n. If ( f j ,D j) : 0 ≤ j ≤ n is a collection of function elements such thatD0,D1, . . . ,Dn is a chainof disks and fj−1(z) = f j(z) for z in Dj−1 ∩ D j , 1 ≤ j ≤ n; then( f j ,D j) : 0 ≤ j ≤ n is called an analyticcontinuation along a chain of disks. We say that( fn,Dn) is obtained by an analytic continuation of( f0,D0)along a chain of disks.(a) Let( f j ,D j) : 0 ≤ j ≤ n be an analytic continuation along a chain of disks and let a and b be the centersof the disks D0 and Dn respectively. Show that there is a pathγ from a to b and an analytic continuation(gt, Bt) alongγ such thatγ ⊂ ⋃n

j=0 D j , [ f0]a = [g0]a and[ fn]b = [g1]b.(b) Conversely, let( ft,Dt) : 0 ≤ t ≤ 1 be an analytic continuation along a pathγ : [0,1] → C and leta = γ(0), b = γ(1). Show that there is an analytic continuation along a chain ofdisks(g j , Bj) : 0 ≤ j ≤ nsuch thatγ ⊂ ⋃n

j=0 Bj , [ f0]a = [g0]a and[ f1]b = [gn]b.


Exercise 2. Let D0 = B(1; 1) and let f0 be the restriction of the principal branch of√

z to D0. Letγ(t) =exp(2πit) andσ(t) = exp(4πit) for 0 ≤ t ≤ 1.(a) Find an analytic continuation( ft,Dt) : 0 ≤ t ≤ 1 of ( f0,D0) alongγ and show that[ f1]1 = [− f0]1.(b) Find an analytic continuation(gt, Bt) : 0 ≤ t ≤ 1 of ( f0,D0) alongσ and show that[g1]1 = [g0]1.


Exercise 3. Let f be an entire function, D0 = B(0; 1), and letγ be a path from 0 to b. Show that if( ft,Dt) : 0 ≤ t ≤ 1 is a continuation of( f ,D0) alongγ then f1(z) = f (z) for all z in D1 (This exercise israther easy; it is actually an exercise in the use of the terminology.)


Exercise 4. Letγ : [0,1] → C be a path and let( ft,Dt) : 0 ≤ t ≤ 1 be an analytic continuation alongγ.Show that( f ′t ,Dt) : 0 ≤ t ≤ 1 is also a continuation alongγ.


Exercise 5. Supposeγ : [0,1] → C is a closed path withγ(0) = γ(1) = a and let( ft,Dt) : 0 ≤ t ≤ 1 bean analytic continuation alongγ such that[ f1]a = [ f ′0]a and f0 , 0. What can be said about( f0,D0)?


Exercise 6. Let D0 = B(1; 1) and let f0 be the restriction to D0 of the principal branch of the logarithm.For an integer n letγ(t) = exp(2πint), 0 ≤ t ≤ 1. Find a continuation( ft,Dt) : 0 ≤ t ≤ 1 alongγ of( f0,D0) and show that[ f1]1 = [ f0 + 2πin].


Exercise 7. Letγ : [0,1] → C be a path and let( ft,Dt) : 0 ≤ t ≤ 1 be an analytic continuation alongγ.Suppose G is a region such that ft(Dt) ⊂ G for all t, and suppose there is an analytic function h: G → Csuch that h( f0(z)) = z for all z in D0. Show that h( ft(z)) = z for all z in Dt and for all t. Hint: Show thatT = t : h( ft(z)) = z for all z in Dt is both open and closed in[0,1].

131


Exercise 8. Let γ : [0,1] → C be a path withγ(0) = 1 andγ(t) , 0 for any t. Suppose that( ft,Dt) : 0 ≤t ≤ 1 is an analytic continuation of f0(z) = logz. Show that each ft is a branch of the logarithm.


9.3 Monodromy Theorem

Exercise 1. Prove that the set T defined in the proof of Lemma 3.2 is closed.


Exercise 2. Let ( f ,D) be a function element and let a∈ D. If γ : [0,1] → C is a path withγ(0) = a andγ(1) = b and( ft,Dt) : 0 ≤ t ≤ 1 is an analytic continuation of( f ,D) alongγ, let R(t) be the radius ofconvergence of the power series expansion of ft at z= γ(t).(a) Show that R(t) is independent of the choice of continuation. That is, if a second continuation(gt, Bt)alongγ is given with[g0]a = [ f ]a and r(t) is the radius of convergence of the power series expansion ofgt

about z= γ(t) then r(t) = R(t) for all t.(b) Suppose that D= B(1; 1), f is the restriction of the principal branch of the logarithm to D, andγ(t) = 1+ at for 0 ≤ t ≤ 1 and a> 0. Find R(t).(c) Let( f ,D) be as in part (b), let0 < a < 1 and letγ(t) = (1− at) exp(2πit) for 0 ≤ t ≤ 1. Find R(t).(d) For each of the functions R(t) obtained in parts (b) and (c), findminR(t) : 0 ≤ t ≤ 1 as a function of aand examine the behavior of this function as a→ ∞ or a→ 0.


Exercise 3. Let Γ : [0,1] × [0,1] → C be a continuous function such thatΓ(0,u) = a, Γ(1,u) = b forall u. Let γu(t) = Γ(t,u) and suppose that( ft,u,Dt,u) : 0 ≤ t ≤ 1 is an analytic continuation alongγu

such that[ f0,u]a = [ f0,v]a for all u and v in [0,1]. Let R(t,u) be the radius of convergence of the powerseries expansion of ft,u about z= Γ(t,u). Show that either R(t,u) ≡ ∞ or R : [0,1] × [0,1] → (0,∞) is acontinuous function.


Exercise 4. Use Exercise 3 to give a second proof of the Monodromy Theorem.


9.4 Topological Spaces and Neighborhood Systems

Exercise 1. Prove the propositions which were stated in this section without proof.


Exercise 2. Let (X,T ) and (Ω,S) be topological spaces and let Y⊂ X. Show that if f : X → Ω is acontinuous function then the restriction of f to Y is a continuous function of(Y,TY) into (Ω,S).


132

Exercise 3. Let X andΩ be sets and letNx : x ∈ X andMω : ω ∈ Ω be neighborhood systems and letT andS be the induced topologies on X andΩ respectively.(a) Show that a function f: X → Ω is continuous iff when x∈ X andω = f (x), for each∆ inMω there isa U inNx such that f(U) ⊂ ∆.(b) Let X= Ω = C and letNz = Mz = B(z; ε) : ε > 0 for each z inC. Interpret part (a) of this exercisefor this particular situation.


Exercise 4. Adopt the notation of Exercise 3. Show that a function f: X → Ω is open iff for each x in Xand U inNx there is a set∆ inMω (whereω = f (x)) such that∆ ⊂ f (U).


Exercise 5. Adopt the notation of Exercise 3. Let Y⊂ X and defineUY = Y ∩ U : U ∈ NY for each y inY. Show thatUY : y ∈ Y is a neighborhood system for Y and the topology it induces on YisTY.


Exercise 6. Adopt the notation of Exercise 3. For each point(x, ω) in X ×Ω let

U(x,ω) = U × ∆ : U ∈ Nx,∆ ∈ Mω

(a) Show thatU(x,ω) : (x, ω) ∈ X×Ω is a neighborhood system on X×Ω and letP be the induced topologyon X×Ω.(b) If U ∈ T and∆ ∈ S, call the set U× ∆ an open rectangle. Prove that a set is inP iff it is the union ofopen rectangles.(c) Define p1 : X × Ω → X and p2 : X × Ω → X by p1(x, ω) = x and p2(x, ω) = ω. Show thatp1 and p2 are open continuous maps. Furthermore if(Z,R) is a topological space show that a functionf : (Z,R)→ (X ×Ω,P) is continuous iff p1 f : Z→ X and p2 f : Z→ Ω are continuous.


9.5 The Sheaf of Germs of Analytic Functions on an Open Set

Exercise 1. Define F: S(C)→ C by F(z, [ f ]z) = f (z) and show that F is continuous.


Exercise 2. LetF be the complete analytic function obtained from the principal branch of the logarithmand let G= C − 0. If D is an open subset of G and f: D → C is a branch of the logarithm show that[ f ]a ∈ F for all a in D. Conversely, if( f ,D) is a function element such that[ f ]a ∈ F for some a in D, showthat f : D→ C is a branch of the logarithm. (Hint: Use Exercise 2.8.)


Exercise 3. Let G = C − 0, let F be the complete analytic function obtained from the principal branchof the logarithm, and let(R, ρ) be the Riemann surface ofF (so that G is the base ofR). Show thatR ishomeomorphic to the graphΓ = (z,ez) : z ∈ G considered as a subset ofC × C. (Use the map h: R → Γdefined by h(z, [ f ]z) = ( f (z), z) and use Exercise 2.) State and prove an analogous result for branches ofz1/n.

133


Exercise 4. Consider the sheafS(C), let B= z : |z− 1| < 1, let l be the principal branch of the logarithmdefined on B, and let l1(z) = l(z) + 2πi for all z in B. (a) Let D= z : |z| < 1 and show that( 1

2 , [l] 12) and

( 12 , [l1] 1

2) belong to the same component ofρ−1(D). (b) Find two disjoint open subsets ofS(C) each of which

contains one of the points( 12 , [l] 1

2) and( 1

2 , [l1] 12).


