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COMPLEX ANALYSIS

PROBLEMS

CLAUDIA TIMOFTE

COMPLEX ANALYSIS

PROBLEMS

To my students

Preface

The present book is a collection of problems in ordinary differential equations.

The book is based on some lectures I delivered for a number of years at the Faculty

of Physics of the University of Bucharest and covers the curriculum on ordinary

differential equations for the students of the first year of this faculty.

The material follows the textbook [19]. Each chapter contains a brief review of

the corresponding theoretical results, worked out examples and proposed problems.

Since the ”learning-by-doing” method is a successful one, the student is encouraged

to solve as many exercises as possible. The basic prerequisites for studying ordinary

differential equations using this book are undergraduate courses in linear algebra

and one-variable calculus.

It is my hope that this book will serve as an useful outlook for the students of

the first year of the Faculty of Physics of the University of Bucharest.

I would like to thank to my students for their continuous questions, comments

and suggestions, which helped me to improve the content of these notes.

Claudia Timofte

7

Contents

1 Complex Numbers 13

1.1 The Complex Number Field . . . . . . . . . . . . . . . . . . . . . . . 13

1.2 Sequences and Series of Complex Numbers . . . . . . . . . . . . . . . 30

2 Complex Functions 39

2.1 Functions of a Complex Variable . . . . . . . . . . . . . . . . . . . . 39

2.2 Limits of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

2.3 Continuous Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 42

3 Differentiable Functions 45

3.1 Holomorphic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 45

3.2 Harmonic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

4 Complex Integration 63

4.1 The Complex Integral . . . . . . . . . . . . . . . . . . . . . . . . . . 63

4.2 Cauchy’s Theorem. Cauchy’s Integral Formula. Applications . . . . 73

5 Taylor and Laurent Series 83

5.1 Sequences and Series of Functions. Power Series . . . . . . . . . . . . 83

5.2 Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

5.3 Laurent Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

5.4 Residues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

9

10 CONTENTS

6 The Residue Theorem. Applications 115

6.1 The Residue Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 115

6.2 Evaluation of Real Integrals . . . . . . . . . . . . . . . . . . . . . . . 119

Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .133

Chapter 1

Complex Numbers

1.1 The Complex Number Field

If R is the set of real numbers, let us consider the Cartesian product C = R×R, i.e.

the set C = (x, y) | x, y ∈ R. We introduce a special algebraic structure on this

set, by defining the sum and the product of two such ordered pairs of real numbers

as follows:

(x1, y1) + (x2, y2) = (x1 + x2, y1 + y2),

(x1, y1) · (x2, y2) = (x1x2 − y1y2, x1y2 + x2y1).

Then, (C, +, ·) becomes a commutative field, called the complex number field (the

proof is straightforward and it is left to the reader). An element of (C, +, ·) is calleda complex number.

Remark 1.1 It is important to mention that it is not possible to define an order

relation on C that is compatible with its ring structure. So, the complex numbers do

not form an ordered field.

If we consider the subfield of C consisting of all the ordered pairs with the

second element zero, i.e. S = (x, 0) | x ∈ R, then the map ϕ : R → S, defined

by ϕ(x) = (x, 0), is an isomorphism of R onto S. So, R and S can be identified. In

other words, the field R can be considered as being a subfield of the field C.

13

14 COMPLEX ANALYSIS

Now, if we identify, through the above mentioned isomorphism, the pair (x, 0)

with the real number x and the pair (y, 0) with y, it is not difficult to see that we

have

(x, y) = (x, 0) + (0, 1) · (y, 0). (1.1)

Let us denote

z = (x, y).

We consider a special pair, denoted by i:

i = (0, 1). (1.2)

This pair is called the unit imaginary number. In this way, (1.1) can be written as

z = x+ iy. (1.3)

This is the algebraic form of the complex number z (sometimes, a complex number

written in the form (1.3) is said to be in rectangular form). So, the set of complex

numbers can be represented as being:

C = z = x+ iy | x, y ∈ R.

Definition 1.2 For a complex number z = x + iy, the number Re z = x is called

the real part of z and the number Im z = y is said to be the its imaginary part. If

x = 0, z is said to be a purely imaginary number.

Definition 1.3 Let z = x+ iy ∈ C. The complex number

z = x− iy

is called the complex conjugate of z and

|z| =√

x2 + y2

is said to be the absolute value or the modulus of the complex number z.

COMPLEX NUMBERS 15

Remark 1.4 Let us notice that, since (0, 1) · (0, 1) = (−1, 0), we have

i2 = −1, (1.4)

which clearly shows that C cannot be an ordered field.

It is often convenient to use a graphical representation of complex numbers. As

we shall see in this section, complex numbers can be geometrically represented either

as points in the Euclidean plane or as two-dimensional vectors on a diagram, called

the Argand diagram or the complex plane. More precisely, in a given rectangular

coordinate system in the Euclidean plane R2, the complex number z = x + iy can

be identified with the point having the coordinates (x, y). The axes Ox and Oy

are called the real axis and, respectively, the imaginary axis. The number z = 0

corresponds to the origin of the plane and i corresponds to the point (0, 1). The plane

is referred to as the complex plane. This plane is also known as the Argand plane or

the Gauss plane. In this way, we establish a one-to-one correspondence between the

set of all the complex numbers and the set of all the points in the complex plane.

Geometrically, the conjugate z of a complex number z is the reflection of z in the

horizontal axis. The symmetric point of z with respect to the imaginary axis is −z.

It is easy to see that the points z, −z, −z and z are the vertices of a rectangle which

is symmetric with respect to the axes Ox and Oy. A complex number z = (x, y)

can be also represented as a two-dimensional vector pointing from the origin O(0, 0)

of the Argand plane to the point P (x, y). The point P (x, y) is the image-point

in the Argand plane of the complex number z and z is said to be the affix of the

point P (x, y). So, for each point in the complex plane, we associate the directed

line segment from the origin to the point, i.e. −→v = (x, y). Operations such as the

addition of complex numbers can be easily visualized by using this representation.

Definition 1.5 Let C∗ = C\(0, 0). For any complex number z ∈ C∗, any solution

θ of the equation

cos θ + i sin θ =z

z(1.5)

is called an argument of the complex number z.