9.6 Analytic Manifolds

Exercise 1. Show that an analytic manifold is locally compact. That is, prove that if a∈ X and U is anopen neighborhood of a then there is an open neighborhood V ofa such that V− ⊂ U and V− is compact.


Exercise 2.Which of the following are analytic manifolds? What is its analytic structure if it is a manifold?(a) A cone inR3. (b) (x1, x2, x3) ∈ R3 : x2

1 + x22 + x2

3 = 1 or x21 + x2

2 > 1 and x3 = 0.


Exercise 3. The following is a generalization of Proposition 6.3(b). Let (X,Φ) be an analytic manifold, letΩ be a topological space, and suppose there is a continuous function h of X ontoΩ that is locally one-one(that is, if x ∈ X there is an open set U such that x∈ U and h is one-one on U). If(U, ϕ) ∈ Φ and h isone-one on U let∆ = h(U) and letψ : ∆→ C be defined byψ(ω) = ϕ (h/U)−1(ω). LetΨ be the collectionof all such pairs(∆, ψ). Prove that(Ω,Ψ) is an analytic manifold and h is an analytic function from X toΩ.


Exercise 4. Let T = z : |z| = 1 × z : |z| = 1; then T is a torus. (This torus is homeomorphic to theusual hollow doughnut inR3.) If ω andω′ are complex numbers such that Im(ω/ω′) , 0 thenω andω′, considered as elements of the vector spaceC overR, are linearly independent. So each z inC can beuniquely represented as z= tω+ t′ω′; t, t′ in R. Define h: C→ T by h(tω+ tω′) = (e2πit ,e2πit′ ). Show that hinduces an analytic structure on T. (Use Exercise 3.) (b) Ifω,ω′ andζ, ζ′ are two pairs of complex numberssuch that Im(ω/ω′) , 0 and Im(ζ/ζ′) , 0, defineσ(sζ+s′ζ′) = (e2πis,e2πis′ ) andτ(tω+t′ω′) = (e2πit ,e2πit′ ).Let G= tω + t′ω′ : 0 < t < 1,0 < t′ < 1 andΩ = sζ + s′ζ′ : 0 < s < 1,0 < s′ < 1; show that bothσandτ are one-one on G andΩ respectively. (Both G andΩ are the interiors of parallelograms.) IfΦτ andΦσ are the analytic structures induced on T byτ andσ respectively, and if the identity map of(T,Φτ) into(T,Φσ) is analytic then show that the function f: G→ Ω defined by f= σ−1 τ is analytic. (To say thatthe identity map of(T,Φτ) into (T,Φσ) is analytic is to say thatΦτ andΦσ are equivalent structures.) (c)Letω = 1, ω′ = i, ζ = 1, ζ = α where Imα , 0; defineσ, τ, G,Ω and f as in part (b). Show thatΦτ andΦσ are equivalent analytic structures if and only ifα = i. (Hint: Use the Cauchy-Riemann equations.) (d)Can you generalize part (c)? Conjecture a generalization?


Exercise 5. (a) Let f be a meromorphic function defined onC and suppose f has two independent periodsω andω′. That is, f(z) = f (z+nω+n′ω′) for all z inC and all integers n and n′, and Im(ω/ω′) , 0. Usingthe notation of Exercise 4(a) show that there is an analytic function F : T → C∞ such that f= F h. (Foran example of a meromorphic function with two independent periods see Exercise VIII. 4.2(g).)(b) Prove that there is no non-constant entire function withtwo independent periods.

134


Exercise 6. Show that an analytic surface is arcwise connected.


Exercise 7. Suppose that f: C∞ → C∞ is an analytic function.(a) Show that either f≡ ∞ or f −1(∞) is a finite set.(b) If f . ∞, let a1, . . . ,an be the points inCwhere f takes on the value∞. Show that there are polynomialsp0, p1, . . . , pn such that

f (z) = p0(z) +n∑

k=1

pk

(

1z− ak

)

for z inC.(c) If f is one-one, show that either f(z) = az+ b (some a, b inC) or f (z) = a

z−c + b (some a, b, c inC).


Exercise 8. Furnish the details of the discussion of the surface for√

z at the end of this section.


Exercise 9. Let G =

z : − π2 < Re z< π2

and define f: G → C by f(z) = sinz. Give a discussion for fsimilar to the discussion of

√z at the end of this section.


9.7 Covering spaces

Exercise 1. Suppose that(X, ρ) is a covering space ofΩ and (Ω, π) is a covering space of Y; prove that(X, π ρ) is also a covering space of Y.


Exercise 2. Let (X, ρ) and(Y, σ) be covering spaces ofΩ andΛ respectively. Defineρ×σ : X×Y→ Ω×Λby (ρ × σ)(x, y) = (ρ(x), σ(y)) and show that(X × Y, ρ × σ) is a covering space ofΩ × Λ.


Exercise 3. Let (Ω, ψ) be an analytic manifold and let(X, ρ) be a covering space ofΩ. Show that there isan analytic structureΦ on X such thatρ is an analytic function from(X,Φ) to (Ω, ψ).


Exercise 4. Let (X, ρ) be a covering space ofΩ and letω ∈ Ω. Show that each component ofρ−1(ω)consists of a single point andρ−1(ω)) has no limit points in X.


Exercise 5. LetΩ be a pathwise connected space and let(X, ρ) be a covering space ofΩ. If ω1 andω2 arepoints inΩ, show thatρ−1(ω1) andρ−1(ω2) have the same cardinality. (Hint: Letγ be a path inΩ fromω1

toω2; if x1 ∈ ρ−1(ω1) and γ is the lifting ofγ with initial point x1 = γ(0), let f(x1) = γ(1). Show that f is aone-one map ofρ−1(ω1) ontoρ−1(ω2).)

135


Exercise 6. In this exercise all spaces are regions in the plane.(a) Let (G, f ) be a covering ofΩ and suppose that f is analytic; show that ifΩ is simply connected thenf is one-one. (Hint: If f(z1) = f (z2) let γ be a path in G from z1 to z2 and consider a certain analyticcontinuation along f γ; apply the Monodromy Theorem.)(b) Suppose that(G1, f1) and (G2, f2) are coverings of the regionΩ such that both f1 and f2 are analytic.Show that if G1 is simply connected then there is an analytic function f: G1 → G2 such that(G1, f ) is acovering of G2 and f1 = f2 f . That is the diagram is commutative.(c) Let (G1, f1), (G2, f2) andΩ be as in part (b) and, in addition, assume that both G1 and G2 are simplyconnected. Show that there is a one-one analytic function mapping G1 onto G2.


Exercise 7. Let G andΩ be regions in the plane and suppose that f: G → Ω is an analytic functionsuch that(G, f ) is a covering space ofΩ. Show that for every regionΩ1 contained inΩ which is simplyconnected there is an analytic function g: Ω1→ G such that f(g1(ω)) = ω for all ω in Ω1.


Exercise 8. What is a simply connected covering space of the figure eight?


Exercise 9. Give two nonhomeomorphic covering spaces of the figure eightthat are not simply connected.


Exercise 10.Prove that the closed curve in Exercise IV.6.8 is not homotopic to zero in the doubly puncturedplane.


136

Chapter 10

Harmonic Functions

10.1 Basic properties of harmonic functions

Exercise 1. Show that if u is harmonic then so are ux =∂u∂x and uy =

∂u∂y .

Solution. Let u be harmonic, that is

uxx + uyy = 0. (10.1)

Since u: G→ R is harmonic, u is infinitely times differentiable by Proposition 1.3 p. 252. Thus, the thirdpartial derivatives of u exist and are continuous (therefore uxyy = uyyx and uyxx = uxxy). Now, we show thatux is harmonic, that is

(ux)xx + (ux)yy = 0.

(ux)xx + (ux)yy = uxxx+ uxyy = uxxx+ uyyx = (uxx + uyy)x =(10.1)

0x = 0.

Thus, ux is harmonic. Finally, we show that uy is harmonic, that is

(uy)xx + (uy)yy = 0.

(uy)xx + (uy)yy = uyxx+ uyyy = uxxy+ uyyy = (uxx + uyy)y =(10.1)

0y = 0.

Thus, uy is harmonic.

Exercise 2. If u is harmonic, show that f= ux − iuy is analytic.

Solution. Since, u is harmonic, we have

uxx + uyy = 0. (10.2)

In addition, u is infinitely times differentiable by Proposition 1.3 p. 252. Thus, all the partialsare continuous(therefore uxy = uyx).To show: f= ux − iuy is analytic which is equivalent to showinga) Re( f ) = ux and Im( f ) = −uy are harmonic andb) ux and−uy have to satisfy the Cauchy Riemann equations (by Theorem III2.29).

137

a) By the previous exercise, we have seen that ux and uy are harmonic. Therefore,−uy is harmonic, too.Thus, part a) is proved.b) We have to show that the Cauchy-Riemann equations are satisfied, that is

(ux)x = −(uy)y and (ux)y = −(−uy)x.

The first is true by (10.2) since(ux)x = −(uy)y ⇐⇒ uxx + uyy = 0.

The second is true since uxy = uyx. Therefore

(ux)y = −(−uy)x ⇐⇒ uxy = uyx ⇐⇒ uxy = uyx.

In summary, f= ux − iuy is analytic provided u is harmonic.

Exercise 3. Let p(x, y) =∑n

k,l=0 aklxkyl for all x, y inR.Show that p is harmonic iff:(a) k(k− 1)ak,l−2 + l(l − 1)ak−2,l = 0 for 2 ≤ k, l ≤ n;(b) an−1,l = an,l = 0 for 2 ≤ l ≤ n;(c) ak,n−1 = ak,n = 0 for 2 ≤ k ≤ n.


Exercise 4. Prove that a harmonic function is an open map. (Hint: Use the fact that the connected subsetsofR are intervals.)


Exercise 5. If f is analytic on G and f(z) , 0 for any z show that u= log | f | is harmonic on G.


Exercise 6. Let u be harmonic in G and supposeB(a; R) ⊂ G. Show that

u(a) =1πR2

"

B(a;R)u(x, y) dx dy.

Solution. Let D be a disk such thatB(a; R) ⊂ D ⊂ G. There exist an f∈ H(D) such that Re( f ) = u on D.From Cauchy’s Integral Formula, we have

f (a) =1

2πi

∫

γ

f (w)w− a

dw=12π

∫ 2π

0f (a+ reiθ) dθ

whereγ(t) = a+ reit , 0 ≤ t ≤ 2π. So

f (a) =12π

∫ 2π

0f (a+ reiθ) dθ.

Multiply by r, we get

f (a)r =12π

∫ 2π

0f (a+ reiθ)r dθ

138

and then integrate with respect to r yields

∫ R

0f (a)r dr =

∫ R

0

12π

∫ 2π

0f (a+ reiθ)r dθ dr

⇐⇒ f (a)r2

2=

12π

∫ R

0

∫ 2π

0f (a+ reiθ)r dθ dr

⇐⇒ f (a) =1πR2

∫ 2π

0

∫ R

0f (a+ reiθ)r dr dθ

⇐⇒ f (a) =1πR2

"

B(a;R)f (x, y) dx dy.

where the last step follows by changing from polar coordinates to rectangular coordinates. Now

u(a) = Re( f (a)) =1πR2

"

B(a;R)Re( f (x, y)) dx dy=

1πR2

"


Thus

u(a) =1πR2

"


Exercise 7. For |z| < 1 let

u(z) = Im

(

1+ z1− z

)2 .

Show that u is harmonic andlimr→1− u(reiθ) = 0 for all θ. Does this violate Theorem 1.7? Why?

Solution. Let D = z ∈ C : |z| < 1. Let f(z) =(

1+z1−z

)2. Clearly, f ∈ H(D) since z= 1 is not in D. By

Theorem III 2.29 we obtain that v= Re( f ) and u= Im( f ) are harmonic. Therefore

u(z) = Im

(

1+ z1− z

)2

is harmonic. Next, we will showlimr→1−

u(reiθ) = 0

139

for all θ. We have

u(reiθ) = Im

(

1+ reiθ

1− reiθ

)2

= Im

(

1+ reiθ

1− reiθ

)2 (

1− reiθ

1− reiθ

)2

= Im

(

(1+ reiθ)(1− reiθ)(1− reiθ)(1− reiθ)

)2

= Im

[

(1+ reiθ − re−iθ − r2)2

(1− reiθ − re−iθ + r2)2

]

= Im

[

(1+ 2ri sinθ − r2)2

(1− 2r cosθ + r2)2

]

= Im

[

((1− r2) + 2ri sinθ)2

(1− 2r cosθ + r2)2

]

= Im

[

(1− r2)2+ 4ri sinθ(1− r2) − 4r2 sin2 θ

(1− 2r cosθ + r2)2

]

=4r(1− r2) sinθ

(1− 2r cosθ + r2)2.

Note that the denominator is zero iff r = 1 andθ = 2πk, k∈ Z ⇐⇒ z= 1.If we assume, that r= 1 andθ , 2πk, k∈ Z, then

limr→1−

u(reiθ) = u(eiθ) =4·1· (1− 12) sinθ

(1− 2·1· cosθ + 12)2=

022(1− cosθ)2

= 0.

If θ = 2πk, k∈ Z, then

limr→1−

u(re2πk) = limr→1−

4r(1− r2) sin(2πk)(1− 2r cos(2πk) + r2)2

= limr→1−

0(1− r)4

= 0.

Thus,limr→1−

u(reiθ) = 0.

This does not violate Theorem 1.7. The reason is the following: If we pick u= u and v= 0, then thecondition

lim supz→a

u(z) ≤ 0

does not hold∀a ∈ ∂D.Claim: lim supz→1 u(z) > 0.Proof of the claim:

u(z) = u(x, y) = Im

(

1+ x+ iy1− x− iy

)2 =

4y(1− x2 − y2)(1− 2x+ x2 + y2)2

.

If we fix x= 1, then

u(1, y) =4y(−y2)

y4=−4y3

y4= −4

y∀y , 0.

But− 4y → ∞ as y→ 0−.

140

Exercise 8. Let u : G → R be a function with continuous second partial derivatives and define U(r, θ) =u(r cosθ, r sinθ).(a) Show that

r2

[

∂2u∂x2+∂2u∂y2

]

= r2∂2U∂r2+ r

∂U∂r+∂2U∂θ2= r

∂

∂r

(

r∂U∂r

)

+∂2U∂θ2

.

So if0 < G then u is harmonic iff

r∂

∂r

(

r∂U∂r

)

+∂2U∂θ2= 0.

(b) Let u have the property that it depends only on|z| and notargz. That is, u(z) = ϕ(|z|). Show that u isharmonic iff u(z) = a log |z| + b for some constants a and b.


Exercise 9. Let u : G→ R be harmonic and let A= z ∈ G : ux(z) = uy(z) = 0; that is, A is the set of zerosof the gradient of u. Can A have a limit point in G?

Solution. Let u be harmonic and not constant. By Exercise 1 p. 255, we know that ux and uy are harmonic.By exercise 2 p. 255, we know that f= ux − iuy is analytic. Define B= z ∈ G : f (z) = 0 = z ∈ G :ux(z) − iuy(z) = 0. Clearly

A = z ∈ G : ux(z) = uy(z) = 0 = B

since ux(z) − iuy(z) = 0 iff ux(z) = uy(z) = 0 (ux(z) and uy(z) are real). Since f is analytic, the zeros areisolated and therefore B cannot have a limit point in G (Corollary 3.10 p. 79). This implies that A cannothave a limit point in G.If u is constant, then A= B = G and thus A can have a limit point in G. (In this case f≡ 0 and by Theorem3.7 p. 78 A has a limit point in G).

Exercise 10.State and prove a Schwarz Reflection Principle for harmonic functions.


Exercise 11.Deduce the Maximum Principle for analytic functions from Theorem 1.6.


10.2 Harmonic functions on a disk

Exercise 1. Let D= z : |z| < 1 and suppose that f: D− → C is a continuous function such that both Re fand Im f are harmonic. Show that

f (reiθ) =12π

∫ π

−πf (eit )Pr (θ − t) dt

for all reiθ in D. Using Definition 2.1 show that f is analytic on D iff∫ π

−πf (eit )eint dt = 0

for all n ≥ 1.

141

Solution. Let D= z : |z| < 1 and suppose f= u+ iv : D → C is a continuous function such that u and vare harmonic. Clearly u and v are continuous functions mapping D toR. By Corollary 2.9 we have

u(reiθ) =12π

∫ π

−πPr (θ − t)u(eit ) dt for 0 ≤ r < 1 and all θ (10.3)

and

v(reiθ) =12π

∫ π

−πPr (θ − t)v(eit ) dt for 0 ≤ r < 1 and all θ. (10.4)

Hence,

f (reiθ) = u(reiθ) + iv(reiθ) =(10.3),(10.4)

12π

∫ π

−πPr (θ − t)u(eit ) dt+ i

12π

∫ π

−πPr (θ − t)v(eit ) dt

=12π

∫ π

−πPr (θ − t)

[

u(eit ) + iv(r iθ)]

dt =12π

∫ π

−πPr (θ − t) f (eit ) dt

for 0 ≤ r < 1 and all θ. Thus, we have proved

f (reiθ) =12π

∫ π


for all reiθ in D.Next, we will prove

f is analytic ⇐⇒∫ π

−πf (eiθ)eint dt = 0

for all n ≥ 1. ⇒: Let f be analytic on D and assume n≥ 1 (n integer). Define g(z) = f (z)zn−1. Clearly g isanalytic on D. Letγ = reit , −π ≤ t ≤ π, 0 < r < 1, then by Cauchy’s Theorem

∫

γ

g(z) dz= 0

⇐⇒∫

γ

f (z)zn−1 dz= 0

⇐⇒∫ π

−πf (reit )ireiteit(n−1)rn−1 dt = 0

⇐⇒ i∫ π

−πf (reit )eintrn dt = 0

⇐⇒∫ π

−πf (reit )eintrn dt = 0

which implies

limr→1−

∫ π

−πf (reit )eintrn dt = 0 ⇒

by Ex 3a p. 262

∫ π

−πf (eit )eint dt = 0 ∀n ≥ 1.