16 COMPLEX ANALYSIS

Therefore, there is no argument for z = 0 and the argument (also called the phase) of

any complex number z 6= 0 is not uniquely determined, being defined only up to the

addition of integer multiples of 2π, i.e. the argument is a so-called multiple-valued

function.

Any two values of the complex argument differ by an exact integer multiple of

2π. To remove this ambiguity and to get a unique representation, a conventional

choice is to limit θ to an interval of length 2π.

We shall denote by arg z such a value of θ and we shall call it the principal or

the reduced argument of the complex number z. Also, we shall denote by

Arg z = arg z + 2kπ | k ∈ Z

the class of all the arguments of a given complex number z ∈ C∗. Alternatively, we

shall use the following notation:

Arg z = arg z (mod 2π).

Definition 1.6 Any complex number z ∈ C∗ can be represented in the following

trigonometric form:

z = |z| (cos θ + i sin θ) , with θ ∈ Arg z.

Here, the argument must be understood as being a multi-valued function, i.e.

Arg : C∗ → P(R).

There is no standard choice for such an interval of length 2π for the reduced ar-

gument, the appropriate one depending on the problem we address. Still, it is

customary to use one of the following two typical conventions for specifying the

principal argument: either we can consider that −π < arg z ≤ π or 0 ≤ arg z < 2π.

COMPLEX NUMBERS 17

If we adopt the first convention, the principal argument arg z is defined by

arg z =

arctany

x, x > 0, y ∈ R,

arctany

x+ π, x < 0, y > 0,

arctany

x− π, x < 0, y < 0,

π/2, x = 0, y > 0,

−π/2, x = 0, y < 0,

π, x < 0, y = 0.

In what follows, unless otherwise mentioned, we shall adopt the first convention.

So, for any complex number z ∈ C∗, its principal argument will be the unique real

number θ ∈ (−π, π] given by (1.5). We shall write

z = |z| (cos θ + i sin θ).

Remark 1.7 Let us notice that arg is not a continuous function: it has a disconti-

nuity along the negative real axis.

Remark 1.8 Using Euler’s formula

ei θ = cos θ + i sin θ, (1.6)

we can write any complex number z ∈ C∗ in the so-called exponential or polar form:

z = r ei θ, (1.7)

where r = |z| and θ ∈ Arg z.

From Euler’s formula, we obtain immediately the following identities:

cos θ =eiθ + e−iθ

2, sin θ =

eiθ − e−iθ

2i.

18 COMPLEX ANALYSIS

If we take θ = π in Euler’s formula (1.6), we obtain a famous and amazing identity

eiπ + 1 = 0,

connecting several important constants in mathematics.

Remark 1.9 The modulus of a complex number is a multiplicative function, i.e.

|z1 z2| = |z1| |z2|,

while the argument is additive:

arg (z1 z2) = arg z1 + arg z2,

up to integer multiples of 2π, i.e. we can write

Arg (z1 z2) = Arg z1 +Arg z2.

Also, the modulus is invariant under conjugation, i.e.

|z| = |z|

and the argument changes sign:

arg z = − arg z,

up to integer multiples of 2π.

Let us recall now two formulas for computing powers and roots of complex num-

bers. We start with the so-called de Moivre’s formula:

zn = rn [cos (nθ) + i sin (nθ)] , for n ∈ Z.

Also, let us turn our attention to the root extraction rule. Let a ∈ C∗ and n ∈ N,

with n ≥ 2. We consider the equation

zn = a, (1.8)

COMPLEX NUMBERS 19

with a = ρ (cos θ + i sin θ) and z = r (cosφ + i sinφ). The equation (1.8) has n

complex roots:

zk = ρ1/n(cos

θ + 2kπ

n+ i sin

θ + 2kπ

n

), k = 0, n − 1.

So, there are n roots of the n-th order of any complex number z ∈ C∗. They have

the same modulus and their arguments are equally spaced. From a geometric point

of view, they represent the vertices of a regular polygon with n sides.

Remark 1.10 In many problems, it is necessary to extend the complex number

system C by introducing a symbol ”∞” to represent infinity. So, adding an improper

number denoted by ”∞”, we define C∞ = C ∪ ∞, ∞ /∈ C. The convention

|∞| = +∞ extends the notion of absolute value from C to C∞. Also, we define:

a+∞ = ∞+ a = ∞, for a ∈ C,

a · ∞ = ∞ · a = ∞, for a ∈ C∞ \ 0,a

0= ∞, for a ∈ C∞ \ 0, a

∞ = 0, for a ∈ C.

We shall not define:∞∞ ,

0

0, 0 · ∞, ∞+∞.

The extended complex numbers do not form a field.

Definition 1.11 A sequence (zn)n is said to converge to infinity (we shall denote

limn→∞

zn = ∞) if

limn→∞

|zn| = ∞,

i.e. for any R > 0, there exists NR ∈ N such that |zn| > R, for any n ≥ NR.

Definition 1.12 If C is identified with the Euclidean plane, then C∞ is called the

extended complex plane. We call z = ∞ the point at infinity.

Remark 1.13 The set C∪∞ can be visualized as a sphere (the Riemann sphere or

the extended complex plane). We consider the Euclidean space R3, with coordinates

(X,Y,Z) and the XY -plane identified with C. Let S be the sphere centered at the

20 COMPLEX ANALYSIS

point (0, 0, 1/2) and with radius 1/2. The sphere S has the diameter equal to one and

touches the complex plane at the point O = (0, 0, 0). We denote by N = (0, 0, 1) the

north pole of the sphere, i.e. the antipode of O. For each point z ∈ C, let M be the

point obtained by the intersection with the sphere of the segment joining z and M . To

a sequence (zn)n converging to the point at infinity it corresponds a sequence of points

on S, converging to N . Therefore, the ”image” of z = ∞ is N . This correspondence

between the points of the extended plane and those on the sphere S is one-to-one

and is called the stereographic projection. S is said to be Riemann’s sphere. The

stereographic projection is a conformal mapping. Therefore, if we identify, via the

stereographic projection, the points in the complex plane with the points on S \ Nand, further, ∞ with N , we get a bijection between the extended complex plane C∞

and S. Such a construction maps the unbounded set C into the compact set S by

adding one point, the point at infinity. The Riemann sphere is also called the one-

point compactification of C. The Riemann sphere has many useful applications in

various fields, such as physics, cartography, crystallography, quantum mechanics or

geology.

Remark 1.14 The usual topology we shall consider on C will be the one induced by

the metric

d : C× C → R+,

defined, for any z1, z2 ∈ C, by

d (z1, z2) = |z1 − z2| .

This topology coincides with the one induced by the Euclidean metric on R2. Here,

| · | : C → R+ is the complex modulus, which is taken to be the standard norm on the

real vector space C. If we consider the Euclidean space (R2, | · |), endowed with the

classical Euclidean norm, then there exists an isomorphism of vector normed spaces

which allows us to identify the normed spaces (R2, ‖ · ‖) and (C, | · |).

Definition 1.15 Let z0 ∈ C and r ∈ R, with r > 0. The set

B(z0, r) = z ∈ C | |z − z0| < r

COMPLEX NUMBERS 21

is called the open ball or the open disk centered at the point z0 and having the radius

r.

Definition 1.16 Let z0 ∈ C and r ∈ R, with r ≥ 0. The set

B(z0, r) = z ∈ C | |z − z0| ≤ r

is called the closed disk centered at the point z0 and having the radius r. If in the

above definition we take r = 0, then B(z0, 0) = z0.

Remark 1.17 We shall sometimes denote the open disk B(z0, r) by U(z0; r). Also,

we shall denote by U(z0; r) the set U(z0; r) \ z0 and we shall call it the punctured

disk centered at z0 and with radius r:

U(z0; r) = z ∈ C | 0 <| z − z0 |< r.

Definition 1.18 For z0 ∈ C and r,R ∈ R, with 0 ≤ r < R ≤ +∞, the set

U(z0; r,R) = z ∈ C | r < |z − z0| < R

is called the open annulus centered at z0 and with radii r and, respectively, R.

Let us notice that

U(z0; 0, R) = U(z0;R), U(z0; r,∞) = C \ U(z0; r), U(z0; 0,∞) = C \ z0.

Definition 1.19 Let z0 ∈ C and r > 0. The set

C(z0, r) = z ∈ C | |z − z0| = r

is called the circle centered at the point z0 and having the radius r.

Example 1.20 As a concrete example, we can consider the circle with center at the

point z0 = (1, 1) and radius two, which is the locus of all the points z = (x, y) ∈ C

such that

(x− 1)2 + (y − 1)2 = 22

or, alternatively, the set of all the points z ∈ C with

|z − z0| = 2.

22 COMPLEX ANALYSIS

Definition 1.21 If z1, z2 ∈ C, the line segment joining z1 and z2 is defined by

[z1, z2] = z ∈ C | z = (1− t)z1 + tz2, 0 ≤ t ≤ 1.

A polygonal line is a piecewise-smooth curve which consists of finitely many straight

line segments. We shall say that a set A ⊆ C is polygonally connected if for any

two given points a, b ∈ A there exists a polygonal path lying in A and having the

endpoints a and, respectively, b.

Remark 1.22 A nonempty open set A ⊆ C is connected if and only if it is polygonally-

connected.

Definition 1.23 A set A ⊆ C is called convex if for any pair of points a, b ∈ A, the

whole segment [a, b] is contained in A.

Definition 1.24 An open connected set in C is called a domain or a region.

Definition 1.25 A domain A ⊆ C is said to be star-shaped or a star-like domain

if there exists a point a ∈ A such that the line segment [a, z] ⊆ A, for all z ∈ A.

Such a point a is called a star-center.

Remark 1.26 Every convex set in C is connected. Also, any convex domain is a

star-like one. The converse is false, in general.

Definition 1.27 A domain D ⊆ C is said to be simply connected if every closed

path lying in D can be continuously deformed into a point without leaving D.

Exercise 1.28 Prove that the set of all matrices of the form(

a b−b a

),

with a, b ∈ R, endowed with the matrix addition and multiplication, is isomorphic

to the field of complex numbers.

Exercise 1.29 Let z1, z2 ∈ C. Prove that

z1 = z2 if and only if Re z1 = Re z2 and Im z1 = Im z2.

COMPLEX NUMBERS 23

Exercise 1.30 Show that Re (iz) = − Im z and Im (iz) = Re z.

Exercise 1.31 Prove that z = z if and only if z is a real number.

Exercise 1.32 Let z ∈ C. Prove that (z) = z.

Exercise 1.33 Show that

(1

z

)=

1

z, for any z ∈ C, with z 6= 0.

Exercise 1.34 Prove that z1 + z2 = z1 + z2 and z1z2 = z1 z2, for any z1, z2 ∈ C.

Exercise 1.35 Show that Re z =z + z

2and Imz =

z− z

2i, for any z ∈ C.

Exercise 1.36 Prove that |z1 z2| = |z1| |z2|, for any z1, z2 ∈ C.

Exercise 1.37 Let z1, z2 ∈ C, with z2 6= 0. Show that |z1/z2| = |z1|/|z2|.

Exercise 1.38 Prove that |z| = |z| and |z|2 = z z, for any z ∈ C.

Exercise 1.39 Let z1, z2 ∈ C. Show that ||z1| − |z2|| ≤ |z1 − z2|.

Exercise 1.40 Prove that |Re z| ≤ |z| ≤ |Re z|+ |Im z| ≤√2 |z|, for any z ∈ C.

Exercise 1.41 Show that |Im z| ≤ |z| ≤ |Re z|+ |Im z| ≤√2 |z|, for all z ∈ C.

Exercise 1.42 Let z1, z2 ∈ C. Prove that

|z1 + z2|2 + |z1 − z2|2 = 2 (|z1|2 + |z2|2).

Exercise 1.43 Find the real and the imaginary parts of the complex number (1 + i)100.

Solution. We notice that (1 + i)2 = 2 i. Thus, (1 + i)100 = 250 i50 = −250.

Exercise 1.44 Compute the absolute value of the complex number z = 1 + 3 i.

Solution. It is easy to see that |z| =√10.

24 COMPLEX ANALYSIS

Exercise 1.45 Find the absolute value and the conjugate of the following complex

numbers:

a) z =3 + i

2 + 3 i; b) z = (1 + i)4 ; c) z = (2 + i) (7− i) .

Exercise 1.46 Express the complex number z = (1 + i) /i in algebraic form.