142

⇐: Assume∫ π

−π f (reit )eintrn dt = 0 for all n ≥ 1. We have for all z∈ D

f (z) = f (reiθ) =Part 1

12π

∫ π


=12π

∫ π

−πf (eit )

∞∑

n=−∞r |n|ein(θ−t) dt

=12π

∞∑

n=−∞r |n|einθ

∫ π

−πf (eit )e−int dt

=12π

−1∑

n=−∞r |n|einθ

∫ π


︸︷︷︸

=0 by assumption

+12π

∞∑

n=0

rneinθ∫ π


=12π

∞∑

n=0

zn∫ π

−πf (eit )e−int dt.

Now, for any closed rectifiable curveγ in D we get

∫

γ

f (z) dz=∫

γ

12π

∞∑

n=0

zn∫ π

−πf (eit )e−int dt dz=

12π

∞∑

n=0

∫

γ

zn dz

︸︷︷︸

=0

∫ π


since zn is analytic for n≥ 0.Thus f is analytic on D.

Exercise 2. In the statement of Theorem 2.4 suppose that f is piecewise continuous on∂D. Is the conclusionof the theorem still valid? If not, what parts of the conclusion remain true?


Exercise 3. Let D= z : |z| < 1, T = ∂D = z : |z| = 1(a) Show that if g: D− → C is a continuous function and gr : T → C is defined by gr (z) = g(rz) thengr (z)→ g(z) uniformly for z in T as r→ 1−.(b) If f : T → C is a continuous function definef : D− → C by f (z) = f (z) for z in T and

f (reiθ) =12π

∫ π


(So Ref and Im f are harmonic in D). Definefr : T → C by fr (z) = f (rz). Show that for each r< 1 thereis a sequencepn, (z, z) of polynomials in z andz such that pn(z, z)→ fr (z) uniformly for z in T. (Hint: UseDefinition 2.1.)(c) Weierstrass approximation theorem for T. If f : T → C is a continuous function then there is asequencepn(z, z) of polynomials in z andz such that pn(z, z)→ f (z) uniformly for z in T.(d) Suppose g: [0,1]→ C is a continuous function such that g(0) = g(1). Use part (c) to show that there isa sequencepn of polynomials such that pn(t)→ g(t) uniformly for t in[0,1].(e) Weierstrass approximation theorem for [0,1]. If g : [0,1] → C is a continuous function then there is asequencepn of polynomials such that pn(t) → g(t) uniformly for t in [0,1]. (Hint: Apply part (d) to thefunction g(t) + (1− t)g(1)+ tg(0).)(f) Show that if the function g in part (e) is real valued then the polynomials can be chosen with realcoefficients.

143


Exercise 4. Let G be a simply connected region and letΓ be its closure inC∞; ∂∞G = Γ − G. Supposethere is a homeomorphismϕ of Γ onto D−(D = z : |z| < 1) such thatϕ is analytic on G.(a) Show thatϕ(G) = D andϕ(∂∞G) = ∂D.(b) Show that if f: ∂∞G→ R is a continuous function then there is a continuous functionu : Γ → R suchthat u(z) = f (z) for z in∂∞G and u is harmonic in G.(c) Suppose that the function f in part (b) is not assumed to becontinuous at∞. Show that there is acontinuous function u: G− → R such that u(z) = f (z) for z in∂G and u is harmonic in G (see Exercise 2).


Exercise 5. Let G be an open set, a∈ G, and G0 = G − a. Suppose that u is a harmonic function on G0

such thatlimz→a u(z) exists and is equal to A. Show that if U: G → R is defined by U(z) = u(z) for z , aand U(a) = A then U is harmonic on G.


Exercise 6. Let f : z : Re z= 0 → R be a bounded continuous function and define u: z : Re z> 0 → Rby

u(x+ iy) =1π

∫ ∞

−∞

x f(it)x2 + (y− t)2

dt.

Show that u is a bounded harmonic function on the right half plane such that for c inR, f (ic) = limz→ic u(z).


Exercise 7. Let D= z : |z| < l and suppose f: ∂D → R is continuous except for a jump discontinuity atz= 1. Define u: D → R by (2.5). Show that u is harmonic. Let v be a harmonic conjugate of u. What canyou say about the behavior of v(r) as r→ 1−? What about v(reiθ) as r→ 1− andθ → 0?


10.3 Subharmonic and superharmonic functions

Exercise 1. Which of the following functions are subharmonic? superharmonic? harmonic? neithersubharmonic nor superharmonic? (a)ϕ(x, y) = x2

+ y2; (b) ϕ(x, y) = x2 − y2; (c) ϕ(x, y) = x2+ y; (d)

ϕ(x, y) = x2 − y; (e)ϕ(x, y) = x+ y2; (f) ϕ(x, y) = x− y2.

Solution. Note that∫ π

−π sinθ dθ = 0,∫ π

−π cosθ dθ = 0,∫ π

−π sin2 θ dθ = π and∫ π

−π cos2 θ dθ = π. For alla = (α, β) ∈ C and any r> 0 we havea)

ϕ(a+ reiθ) = ϕ(α + r cosθ, β + r sinθ) = (α + r cosθ)2+ (β + r sinθ)2

= α2+ 2αr cosθ + r2 cos2 θ + β2

+ 2βr sinθ + r2 sin2 θ

= α2+ β2+ 2αr cosθ + 2βr sinθ + r2.

Thus

12π

∫ π

−πϕ(a+ reiθ) dθ = α2

+ β2+αrπ

∫ π

−πcosθ dθ

︸︷︷︸

=0

+βrπ

∫ π

−πsinθ dθ

︸︷︷︸

=0

+r2

= α2+ β2+ r2 ≥ α2

+ β2= ϕ(a).

144

Hence,ϕ ∈ Subhar(G).b)

ϕ(a+ reiθ) = ϕ(α + r cosθ, β + r sinθ) = (α + r cosθ)2 − (β + r sinθ)2

= α2+ 2αr cosθ + r2 cos2 θ − β2 − 2βr sinθ − r2 sin2 θ.

Thus

12π

∫ π

−πϕ(a+ reiθ) dθ = α2 − β2

+αrπ

∫ π

−πcosθ dθ

︸︷︷︸

=0

+r2

2π

∫ π

−πcos2 θ dθ

︸︷︷︸

=π

−βrπ

∫ π

−πsinθ dθ

︸︷︷︸

=0

− r2

2π

∫ π

−πsin2 θ dθ

︸︷︷︸

=π

= α2 − β2+

r2

2− r2

2= α2 − β2

= ϕ(a).

Hence,ϕ ∈ Har(G), therefore alsoϕ ∈ Subhar(G) andϕ ∈ Superhar(G).c)

ϕ(a+ reiθ) = ϕ(α + r cosθ, β + r sinθ) = (α + r cosθ)2+ (β + r sinθ)

= α2+ 2αr cosθ + r2 cos2 θ + β + r sinθ.

Thus

12π

∫ π

−πϕ(a+ reiθ) dθ = α2

+ β +αrπ

∫ π

−πcosθ dθ

︸︷︷︸

=0

+r2

2π

∫ π

−πcos2 θ dθ

︸︷︷︸

=π

+r

2π

∫ π

−πsinθ dθ

︸︷︷︸

=0

= α2+ β +

r2

2≥ α2

+ β = ϕ(a).

Hence,ϕ ∈ Subhar(G).d)

ϕ(a+ reiθ) = ϕ(α + r cosθ, β + r sinθ) = (α + r cosθ)2 − (β + r sinθ)

= α2+ 2αr cosθ + r2 cos2 θ − β − r sinθ.

Thus

12π

∫ π

−πϕ(a+ reiθ) dθ = α2 − β + αr

π

∫ π

−πcosθ dθ

︸︷︷︸

=0

+r2

2π

∫ π

−πcos2 θ dθ

︸︷︷︸

=π

− r2π

∫ π

−πsinθ dθ

︸︷︷︸

=0

= α2 − β + r2

2≥ α2 − β = ϕ(a).

Hence,ϕ ∈ Subhar(G).e)

ϕ(a+ reiθ) = ϕ(α + r cosθ, β + r sinθ) = (α + r cosθ) + (β + r sinθ)2

= α + r cosθ + β2+ 2βr sinθ + r2 sin2 θ.

145

Thus

12π

∫ π

−πϕ(a+ reiθ) dθ = α + β2

+r

2π

∫ π

−πcosθ dθ

︸︷︷︸

=0

+βrπ

∫ π

−πsinθ dθ

︸︷︷︸

=0

+r2

2π

∫ π

−πsin2 θ dθ

︸︷︷︸

=π

= α + β2+

r2

2≥ α + β2

= ϕ(a).

Hence,ϕ ∈ Subhar(G).f)

ϕ(a+ reiθ) = ϕ(α + r cosθ, β + r sinθ) = (α + r cosθ) − (β + r sinθ)2

= α + r cosθ − β2 − 2βr sinθ − r2 sin2 θ.

Thus

12π

∫ π

−πϕ(a+ reiθ) dθ = α − β2

+r

2π

∫ π

−πcosθ dθ

︸︷︷︸

=0

−βrπ

∫ π

−πsinθ dθ

︸︷︷︸

=0

− r2

2π

∫ π

−πsin2 θ dθ

︸︷︷︸

=π

= α − β2 − r2

2≤ α − β2

= ϕ(a).