Solution. It is easy to see that z = 1− i.

Exercise 1.47 Write the complex number z = 1− i in trigonometric form.

Solution. If we adopt the convention that the principal argument of a complex

number z ∈ C∗ is the unique argument that lies on the interval (−π, π], it is not

difficult to see that the complex number z = 1− i can be written as:

1− i =√2[cos(−π

4

)+ i sin

(−π

4

)].

Exercise 1.48 Write the number

z =1 + i√3 + i

in trigonometric form.

Solution. It is easy to see that the trigonometric forms of 1 + i and√3 + i are

1 + i =√2[cos

π

4+ i sin

π

4

]

and, respectively, √3 + i = 2

[cos

π

6+ i sin

π

6

].

It follows that

z =

√2

2

[cos

π

12+ i sin

π

12

].

Exercise 1.49 Let a ∈ R and n ∈ N. Prove that

z = (1 + ai)n + (1− ai)n

is real.

COMPLEX NUMBERS 25

Exercise 1.50 Compute the absolute values of

z1 = i(2 + i)(1 − 4i), z2 =(3 + 2i)(1 − i)

(2 + i)(3 − 4i).

Exercise 1.51 Compute (1 + i)16 and (1 +√3i)5.

Exercise 1.52 Show that if the sum and the product of two given complex numbers

are both real, then either the numbers are real or they are complex conjugated.

Solution. Let

z1 = a+ i b, z2 = c+ i, d.

Then, we get b = −d and ad+ bc = 0, which implies that either b = d = 0 or a = c

and b = −d.

Exercise 1.53 Find all the roots of the equation z4 + 1 = 0.

Exercise 1.54 Let z = 3 + 4i. Compute z2 + 3z + 7.

Exercise 1.55 Find the cube roots of z = −27 and represent them on the Argand

diagram.

Exercise 1.56 Solve the equation

z4 − 4z2 + 4− 2i = 0.

Exercise 1.57 Solve the equation

z4 = −16.

Exercise 1.58 Sketch the following sets in the complex plane:

a) A =z ∈ C : |z − 2 + i| ≤ 3

; b) B = z ∈ C : |z| = |z + 1|.

Exercise 1.59 Prove that the sets D1 = z ∈ C | |z| < 1 and D2 = z ∈ C | 1 <

|z| < 2 are open sets.

26 COMPLEX ANALYSIS

Exercise 1.60 Show that the upper half-plane, i.e. the set of all the complex

numbers with positive imaginary part,

D = z ∈ C | Im z > 0,

is an open set.

Exercise 1.61 Let A1, A2, . . . , An be open sets in C. Prove thatn⋂

i=1Ai is an open

set, too.

Exercise 1.62 Let Aii∈I be an arbitrary family of open sets in C. Prove that⋃i∈I

Ai is also an open set.

Exercise 1.63 Show that any closed ball B(z0, r) is a closed set.

Exercise 1.64 Let B1, B2, . . . , Bn be closed sets in C. Prove thatn⋃

i=1Bi is also a

closed set.

Exercise 1.65 Let Bii∈I be an arbitrary family of closed sets in C. Prove that⋂i∈I

Bi is also a closed set.

Exercise 1.66 Any nonempty set X can be endowed with a metric. In particular,

on C, we can define the so-called discrete metric by putting, for any z, w ∈ C,

d0(z, w) =

1, z 6= w,

0, z = w.

Show that d0 is indeed a metric on C.

Exercise 1.67 Prove that the map d1 : C× C → R+, given by

d1(z1, z2) =|z1 − z2|

1 + |z1 − z2|,

is a distance on C.

COMPLEX NUMBERS 27

Exercise 1.68 If z1, . . . , zn and w1, . . . , wn ∈ C, prove that

∣∣∣∣∣

n∑

i=1

ziwi

∣∣∣∣∣

2

≤(

n∑

i=1

|zi|2)(

n∑

i=1

|wi|2)

(Cauchy’s inequality).

Exercise 1.69 If z = (z1, . . . , zn), show that the following maps are norms on Cn:

a) ‖z‖1 =

n∑

i=1

|zi|;

b) ‖z‖2 =

√√√√n∑

i=1

|zi|2;

c) ‖z‖∞ = max1≤i≤n

|zi|.

The norm ‖ · ‖2 is usually denoted by ‖ · ‖.

Exercise 1.70 Let z, w ∈ C. Prove that

‖z + w‖2 + ‖z − w‖2 = 2(‖z‖2 + ‖w‖2

)

and

‖z − w‖2 + ‖z + w‖2 = −4 (Re z) (Re w) .

Exercise 1.71 Prove that the map d : C× C → R+, given by

d(z1, z2) =2 |z1 − z2|√

1 + |z1|2√

1 + |z2|2, (1.9)

is a distance. If z1 and z2 are finite points in C, the distance between their stere-

ographic projection is given by (2.9). This distance is called the chordal or the

spherical distance between the points z1 and z2. If z2 = ∞, the corresponding

distance is given by

d(z1,∞) =2√

1 + |z1|2.

Exercise 1.72 Describe the sets of the points z ∈ C∗ satisfying the relation 1/z = z.

28 COMPLEX ANALYSIS

Solution. It is not difficult to see that we get the unit circle.

Exercise 1.73 Describe the sets of the points z ∈ C satisfying the relation |z| =Re z + 1.

Solution. Since√

x2 + y2 = x + 1, we are led to y2 = 2x + 1. Thus, we get a

parabola.

Exercise 1.74 Draw the set of all the points z ∈ C satisfying the condition Re (z) >

3.

Exercise 1.75 Draw the set of all the points z ∈ C satisfying the condition 1 <

Im (z) < 3.

Exercise 1.76 Draw the set of all the points z ∈ C satisfying the condition Re (z) >

3.

Exercise 1.77 Draw the set of all the points z ∈ C satisfying the condition |z−2| ≤|z + 2|.

Exercise 1.78 Solve the following equations:

a) z3 = 1; b) z4 = 81 i.

Exercise 1.79 Describe the following sets:

a) A = z ∈ C : Im(z) > 2; b) B = i

n: n ∈ N

∗.

Exercise 1.80 Decide if the following sets are open, closed, bounded or compact:

a) A = z ∈ C : 0 < Re(z) ≤ 2, Im(z) = 0; b) B = z ∈ C : 0 ≤ Re(z) ≤ 2, Im(z) = 0.