Hence,ϕ ∈ Superhar(G).

Exercise 2. Let Subhar(G) and Superhar(G) denote, respectively, the sets of subharmonic and superhar-monic functions on G.(a) Show that Subhar(G) and Superhar(G) are closed subsets of C(G;R).(b) Does a version of Harnack’s Theorem hold for subharmonicand superharmonic functions?


Exercise 3. (This exercise is difficult.) If G is a region and if f: ∂∞G → R is a continuous function letuf be the Perron Function associated with f . This defines a map T: (∂∞G;R) → Har(G) by T( f ) = uf .Prove:(a) T is linear (i.e., T(a1 f1 + a2 f2) = a1T( f1) + a2T( f2)).(b) T is positive (i.e., if f(a) ≥ 0 for all a in ∂∞G then T( f )(z) ≥ 0 for all z in G).(c) T is continuous. Moreover, if fn is a sequence in C(∂∞G;R) such that fn→ f uniformly then T( fn)→T( f ) uniformly on G.(d) If the Dirichlet Problem can be solved for G then T is one-one. Is the converse true?


Exercise 4. In the hypothesis of Theorem 3.11, suppose only that f is a bounded function on∂∞G; provethat the conclusion remains valid. (This is useful if G is an unbounded region and g is a bounded continuousfunction on∂G. If we define f: ∂∞G → R by f(z) = g(z) for z in ∂G and f(∞) = 0 then the conclusionof Theorem 3.11 remains valid. Of course there is no reason toexpect that the harmonic function will havepredictable behavior near∞ — we could have assigned any value to f(∞). However, the behavior nearpoints of∂G can be studied with hope of success.)


Exercise 5. Show that the requirement that G1 is bounded in Corollary 3.5 is necessary.

146


Exercise 6. If f : G→ Ω is analytic andϕ : Ω → R is subharmonic, show thatϕ f is subharmonic if fis one-one. What happens if f′(z) , 0 for all z in G?

Solution. Clearly f is continuous, since f: G → Ω is analytic andϕ is continuous, sinceϕ : Ω → R issubharmonic. Define u= ϕ f , then u : G → R is continuous. If we can show that for every boundedregion G1 such thatG1 ⊂ G and for every continuous function u1 : G1 → R that is harmonic in G1 andsatisfies u(z) ≤ u1(z) for z in∂G1, we have

u(z) ≤ u1

for z in G1, then u is subharmonic (by Corollary 3.5 p. 265).Hence, let G1 be a bounded region G1 such thatG1 ⊂ G and let u1 : G1 → R that is harmonic in G1 andsatisfies u(z) ≤ u1(z) for z on∂G1. We have to show u(z) ≤ u1(z) ∀z ∈ G1.Since f is one-one, we know that f−1 exists and is analytic. So define

ϕ1 = u1 f −1 : Ω→ R

with f−1 : Ω1→ G1. Clearlyϕ1 is harmonic (harmonic composite analytic is harmonic). Sinceϕ : Ω→ Ris subharmonic, we can use Theorem 3.4 to obtain: For every regionΩ1 contained inΩ and every harmonicfunctionϕ onΩ1, ϕ−ϕ satisfies the Maximum Principle onΩ1. In factϕ−ϕ1 satisfies the Maximum PrincipleonΩ1. By assumption we have

u(z) ≤ u1(z) ∀z ∈ G1

which is the same asu(z) = ϕ( f (z)) ≤ u1(z) = ϕ1( f (z)) ∀z ∈ ∂G1

since f maps G1 ontoΩ1, and thus∂G1 onto∂Ω1 (we choose this by assumption). Thus,

ϕ( f (z)) − ϕ1( f (z)) ≤ 0 ∀z ∈ ∂G1.

Sinceϕ − ϕ1 satisfies the Maximum Principle onΩ1, this implies

ϕ( f (z)) − ϕ1( f (z)) ≤ 0 ∀z ∈ G1,

that isu(z) − u1(z) ≤ 0 ∀z ∈ G1

and we are done. Therefore, u= ϕ f is subharmonic on G.

If f ′(z) , 0 for all z in G, then the previous result holds locally for someneighborhood in G. ( f is lo-cally one-one and onto and f−1 is analytic and hence u= ϕ f is locally subharmonic). Here is the claim:Let f : G → C be analytic, let z0 ∈ A, and let f′(z0) , 0. Then there is a neighborhood U of z0 and aneighborhood V of w0 = f (z0) such that f: U → V is one-one and onto and f−1 : V → U is analytic.Proof of the claim: f(z) − w0 has a simple zero at z0 since f′(z0) , 0. We can use Theorem 7.4 p. 98 tofind ε > 0 andδ > 0 such that each w with|w− w0| < δ has exactly one pre-image z with|z− z0| < ε. LetV = w| |w − w0| < δ and let U be the inverse image of V under the map f restricted toz| |z− z0| < ε.By Theorem 7.4 p. 98, f maps U one-one onto V. Since f is continuous, U is a neighborhood of z0. By theOpen Mapping Theorem (Theorem 7.5 p. 99), f= ( f −1)−1 is an open map, so f−1 is continuous from V toU. To finally show that it is analytic, we can use Proposition 3.7 p. 125. We have

f −1(w) =1

2πi

∫

|z−z0|=ε

z f′(w)f (z) − w

dz.

Note, if we assume thatϕ is twice differentiable, then the result holds globally.

147

10.4 The Dirichlet Problem

Exercise 1. Let G= B(0; 1) and find a barrier for G at each point of the boundary.


Exercise 2. Let G= C − (∞,0] and construct a barrier for each point of∂∞G.


Exercise 3. Let G be a region and a a point in∂∞G such that there is a harmonic function u: G→ R withlimz→a u(z) = 0 and lim inf z→w u(z) > 0 for all w in ∂∞G, w, a. Show that there is a barrier for G at a.


Exercise 4. This exercise asks for an easier proof of a special case of Theorem 4.9. Let G be a boundedregion and let a∈ ∂G such that there is a point b with[a,b] ∩G− = a. Show that G has a barrier at a.(Hint: Consider the transformation(z− a)(z− b)1.)


10.5 Green’s Function

Exercise 1. (a) Let G be a simply connected region, let a∈ G, and let f : G → D = z : |z| < 1 be aone-one analytic function such that f(G) = D and f(a) = 0. Show that the Green’s Function on G withsingularity at a is ga(z) = − log | f (z)|.(b) Find the Green’s Functions for each of the following regions: (i) G= C− (∞,0]; (ii) G = z : Re z> 0;(iii) G = z : 0 < Im z< 2π.

Solution. Solution to part a):Let G be simply connected, let a∈ G, and let f : G→ D = z : |z| < 1 be a one-one analytic function suchthat f(G) = D and f(a) = 0.To show: ga(z) = − log | f (z)| is a Green’s Function on G with singularity at a. Clearly, ga(z) has a singu-larity if | f (z)| = 0. This happens if f(z) = 0. By assumption f(z) = 0 iff z = a. (Note that there is no otherpoint z∈ G such that f(z) = 0 since f is assumed to be one-one). This implies ga(z) has a singularity at a.It remains to verify a), b) and c) of Definition 5.1.a) ga is harmonic in G− a. This follows directly by Exercise 5 p. 255 in G− a.b) G(z) = ga(z)+log |z−a| is harmonic in a disk about a. Let the disk be B(a; r). Clearly ga(z) = − log | f (z)| isharmonic on B(a; r)−a. By Exercise 5 p. 255, we can argue again thatlog |z−a| is harmonic on B(a; r)−a(choose f(z) = z− a and f(z) = 0 iff z = a). Hence G(z) = ga(z) + log |z− a| = − log | f (z)| + log |z− a| isharmonic on B(a; r) − a as a sum of two harmonic functions. But

G(z) = − log | f (z)| + log |z− a| = − log

( | f (z)||z− a|

)

= − log

( |z− a|· | f (z)||z− a|

)

= − log | f (z)|

where f(z) = (z− a) f (z) by assumption andf (z) is analytic on G with no zero. So G(z) has a removablesingularity at a and therefore G(z) is harmonic on B(a; r).c) limz→w ga(z) = 0 for each w in∂∞G. We have| f (z)| = 1∀z ∈ ∂∞G by assumption ( f(∂∞G) = ∂D). Hence

limz→w

ga(z) = limw→a− log | f (w)| = − log 1= 0

148

for each a in∂∞G. This implies ga(z) = − log | f (z)| is a Green’s Function on G with singularity at a.Solution to part b) ii):Clearly, G is a simply connected region (not the whole plane). If we can find a map, say h: G → D, thenthe Green’s Function, say ga, is given by

ga(z) = − log |h(z)|

by Part 1 a). Note that h has to satisfy the following assumptions to use Part 1 a):a) h : G→ D one-one and analyticb) h(a) = 0c) h(G) = D.The existence and uniqueness of this function h with the assumptions a), b) and c) is guaranteed by theRiemann Mapping Theorem p. 160 Theorem 4.2. It remains to findh:Claim: h : G→ D given by

h(z) = f (g(z))

where

g(z) =z− 1z+ 1

and f(z) =z− g(a)

1− g(a)z

works. Thus, the Green’s Function for the region G= z : Re(z) > 0 is given by ga(z) = − log |h(z)| whereh(z) is given by the Claim.Proof of the Claim:We will invoke the Orientation Principle (p. 53) to find g: G→ D where g is one-one and analytic. Then

g(z) =z− 1z+ 1

(see p. 53 in the book for the derivation).From p. 162 we have seen that the Möbius Transformation (which is one-one and analytic)

f (z) =z− g(a)

1− g(a)z

maps D onto D such that f(g(a)) = 0. Hence h(z) = f (g(z)) maps G onto D. In addition h is one-one,analytic and h(a) = 0 and therefore satisfies a), b) and c).

Exercise 2. Let ga be the Green’s Function on a region G with singularity at z= a. Prove that ifψ is apositive superharmonic function on G− a with lim inf z→a[ψ(z) + log |z− a|] > −∞, then ga(z) ≤ ψ(z) forz, a.


Exercise 3. This exercise gives a proof of the Riemann Mapping Theorem where it is assumed that if G isa simply connected region, G, C, then: (i)C∞ −G is connected, (ii) every harmonic function on G has aharmonic conjugate, (iii) if a< G then a branch oflog(z− a) can be defined.(a) Let G be a bounded simply connected region and let a∈ G; prove that there is a Green’s Function ga onG with singularity at a. Let u(z) = ga(z) + log |z− a| and let v be the harmonic conjugate of u. Ifϕ = u+ ivlet f(z) = eiα(z− a)e−ϕ(z) for a real numberα. (So f is analytic in G.) Prove that| f (z)| = exp(−ga(z))and thatlimz→w | f (z)| = 1 for each w in∂G (Compare this with Exercise 1). Prove that for0 < r < 1,Cr = z : | f (z)| = r consists of a finite number of simple closed curves in G (see Exercise VI.1.3). Let Grbe a component ofz : | f (z)| < r and apply Rouche’s Theorem to get that f(z) = 0 and f(z) − w0 = 0,

149

|w0| < r, have the same number of solutions in Gr . Prove that f is one-one on Gr . From here conclude thatf (G) = D = z : |z| < 1 and f′(a) > 0, for a suitable choice ofα.(b) Let G be a simply connected region with G, C, but assume that G is unbounded and0,∞ ∈ ∂∞G. Let lbe a branch oflogz on G, a∈ G, andα = l(a). Show that l is one-one on G and l(z) , α + 2πi for any z inG. Prove thatϕ(z) = [l(z) − α − 2πi]−1 is a conformal map of G onto a bounded simply connected regioninthe plane. (Show that l omits all values in a neighborhood ofα + 2πi.)(c) Combine parts (a) and (b) to prove the Riemann Mapping Theorem.


Exercise 4. (a) Let G be a region such that∂G = γ is a simple continuously differentiable closed curve. Iff : ∂G→ R is continuous and g(z,a) = ga(z) is the Green’s Function on G with singularity at a, show that

h(a) =∫

γ

f (z)∂g∂n

(z,a) ds (5.5)

is a formula for the solution of the Dirichlet Problem with boundary values f ; where∂g∂n is the derivative of

g in the direction of the outward normal toγ and ds indicates that the integral is with respect to arc length.(Note: these concepts are not discussed in this book but the formula is sufficiently interesting so as to meritpresentation.) (Hint: Apply Green’s formula

"

Ω

[u∆v− v∆u] dx dy=∫

∂Ω

[

u∂v∂n− v

∂u∂n

]

ds

withΩ = G − z : |z− a| ≤ δ, δ < d(a, γ), u = h, v= ga(z) = g(z,a).)(b) Show that if G= z : |z| < 1 then (5.5) reduces to equation (2.5).


150

Chapter 11

Entire Functions

11.1 Jensen’s Formula

Exercise 1. In the hypothesis of Jensen’s Formula, do not suppose that f(0) , 0. Show that if f has a zeroat z= 0 of multiplicity m then

log

∣∣∣∣∣∣

f (m)(0)m!

∣∣∣∣∣∣+mlog r = −

n∑

k=1

log

(

r|ak|

)

+12π

∫ 2π

0log | f (reiθ)| dθ.

Solution. We haver(z−ak)r2−akz maps B(0; r) onto itself and takes the boundary to the boundary. Let

F(z) = f (z)rm

zm

n∏

k=1

r2 − akzr(z− ak)

.

Then F∈ H(G) and F has no zeros in B(0; r) and |F(z)| = | f (z)| if |z| = r. Thus, by a known result, we have

log |F(0)| = 12π

∫ 2π

0log |F(reiθ)| dθ = 1

2π

∫ 2π


Also

g(z) ≡ f (z)zm=

∑∞i=0

f (i)(0)zi

i!

zm

=

∑∞i=m

f (i)(0)zi

i!

zm

=

∞∑

i=m

f (i)(0)zi−m

i!=

f (m)(0)m!

z0+

f (m+1)(0)(m+ 1)!

z1+ . . .

( f has a zero at z= 0 of multiplicity m) which implies g(0) = f (m)(0)m! . Thus,

F(0) = g(0)rmn∏

k=1

(

− rak

)

.

151

Therefore

|F(0)| = |g(0)|rmn∏

k=1

r|ak|

implies

log |F(0)| = log |g(0)| +mlog r +n∑

k=1

logr|ak|= log

∣∣∣∣∣∣

f (m)(0)m!

∣∣∣∣∣∣+mlog r +

n∑

k=1

logr|ak|

.

Hence,

log |F(0)| = log

∣∣∣∣∣∣

f (m)(0)m!

∣∣∣∣∣∣+mlog r +

n∑

k=1

logr|ak|=

12π

∫ 2π

0log | f (reiθ)| dθ


log

∣∣∣∣∣∣

f (m)(0)m!

∣∣∣∣∣∣+mlog r = −

n∑

k=1

logr|ak|+

12π

∫ 2π

0log | f (reiθ)| dθ

which proves the assertion.

Exercise 2. Let f be an entire function, M(r) = sup| f (reiθ)| : 0 ≤ θ ≤ 2π, n(r) = the number of zeros of fin B(0; r) counted according to multiplicity. Suppose that f(0) = 1 and show that n(r) log 2≤ log M(2r).

Solution. Since f is an entire function, it is analytic onB(0; 2r). Suppose a1, . . . ,an are the zeros of f inB(0; r) repeated according to multiplicity and b1, . . . ,bm are the zeros of f in B(0; 2r) − B(0; r). That is,a1, . . . ,an,b1, . . . ,bm are the zeros of f in B(0; 2r) repeated according to multiplicity. Since f(0) = 1 , 0,we have by Jensen’s formula

0 = log 1= log | f (0)| = −n∑

k=1

log

(

2r|ak|

)

−m∑

k=1

log

(

2r|bk|

)

+12π

∫ 2π

0log | f (2reiθ)| dθ

which impliesn∑

k=1

log

(

2r|ak|

)

= −m∑

k=1

log

(

2r|bk|

)

︸︷︷︸

≥ 0 since r≤ |bk| < 2r

+12π

∫ 2π


≤ 12π

∫ 2π


≤Def of M(r)

12π

log |M(2r)|∫ 2π

0dθ = log |M(2r)|.

Hence,n∑

k=1

log

(

2r|ak|

)

≤ log |M(2r)|.

Butn∑

k=1

log

(

2r|ak|

)

=

n(r)∑

k=1

log

(

2r|ak|

)

= logn(r)∏

k=1

(

2r|ak|

)

= log

2n

n(r)∏

k=1

r|ak|

= log

(

2n(r))

+ logn(r)∏

k=1

2r|ak|

≥ n(r) log(2)

152

since0 < |ak| < r implies r|ak| > 1 ∀k which means

∏n(r)k=1

r|ak| > 1 so the logarithm is greater than zero. Thus,

n(r) log(2)≤ log |M(2r)|

which proves the assertion.

Exercise 3. In Jensen’s Formula do not suppose that f is analytic in a region containingB(0; r) but onlythat f is meromorphic with no pole at z= 0. Evaluate

12π

∫ 2π



Exercise 4. (a) Using the notation of Exercise 2, prove that

∫ r

0

n(t)t

dt =n∑

k=1

log

(

r|ak|

)

where a1, . . . ,an are the zeros of f in B(0; r).(b) Let f be meromorphic without a pole at z= 0 and let n(r) be the number of zeros of f in B(0; r) minusthe number of poles (each counted according to multiplicity). Evaluate

∫ r

0

n(t)t

dt.


Exercise 5. Let D= B(0; 1)and suppose that f: D→ C is an analytic function which is bounded.(a) If an are the non-zero zeros of f in D counted according to multiplicity, prove that

∑

(1 − |an|) < ∞.(Hint: Use Proposition VII. 5.4).(b) If f has a zero at z= 0 of multiplicity m≥ 0, prove that f(z) = zmB(z) exp(−g(z)) where B is a BlaschkeProduct (Exercise VII. 5.4) and g is an analytic function on Dwith Re g(z) ≥ − log M (M = sup| f (z)| :|z| < 1).


11.2 The genus and order of an entire function

Exercise 1. Let f(z) =∑

cnzn be an entire function of finite genusµ; prove that

limn→∞

cn(n!)1/(µ+1)= 0.

(Hint: Use Cauchy’s Estimate.)