Exercise 1.81 Let A,B ⊆ C. Prove the following properties of usual operations

over sets:

1. Int A ⊆ A;

COMPLEX NUMBERS 29

2. if A ⊆ B, then Int A ⊆ IntB;

3. Int (A ∩B) = ( Int A) ∩ ( Int B);

4. Int (A ∪B) ⊇ ( Int A) ∪ ( Int B);

5. the interior of A is the largest open set contained in A, i.e.

Int A =⋃

D | D ⊆ A, D open;

6. A is open if and only if A = Int A;

7. Int A is an open set;

8. A ⊆ A and A′ ⊆ A;

9. if A ⊂ B ⇒ A ⊆ B and A′ ⊆ B

′

;

10. A ∪B = A ∪B and (A ∪B)′

= A′ ∪B

′

;

11. A ∩B ⊆ A ∩B and (A ∩B)′ ⊆ A

′ ∩B′

;

12. A = A;

13. A = A ∪A′

;

14. the set A is the smallest closed set which contains A, i.e.

A =⋂

F | F closed, F ⊇ A;

15. A is a closed set;

16. A = A ∪ ∂A;

17. ∂A = A \ IntA;18. IntA = A \ ∂A;19. A is closed if and only if A = A;

20. ∂A ∩ Int A = ∅.

Exercise 1.82 Show that D1 = z ∈ C | |z| < 2 is a connected set.

Exercise 1.83 Prove that an open set is pathwise connected if and only if it is

connected.

30 COMPLEX ANALYSIS

Exercise 1.84 Let A ⊆ C be a connected set and B ⊆ C a set such that A ⊆ B ⊆ A.

Prove that B is connected, too.

Exercise 1.85 If A ⊆ C is a connected set, then A is also connected.

Exercise 1.86 Let (Ai)i∈I be a family of connected sets in C such that⋂i∈I

Ai 6= ∅.Prove that

A =⋃

i∈I

Ai

is a connected set.

Exercise 1.87 Show that the set of all the points z ∈ C with 2 < |z| < 3 is open

and connected, but not simply connected.

Exercise 1.88 Show that the punctured plane C\0 is not simply connected, but

the slit plane D = C \ (−∞, 0] is a simply connected domain.

Exercise 1.89 Prove that a star-shaped domain in C is simply connected.

Exercise 1.90 Let z0 ∈ C and 0 < r < R. Prove that the annulus U(z0; r,R) is

not a star-domain.

Exercise 1.91 Show that every path-connected set is connected. Find a connected

set which is not path-connected.

1.2 Sequences and Series of Complex Numbers

Definition 1.92 Let (zn)n ⊆ C. The sequence (zn)n is called convergent if there

exists z ∈ C such that, for any V ∈ V(z), there exists nV ∈ N such that zn ∈ V , for

any n ≥ nV . We shall use the standard notation z = limn→∞

zn.

Theorem 1.93 A convergent sequence in C has only one limit.

Proposition 1.94 Let (zn)n be a convergent sequence in C and A ⊆ C. If there

exists n0 ∈ N such that zn ∈ A, for any n ≥ n0, then

limn→∞

zn ∈ A.

COMPLEX NUMBERS 31

Theorem 1.95 Let (zn)n be a sequence in C and z ∈ C. The following two asser-

tions are equivalent:

(i) z = limn→∞

zn;

(ii) for any ε > 0, there exists nε ∈ N such that |zn − z| < ε, for any n ≥ nε.

Definition 1.96 A sequence (zn)n in C is called a fundamental or a Cauchy se-

quence if, for any ε > 0, there exists nε ∈ N such that

|zn − zm| < ε, for any n,m ≥ nε.

Proposition 1.97 A convergent sequence in C is also a Cauchy sequence.

Remark 1.98 The converse implication in Proposition 2.51 is, in a general metric

space, not true. However, in (C, | · |), any Cauchy sequence is convergent. So, (C, | · |)is a complete space.

Remark 1.99 Any Cauchy sequence in C is bounded.

Remark 1.100 If a Cauchy sequence (zn)n possesses a subsequence (znk) which is

convergent to z ∈ C, then (zn)n is convergent to z.

Definition 1.101 Let (zn)n≥0 ⊆ C. If

Sn =

n∑

i=0

zi, (1.10)

then the pair ((zn)n≥0, (Sn)n≥0) is called a series (the series associated to the se-

quence (zn)n≥0).

Instead of the above pair notation, we shall use a simpler one:

∑

n≥0

zn. (1.11)

Sn is called the nth partial sum of the series (2.11), while zn is said to be its general

term.

32 COMPLEX ANALYSIS

Definition 1.102 If there exists S ∈ C such that

S = limn→∞

Sn,

then the series (2.11) is called convergent. In this case, S is said to be its sum.

If the series∑

n≥0

zn converges to S, we shall write

S =

∞∑

n=0

zn.

Definition 1.103 A series which is not convergent is called divergent.

Definition 1.104 A series∑

n≥0

zn is called absolutely convergent if the number se-

ries∑

n≥0

|zn| is convergent (in R).

Definition 1.105 A series which is convergent, but not absolutely convergent, is

called semi-convergent or conditionally convergent.

Proposition 1.106 If the series∑

n≥0

zn is convergent, then the sequence (zn)n is

convergent and

limn→∞

zn = 0.

Proposition 1.107 (Cauchy’s Criterion) The series∑

n≥0

zn is convergent in C if

and only if, for any ε > 0, there exists Nε ∈ N such that

|zn+1 + · · ·+ zn+p| < ε, for any n ≥ Nε and p ≥ 1. (1.12)

Proposition 1.108 (The Geometric Series) For z ∈ C with |z| < 1, the series∑n≥0

zn is convergent and

∞∑

n=0

zn =1

1− z.

COMPLEX NUMBERS 33

Proposition 1.109 (Comparison Test) Let (zn)n ⊆ C and (wn)n ⊆ R+. We as-

sume that there exists n0 ∈ N such that

|zn| ≤ wn, for any n ≥ n0.

If the series∑

n≥0

wn is convergent, then the series∑

n≥0

zn is absolutely convergent.