Solution. Define M(r) = max f (reiθ) : 0 ≤ θ ≤ 2π = max|z|=r | f (z)|. For sufficiently large r= |z|, we have

M(r) < eαrµ+1(11.1)

by Theorem 2.6 p. 283. By Cauchy’s estimate p. 73, we have

|cn| ≤M(r)rn

153

where cn =f (n)(0)

n! . Hence,

|cn| ≤M(r)rn

<(11.1)

eαrµ+1. (11.2)

Define, r= n1µ+1λ, whereλ is a positive constant, and chooseλ > e for convenience. Then (11.2) can be

written in the form

|cn| <eαnλµ+1

(nn)1µ+1 λn

⇐⇒ (nn)1µ+1 |cn| <

eαnλµ+1

λn=

eαλµ+1

λ

n

.

Now, if we chooseα such thatαλµ+1= 1, the last inequality becomes

(nn)1µ+1 |cn| <

(eλ

)n< ε

for anyε > 0, provided n is taken large enough, n> N say. Since n! < nn, for n ≥ 2, we have

(n!)1µ+1 |cn| < ε

for n > N orlimn→∞

cn(n!)1/(µ+1)= 0.

Exercise 2. Let f1 and f2 be entire functions of finite ordersλ1 andλ2 respectively. Show that f= f1 + f2has finite orderλ andλ ≤ max(λ1, λ2). Show thatλ = max(λ1, λ2) if λ1 , λ2 and give an example whichshows thatλ < max(λ1, λ2) with f , 0.


Exercise 3. Suppose f is an entire function and A, B, a are positive constants such that there is a r0 with| f (z)| ≤ exp(A|z|a + B) for |z| > r0. Show that f is of finite order≤ a.


Exercise 4. Prove that if f is an entire function of orderλ then f′ also has orderλ.

Solution. Let f be an entire function of orderλ, then f has power series

f (z) =∞∑

n=0

cnzn.

By Exercise 5 e) p. 286 we have

α = lim infn→∞

− log |cn|n logn

=1λ.

Differentiating f yields

f ′(z) =∞∑

n=1

ncnzn−1=

∞∑

n=0

(n+ 1)cn+1︸︷︷︸

dn

zn.

154

Now, letλ′ be the order of f′, then

1λ′= lim inf

n→∞− log |dn|n logn

= lim infn→∞

− log |(n+ 1)cn+1|n logn

= lim infn→∞

− log(n+ 1)− log |cn+1|n logn

= lim infn→∞

[− log(n+ 1)n logn

− log |cn+1|n logn

· (n+ 1) log(n+ 1)(n+ 1) log(n+ 1)

]

= lim infn→∞

− log(n+ 1)n logn

︸︷︷︸

→0

+− log |cn+1|

(n+ 1) log(n+ 1)︸︷︷︸

→ 1λ

· n+ 1n

︸︷︷︸

→1

· log(n+ 1)log(n)

︸︷︷︸

→1

=1λ.

Hence,λ′ = λ and therefore f′ also has orderλ.

Exercise 5. Let f(z) =∑

cnzn be an entire function and define the numberα by

α = lim infn→∞

− log |cn|n logn

(a) Show that if f has finite order thenα > 0. (Hint: If the order of f isλ and β > λ show that|cn| ≤r−n exp(rβ) for sufficiently large r, and find the maximum value of this expression.)(b) Suppose that0 < α < ∞ and show that for anyε > 0, ε < α, there is an integer p such that|cn|1/n <n−(α−ε) for n > p. Conclude that for|z| = r > 1 there is a constant A such that

| f (z)| < Arp+

∞∑

n=1

( rnα−ε

)n

(c) Let p be as in part (b) and let N be the largest integer≤ (2r)1/(α−ε). Take r sufficiently large so thatN > p and show that

∞∑

n=N+1

( rnα−ε

)n< 1 and

N∑

n=p+1

( rnα−ε

)n< Bexp

(

(2r)1/(α−ε))

where B is a constant which does not depend on r.(d) Use parts (b) and (c) to show that if0 < α < ∞ then f has finite orderλ andλ ≤ α−1.(e) Prove that f is of finite order iff α > 0, and if f has orderλ thenλ = α−1.


Exercise 6. Find the order of each of the following functions: (a)sinz; (b) cosz; (c) cosh√

z; (d)∑∞

n=1 n−anzn where a> 0. (Hint: For part (d) use Exercise 5.)

Solution. c) The order isλ = 12. Note thatsinh2 z= cosh2 z− 1 and thatcosx is bounded for real values x,

so| cosz|2 = cos2 x+ sinh2 y ≤ cosh2 y ≤ cosh2 |z|.

155

Thus

| cosz| ≤ cosh|z| = 12

(e|z| + e−|z|) ≤ 12

e|z| ≤ e|z|,

and therefore| cos

√z| ≤ exp(|z|1/2).

This implies that the orderλ ≤ 12. Next we want to show thatλ = 1

2. Seeking contradiction assume thatthere isε > 0 and r> 0 such that

| cos√

z| ≤ exp(|z| 12−ε) (11.3)

for all |z| > r.Let z= −n2 for an n∈ N with n > max √r ,4,21/(2ε) then in particular (i)|z| > r, (ii) − ln 2+ n > n

2, and(iii) 1

2 >1

n2ε and

| f (z)| = | cos√

z| = 12|en+ e−n| > 1

2en= e− log 2+n >

(ii )e

n2 >

(iii )exp(n−2εn) = exp(n2(1

2−ε)) = exp(|z| 12−ε).

This shows that noε > 0 can be found that satisfies equation (11.3). This impliesλ = 12.

Exercise 7. Let f1 and f2 be entire functions of finite orderλ1, λ2; show that f = f1 f2 has finite orderλ ≤ max(λ1, λ2).

Solution. Define M(r) = max|z|=r | f (z)|, M1(r) = max|z|=r | f1(z)| and M2(r) = max|z|=r | f2(z)|. Since f1 andf2 are entire functions of finite orderλ1 andλ2, respectively, we have forε > 0,

M1(r) < erλ1+ε/2 (11.4)

and

M2(r) < erλ2+ε/2 (11.5)

for sufficiently large r= |z| by Proposition 2.14. Hence

M(r) = max|z|=r| f (z)| = max

|z|=r| f1(z)· f2(z)|

≤ max|z|=r| f1(z)|·max

|z|=r| f2(z)|

= M1(r)M2(r) <by (11.4),(11.5)

erλ1+ε/2erλ2+ε/2

= erλ1+ε/2+rλ2+ε/2 ≤λ≤max(λ1,λ2)

e2rλ+ε/2 < erλ+ε

provided r is sufficiently large. Hence, the order of f(z) = f1(z)· f2(z) has finite orderλ ≤ max(λ1, λ2).

Exercise 8. Let an be a sequence of non-zero complex numbers. Letρ = inf a :∑ |an|−a < ∞; the

number called the exponent of convergence ofan.(a) If f is an entire function of rank p then the exponent of convergenceρ of the non-zero zeros of f satisfies:p ≤ ρ ≤ p+ 1.(b) If ρ′ = the exponent of convergence ofan then for everyε > 0,

∑ |an|−(ρ+ε) < ∞ and∑ |an|−(ρ−ε)

= ∞.(c) Let f be an entire function of orderλ and leta1,a2, . . . be the non-zero zeros of f counted accordingto multiplicity. If ρ is the exponent of convergence ofan prove thatρ ≤ λ. (Hint: See the proof of (3.5) inthe next section.)

156

(d) Let P(z) =∏∞

n=1 Ep(z/an) be a canonical product of rank p, and letρ be the exponent of convergence ofan. Prove that the order of P isρ. (Hint: If λ is the order of P,ρ ≤ λ; assume that|al | ≤ |a2| ≤ . . . and fixz, |z| > 1. Choose N such that|an| ≤ 2|z| if n ≤ N and|an| > 2|z| if n ≤ N + 1. Treating the casesρ < p+ 1andρ = p+ 1 separately, use(2.7) to show that for someε > 0

∞∑

n=N+1

log

∣∣∣∣∣∣Ep

(

zan

)∣∣∣∣∣∣< A|z|ρ+ε .

Prove that for|z| ≥ 12, log |Ep(z)| < B|z|p where B is a constant independent of z. Use this to prove that

N∑

n=1

log

∣∣∣∣∣∣Ep

(

zan

)∣∣∣∣∣∣< C|z|ρ+ε

for some constant C independent of z.)


Exercise 9. Find the order of the following entire functions:(a)

f (z) =∞∏

n=1

(1− anz), 0 < |a| < 1

b)

f (z) =∞∏

n=1

(

1− zn!

)

c)f (z) = [Γ(z)]−1 .

Solution. b) Considering g(z) :=∏∞

m=1

(

1− zn!

)

one notes that g(z) is already in form of the canonicalproduct with the entire function in the exponent being constantly zero and p= 0. Hence also the genusµ = 0. The simple zeros of g(z) are at z= n! and from Exercise 8a) it follows thatλ ≤ 1. In fact we willshow thatλ = 0. It suffices to show that the exponent of convergenceρ = 0 and to employ Exercise 8d).Let m≤ N, let n> 4m then

n! ≥ n· (n− 1)·2︸︷︷︸

≥n

· (n− 2)·3︸︷︷︸

≥n

. . . (n− (2k− 1))·2m︸︷︷︸

≥n

.