Proposition 1.110 (Abel-Dirichlet) Let (zn)n ⊆ C and (wn)n ⊆ R+. We assume

that there exists M > 0 such that∣∣∣∣∣

n∑

i=0

zi

∣∣∣∣∣ ≤ M, for any n ≥ 0.

If the sequence (wn)n is nonincreasing and

limn→∞

wn = 0,

then the series∑

n≥0

znwn is convergent.

Let us give now a very powerful convergence test, called Cauchy’s Test or the

Root Test.

Theorem 1.111 Let (zn)n≥1 ⊆ C and

L = lim supn→∞

n√

|zn|.

Then, we have:

(i) if L < 1, the series∑

n≥1

zn is absolutely convergent;

(ii) if L > 1, the series∑

n≥1

zn is divergent.

Corollary 1.112 If there exists

L = limn→∞

n√

|zn|,

34 COMPLEX ANALYSIS

then:


n≥1



n≥1

zn is divergent;

(iii) if L = 1, we cannot decide upon the nature of the series∑

n≥1

zn.

We recall now another powerful test, called D’Alembert’s Test or the Ratio Test.

Theorem 1.113 Let (zn)n≥0 ⊆ C.

(i) If lim supn→∞

∣∣∣∣zn+1

zn

∣∣∣∣ < 1, then the series∑

n≥0

zn is absolutely convergent.

(ii) If lim infn→∞

∣∣∣∣zn+1

zn

∣∣∣∣ > 1, then the series∑

n≥0

zn is divergent.

Corollary 1.114 If there exists

L = limn→∞

∣∣∣∣zn+1

zn

∣∣∣∣,

then:


n≥0



n≥0

zn is divergent;

(iii) if L = 1, we cannot decide upon the nature of the series∑

n≥0

zn.

Definition 1.115 Let∑

n≥0

zn and∑

n≥0

wn be series of complex numbers. Let us con-

sider the series∑

n≥0

tn, with

tn =∑

i+j=n

ziwj.

This series is called the Cauchy product series of the series∑

n≥0

zn and∑

n≥0

wn.

COMPLEX NUMBERS 35

Theorem 1.116 (Mertens) Let∑

n≥0

zn and∑

n≥0

wn be convergent series of complex

numbers. If at least one of the two series is absolutely convergent, then the Cauchy

product series∑

n≥0

tn is convergent and

∞∑

n=0

tn =

(∞∑

n=0

zn

) (∞∑

n=0

wn

).

Theorem 1.117 (Cauchy) If∑

n≥0

zn and∑

n≥0

wn are absolutely convergent series of

complex numbers, then their Cauchy product series∑

n≥0

tn is absolutely convergent,

too.

Exercise 1.118 Let (zn)n ⊆ C. Prove that

zn → z if and only if

Re zn → Im z,Im zn → Im z.

Exercise 1.119 Let (zn)n, (wn)n ⊆ C such that

zn → z, wn → w,

with z, w ∈ C. If wn, w 6= 0, show that

znwn

→ z

w.

Exercise 1.120 If the sequence (zn)n converges to a limit z, prove that (zn)n con-

verges to z.

Exercise 1.121 Find the limit of the sequence zn = in!+1.

Exercise 1.122 Prove that limn→∞

(1 + i

2

)n

= 0.

Exercise 1.123 Let (zn)n ⊆ C, with zn = xn + iyn. Show that

∑

n≥0

zn is convergent if and only if∑

n≥0

xn and∑

n≥0

yn are convergent.

36 COMPLEX ANALYSIS

Exercise 1.124 Show that the sequence zn = (1 + i)n is divergent.

Exercise 1.125 Find the limit of the sequence

zn =

√n+ 2 (n+ 1) i

n, n ≥ 1.

Solution. Since zn = xn + i yn, with

xn =

√n

n, yn =

2(n + 1)

n,

we get limn→∞

zn = 2 i.

Exercise 1.126 Prove that

limn→∞

(n+ i)(1 + n i)

n2= i.

Exercise 1.127 Prove that for any z ∈ C, the series

∑

n≥0

(−1)nz2n

(2n)!

is absolutely convergent.

Exercise 1.128 Prove that for any z ∈ C, the series

∑

n≥0

(−1)nz2n+1

(2n+ 1)!

is absolutely convergent.

Exercise 1.129 Compute the sum of the series∑

n≥1

(i

4

)n−1

.

Exercise 1.130 Test the series∑

n≥0

in

3nfor convergence.

COMPLEX NUMBERS 37

Solution. Using the ratio test, we get

limn→∞

∣∣∣zn+1

zn

∣∣∣ =1

3.

Hence, by the ratio test, the series converges absolutely.

Exercise 1.131 Test the following series for convergence:

a)∑

n≥1

1

2 + in;

b)∑

n≥1

1

n2 + in;

c)∑

n≥1

(3 + i√10

)n

.

Exercise 1.132 Show that the series∑

n≥1

1 + i n (−1)n

n2is convergent.

Solution. If we recall that the real series∑

n≥1

1

n2and

∑

n≥1

(−1)n

nare convergent, it

follows that the given complex series is convergent, as well.


n≥1

1 + i n

n2is divergent.

Solution. If we recall that the real series∑

n≥1

1

nis divergent, it follows that the given

complex series is divergent, too.


n≥1

(1 + i)n is divergent.

Solution. Notice that limn→∞

zn 6= 0.

Exercise 1.135 Prove that the series∑

n≥1

(3 + 4 i)n

5n n4is convergent.

38 COMPLEX ANALYSIS

Solution. Since |zn| = 1/n4, using the comparison test and the fact that the series∑

n≥1

1

n4is convergent, it follows that the given series is convergent.

Exercise 1.136 Test the following series for convergence:

a)∑

n≥1

(1 + i√

3

)n

;

b)∑

n≥1

n

(1

i

)n

.

Chapter 2

Complex Functions

2.1 Functions of a Complex Variable

Definition 2.1 Let D be a nonempty set in C. A single-valued complex function

or, simply, a complex function f : D → C is a map that assigns to each complex

argument z = x+ iy in D a unique complex number w = u+ iv. We write

w = f(z).

The set D is called the domain of the function f and the set f(D) is the range or

the image of f .

So, a complex-valued function f of a complex variable z is a rule that assigns to

each complex number z in a set D one and only one complex number w. We call w

the image of z under f . If z = x+ iy ∈ D, we shall write

f(z) = u(x, y) + iv(x, y)

or

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

The real functions u and v are called the real and, respectively, the imaginary part

of the complex function f . Therefore, we can describe a complex function with the

aid of two real functions depending on two real variables.

39

40 COMPLEX ANALYSIS

Example 2.2 The function f : C → C, defined by

f(z) = z3,

can be written as f(z) = u(x, y) + iv(x, y), with u, v : R2 → R given by

u(x, y) = x3 − 3xy2, v(x, y) = 3x2y − y3.

Example 2.3 For the function f : C → C, defined by

f(z) = ez,

we have

u(x, y) = ex cos y, v(x, y) = ex sin y, for any (x, y) ∈ R2.

Example 2.4 For the function

f(z) = |z|,

we have

u(x, y) =√x2 + y2, v(x, y) = 0,

which means that f is a real-valued function of a complex variable, its domain of

definition being C.

2.2 Limits of Functions

Definition 2.5 Let D ⊆ C, a ∈ D′

and f : D → C. A number l ∈ C is called a

limit of the function f at the point a if for any V ∈ V(l), there exists U ∈ V(a) suchthat, for any z ∈ U ∩D \ a, it follows that f(z) ∈ V . We shall use the notation

l = limz→a

f(z).

Remark 2.6 If a complex function f : D → C possesses a limit l at a given point

a, then this limit is unique.

COMPLEX FUNCTIONS 41

Theorem 2.7 Let D ⊆ C, f : D → C and a ∈ D′

. The following statements are

equivalent:

(i) l = limz→a

f(z);

(ii) for any ε > 0, there exists δε > 0 such that for any z ∈ D, z 6= a, with

|z − a| < δε, it follows that |f(z)− l| < ε;

(iii) for any (zn)n ⊆ D, zn 6= a, such that zn → a, it follows that f(zn) → l.

Let us discuss now limits involving ∞. We notice that z → ∞ means |z| → ∞and f(z) → ∞ means |f(z)| → ∞.

Definition 2.8 a) Let z0 ∈ C. We say that

limz→z0

f(z) = ∞

if, for any M > 0, there exists δ > 0 such that, for any z with 0 < |z − z0| < δ, we

have |f(z)| > M .

b) We say that

limz→∞

f(z) = ∞

if, for any M > 0, there exists R > 0 such that, for |z| > R, we have |f(z)| > M .

c) For a given l ∈ C,

limz→∞

f(z) = l

means that for any ε > 0, there exists R > 0 such that, for any z with |z| > R, we

have |f(z)− l| < ε.

Exercise 2.9 Let f, g : D ⊆ C → C and a ∈ D′

. If limz→a

f(z) and limz→a

g(z) exist,

prove that

limz→a

(f ± g)(z) = limz→a

f(z)± limz→a

g(z);

limz→a

(fg)(z) = limz→a

f(z) limz→a

g(z);

42 COMPLEX ANALYSIS

limz→a

f(z)

g(z)=

limz→a

f(z)

limz→a

g(z),

provided that g(z) 6= 0 on D and limz→a

g(z) 6= 0.

Exercise 2.10 Let f : D ⊆ C → C and a ∈ D′

. Show that

limz→a

f(z) = l if and only if

limz→a

Re f(z) = Re l,

limz→a

Im f(z) = Im l.

Also, prove that

limz→a

f(z) = l if and only if limz→a

f(z) = l.

Exercise 2.11 Prove that limz→0

z

zdoes not exist.

Solution. To prove that the above limit does not exist, we compute this limit as

z → 0 on the real and on the imaginary axis, respectively. In the first situation, i.e.

for z = x ∈ R, the value of the limit is 1. In the second situation, i.e. for z = i y,

with y ∈ R, the limit is −1. Thus, the limit depends on the direction from which we

approach 0, which implies that the limit does not exist.

Exercise 2.12 Compute the following limits:

a) limz→4i

z2 + 16

z − 4i; b) lim

z→i

z4 − 1

z2 + 1; c) lim

z→1−i(z2 + 2).

Exercise 2.13 Find the value of the following limits:

a) limz→∞

z2 + 2

z2 + z − i; b) lim

z→∞

z3 + iz + 2

z2 + i.

2.3 Continuous Functions

Definition 2.14 Let f : C → C and a ∈ C. The function f is said to be continuous

at the point a if for any V ∈ V(f(a)), there exists U ∈ V(a) such that f(U) ⊆ V .

COMPLEX FUNCTIONS 43

Definition 2.15 The function f is said to be continuous on C if it is continuous at

each point a ∈ C.

Theorem 2.16 f : C → C and a ∈ C. The following statements are equivalent:

(i) f is continuous at the point a;

(ii) for any ε > 0, there exists δε > 0 such that for any z ∈ C with |z − a| < δε,

it follows that |f(z)− f(a)| < ε;

(iii) for any (zn)n ⊆ C such that zn → a, it follows that f(zn) → f(a).

Definition 2.17 The function f : D ⊆ C → C is said to be uniformly continuous on

D if, for any ε > 0, there exists δε > 0 such that, for any z, w ∈ D, with |z−w| < δε,

it follows that |f(z)− f(w)| < ε.

Exercise 2.18 (Heine-Cantor) Let K ⊆ C be a compact set and f : K → C be a

continuous map on K. Show that f is uniformly continuous on K.

Exercise 2.19 Let f, g : C → C. If f and g are continuous and α, β ∈ C, prove that

αf + βg, fg, f/g, for g 6= 0, are continuous, too.

Exercise 2.20 If f : D ⊆ C → C and a ∈ D′

, then f is continuous at the point a

if and only if the limit limz→a

f(z) exists and equals f(a).

Exercise 2.21 If a is an isolated point for D, then f is continuous at a.

Exercise 2.22 Show that the function f : C → C, defined by f(z) = z2, is contin-

uous, but it is not uniformly continuous on C.

Exercise 2.23 Prove that the function f : C → C, defined by f(z) = 3z + 4, is

uniformly continuous on C.

Exercise 2.24 Let f : K ⊆ C → C be a continuous function. If K is compact,

show that f(K) is also compact. If K is connected, prove that f(K) is connected,

too.