Each product of consecutive numbers is of the form(n− (k − 1))k, for 1 ≤ k ≤ 2m. By choice of n,kn ≤ 12

and n− k+ 1 > 2m which justifies the individual estimates by n. In total thisyields n! > n2m for n large, orequivalently

∞∑

n=4m

(n!)−1m ≤

∞∑

n=4m

n−2

and the sum on the left converges for every positive integer m. We conclude thatρ = inf m : the left sum converges =0. By Part d) of Exercise 8 we conclude that the order of g(z) is zero.

157

11.3 Hadamard Factorization Theorem

Exercise 1. Let f be analytic in a region G and suppose that f is not identically zero. Let G0 = G − z :f (z) = 0 and define h: G0→ R by h(z) = log | f (z)|. Show that∂h

∂x − i ∂h∂y =

f ′

f on G0.

Solution. Let f be analytic in a region G and suppose that f is not identically zero. Let G0 = G − z :f (z) = 0, then h: G0→ R given by h(z) = log | f (z)| is well defined as well asf

′

f is well defined on G0.Let f = u(x, y) + iv(x, y) = u + iv. Since f is analytic, the Cauchy-Riemann (C-R) equationsux = vy anduy = −vx are satisfied. We have by p. 41 Equation 2.22 and 2.23

f ′ = ux + ivx and f′ = −iuy + vy

and thus

2 f ′ = ux + ivx − iuy + vy

implies

f ′ =12

(

ux + ivx − iuy + vy

)

=(C-R)

12

(

ux − iuy − iuy + ux

)

=12

(

2ux − 2iuy

)

= ux − iuy.

Therefore,

f ′

f=

ux − iuy

u+ iv. (11.6)

Next, we calculate∂h∂x = hx and ∂h

∂y = hy where h(z) = log | f (z)| = 12 log(u2

+ v2). Using the chain rule, weget

hx = huux + hvvx =u

u2 + v2ux +

vu2 + v2

vx

andhy = huuy + hvvy =

uu2 + v2

uy +v

u2 + v2vy

and hence

∂h∂x− i

∂h∂y

= hx − ihy

=u

u2 + v2ux +

vu2 + v2

vx − iu

u2 + v2uy − i

vu2 + v2

vy

=u

u2 + v2(ux − iuy) +

vu2 + v2

(vx − ivy)

=(C-R)

uu2 + v2

(ux − iuy) +v

u2 + v2(−uy − iux)

=u

u2 + v2(ux − iuy) −

ivu2 + v2

(ux − iuy)

=u− ivu2 + v2

(ux − iuy)

=u− iv

(u− iv)(u+ iv)(ux − iuy)

=ux − iuy

u+ iv.

158

Compare this with (11.6) and we obtain the assertion

∂h∂x− i

∂h∂y=

f ′

f.

Exercise 2. Refer to Exercise 2.7 and show that ifλ1 , λ2 thenλ = max(λ1; λ2).


Exercise 3. (a) Let f and g be entire functions of finite orderλ and suppose that f(an) = g(an) for asequencean such that

∑ |an|−(λ+1)= ∞. Show that f= g.

(b) Use Exercise 2.8 to show that if f , g andan are as in part (a) with∑ |an|−(λ+ε)

= ∞ for someε > 0then f = g.(c) Find all entire functions f of finite order such that f(logn) = n.(d) Give an example of an entire function with zeroslog 2, log 3, . . . and no other zeros.

Solution. a) Since f and g are entire functions of finite orderλ, also F= f −g is an entire function of finiteorderλ (see Exercise 2 p. 286). Suppose that f(an) = g(an) for a sequencean such that

∑ |an|−(λ+1)= ∞.

Assume F. 0 ( f . g), we will derive a contradiction. Since F. 0, the sequencean are the zeros of F,because

F(an) = f (an) − g(an) = 0

since f(an) = g(an) by assumption. By Hadamard’s Factorization Theorem (p. 289), F has finite genusµ ≤ λ since F∈ H(C) with finite orderλ. Since by definitionµ = max(p,q) where p is the rank of F, wecertainly have

p ≤ λ.By the definition of the rank (p. 281 Definition 2.1), we have

∞∑

n=1

|an|−(p+1) < ∞,

where an’s are the zeros of F. Therefore by the comparison test, we also have

∞∑

n=1

|an|−(λ+1) ≤p≤λ

∞∑

n=1

|an|−(p+1) < ∞

so∑∞

n=1 |an|−(λ+1) < ∞ contradicting the assumption∑∞

n=1 |an|−(λ+1)= ∞. Hence F≡ 0, and therefore we

obtain f = g.b) Since f and g are entire functions of finite orderλ, we also have F= f − g is an entire function of finiteorderλ (see Exercise 2 p. 286). Suppose that f(an) = g(an) for a sequencean such that

∑ |an|−(λ+ε)= ∞.

Assume F. 0 ( f . g), we will derive a contradiction. Since F. 0, the sequencean are the zeros of F,because

F(an) = f (an) − g(an) = 0

since f(an) = g(an) by assumption. Letρ be the exponent of convergence ofan, then

ρ ≤ λ

by Exercise 8 c) p. 286-287. Hence, according to Exercise 8 b)p. 286-287, we have

∞∑

n=1

|an|−(ρ+ε) < ∞.

159

Therefore by the comparison test, we also have

∞∑

n=1

|an|−(λ+ε) ≤ρ≤λ

∞∑

n=1

|an|−(ρ+ε) < ∞.

Hence∑∞

n=1 |an|−(λ+ε) < ∞ contradicting the fact∑∞

n=1 |an|−(λ+ε)= ∞. Thus, F≡ 0 and therefore we get the

assertion f= g.

160

Chapter 12

The Range of an Analytic Function

12.1 Bloch’s Theorem

Exercise 1. Examine the proof of Bloch’s Theorem to prove that L≥ 1/24.

Solution. Mimic the proof of Block’s Theorem up to the application of Schwarz’s Lemma, i.e. set K(r) =max| f ′(z)|||z| = r,h(r) = (1− r)K(r), r0 = supr : h(r) = 1. Choose a with|a| = r0, | f ′(a)| = K(r0) = 1

1−r0.

Also chooseρ0 =12(1− r0) and infer that for all z∈ B(a, ρ)

| f ′(z)| ≤ 1ρ0, and | f ′(z) − f ′(a)| < 3|z− a|

2ρ20

.

For z ∈ B(0, ρ0) define g(z) = f (z+ a) − f (a). By convexity of B(a, ρ0) the line segmentγ = [a,a + z] iscompletely contained in B(a, ρ0). It follows that

|g(z)| = |∫

γ f ′(w) dw| ≤ 1ρ0|z| ≤ 1 =: M.

The function g is not necessarily injective on B(0, ρ0) but the problem only asks for an estimate on theLandau constant for which injectivity is not required. In this setting Lemma 1.2 gives

g(B(0, ρ0)) ⊃ B

(

0,ρ2|g′(0)|2

6M

)

= B

(

0,1

4 · 6

)

= B

(

0,124

)

.

This statement rewritten for f yields

f (B(a, ρ0)) ⊃ B

(

f (a),124

)

.

Exercise 2. Suppose that in the statement of Bloch’s Theorem it is only assumed that f is analytic on D.What conclusion can be drawn? (Hint: Consider the functionsfs(z) = s−1 f (sz), 0 < s < 1.) Do the samefor Proposition 1.10.


161

12.2 The Little Picard Theorem

Exercise 1. Show that if f is a meromorphic function onC such thatC∞ − f (C) has at least three pointsthen f is a constant. (Hint: What if∞ , f (C)?)

Solution. First consider an entire function f that misses three pointsin C∞, one on which is∞. Then thereare two distinct numbers a,b ∈ C that are not in the image of f . By Little Picard’s Theorem the function fmust be a constant.Next consider a meromorphic function onC that is not entire. Hence∞ ∈ f (C) and by assumption thereare distinct a,b, c ∈ C − f (C). Define a function

g(z) :=1

f (z) − a, z ∈ C

then g is entire because f(z) − a , 0 for all z ∈ C and 1b−a ,

1c−a ∈ C − g(C). Again using Little Picard’s

Theorem conclude that g is a constant function. It cannot be the zero-function, because f was assumedto be meromorphic, hence f(z) , ∞ for at least some z∈ C. Thus g is a nonzero constant function andf (z) = g(z)−1

+ a is also a constant function and the result is established.

Exercise 2. For each integer n≥ 1 determine all meromorphic functions f and g onC with a pole at∞such that fn + gn

= 1.


12.3 Schottky’s Theorem


12.4 The Great Picard Theorem

Exercise 1. Let f be analytic in G= B(0;R) − 0 and discuss all possible values of the integral

12πi

∫

γ

f ′(z)f (z) − a

dz

whereγ is the circle|z| = r < R and a is any complex number. If it is assumed that this integral takes oncertain values for certain numbers a, does this imply anything about the nature of the singularity at z= 0?


Exercise 2. Show that if f is a one-one entire function then f(z) = az+b for some constants a and b, a, 0.


Exercise 3.Prove that the closure of the setF in Theorem 4.1 equalsF together with the constant functions∞, 0, and1.


162

solutions functions of one complex variable i, second edition by john b. conway

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