44 COMPLEX ANALYSIS

Exercise 2.25 Compute the following limits:

a) limz→i

z5 + 1

z2 + 1; b) lim

z→0

1− cos z

2 z2.

Exercise 2.26 Show that:

a) limz→0

2 z2

z= 0; b) lim

z→0

(zz

)2doesn’t exist.

Exercise 2.27 Test the following function for continuity

f(z) =

z2 + 9

z − 3i, z 6= 3i,

4 + 6i, z = 3i.

Exercise 2.28 Test the following function for continuity

f(z) =

z3

zRe z, z 6= 0,

0, z = 0.

Chapter 3

Differentiable Functions

3.1 Holomorphic Functions

Let D ⊆ C be an open nonempty set, z = x+ iy ∈ D and f : D → C be a complex-

valued function of a single complex variable z. In what follows, we shall use the

following notation:

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

or

f(z) = u(x, y) + iv(x, y),

where u and v are the real and, respectively, the imaginary part of the complex

function f . The function f can be considered either as a complex function depending

on a complex variable or as a complex function of two real variables.

Definition 3.1 Let D ⊆ C be an open nonempty set, f : D → C and z0 ∈ D. The

complex function f is said to be C-derivable at the point z0 (or, simply, derivable at

z0) if there exists

limz→z0

f(z)− f(z0)

z − z0∈ C. (3.1)

This limit is called the derivative of f at the point z0 and is denoted by f′

(z0).

Definition 3.2 Let D ⊆ C be a nonempty open set, f : D → C and z0 ∈ D. The

function f is said to be R-differentiable at the point z0 = x0 + iy0 if there exist two

45

46 COMPLEX ANALYSIS

complex numbers A1, A2 and a complex function f1 : D \ z0 → C such that

limz→z0

f1(z) = 0

and, for all z ∈ D \ z0, we have

f(z) = f(z0) +A1(x− x0) +A2(y − y0) + f1(z) |z − z0|. (3.2)

Remark 3.3 If f = u + iv is R-differentiable at z0 ∈ D, then it is not difficult to

see that f possesses first-order partial derivatives at z0 and we have

∂f

∂x(z0) =

∂u

∂x(z0) + i

∂v

∂x(z0) = A1,

∂f

∂y(z0) =

∂u

∂y(z0) + i

∂v

∂y(z0) = A2.

Also, we can see that f is R-differentiable at z0 = x0+ iy0 if and only if u and v are

R-differentiable at (x0, y0).

Remark 3.4 If we take

f1(z) =

f(z)− f(z0)−A1(x− x0)−A2(y − y0)

|z − z0|, for z ∈ D \ z0,

0, for z = z0,

then we can obtain an equivalent definition of R-differentiability of a function f , but

with f1 continuous at z0 and f1(z0) = 0.

Definition 3.5 Let D ⊆ C be a nonempty open set, f : D → C and z0 ∈ D. The

function f is said to be C-differentiable at the point z0 if there exists α ∈ C and a

complex function f1 : D\z0 → C such that limz→z0

f1(z) = 0 and, for all z ∈ D\z0,we have

f(z) = f(z0) + α(z − z0) + f1(z)(z − z0). (3.3)

DIFFERENTIABLE FUNCTIONS 47

Remark 3.6 If we take

f1(z) =

f(z)− f(z0)

z − z0− α, for z ∈ D \ z0,

0, for z = z0,

then we can obtain an equivalent definition of C-differentiability of a function f , but

with f1 continuous at z0 and f1(z0) = 0.

Remark 3.7 If f : D → C is derivable at z0 ∈ D, then f is continuous at z0.

Proposition 3.8 A function f : D → C is derivable at z0 ∈ D if and only if it is

C-differentiable at z0.

Theorem 3.9 (Cauchy-Riemann) Let D ⊆ C be a nonempty open set and let f :

D → C. The complex function f = u+ iv is derivable at a point z0 ∈ D if and only

if f is R-differentiable at z0 and

∂u

∂x(z0) =

∂v

∂y(z0),

∂u

∂y(z0) = −∂v

∂x(z0).

(3.4)

Under these conditions,

f′

(z0) =∂f

∂x(z0). (3.5)

The relations (3.4) can be also written in the form:

∂u

∂x(x0, y0) =

∂v

∂y(x0, y0),

∂u

∂y(x0, y0) = −∂v

∂x(x0, y0).

(3.6)

48 COMPLEX ANALYSIS

Remark 3.10 If f is derivable at z0, using the Cauchy-Riemann equations, we get

the following expressions for f′

(z0):

f′

(z0) =∂f

∂x(z0) =

∂u

∂x(z0) + i

∂v

∂x(z0) =

∂u

∂x(z0)− i

∂u

∂y(z0) =

∂v

∂y(z0) + i

∂v

∂x(z0) =

∂v

∂y(z0)− i

∂u

∂y(z0) = −i

∂f

∂y(z0).

Also, we have:

∣∣∣f′

(z0)∣∣∣2=

(∂u

∂x(z0)

)2

+

(∂v

∂x(z0)

)2

=

(∂u

∂x(z0)

)2

+

(∂u

∂y(z0)

)2

=

=

(∂v

∂y(z0)

)2

+

(∂v

∂x(z0)

)2

=

(∂v

∂y(z0)

)2

+

(∂u

∂y(z0)

)2

and ∣∣∣f ′

(z0)∣∣∣2=

∂u

∂x(z0)

∂v

∂y(z0)−

∂u

∂y(z0)

∂v

∂x(z0) = J(z0).

where J is the Jacobian, defined by

J(z0) =∂(u, v)

∂(x, y)(z0).

Remark 3.11 The Cauchy-Riemann equations can be written in other coordinate

systems. For instance, it is not difficult to see that in the system of coordinates given

by the polar representation z = r eiθ these equations take the following form:

∂u

∂r=

1

r

∂v

∂θ,

∂v

∂r= −1

r

∂u

∂θ.

(3.7)

Alternatively, we can write these equations in the complex form

∂f

∂r=

1

ir

∂f

∂θ.


Proposition 3.12 (Goursat) Let D be an open set in C, f : D → C, f = u + iv

and z0 ∈ D. If all the first order partial derivatives of u and v exist, are continuous

and satisfy the Cauchy-Riemann relations at z0, then f is C-derivable at z0.

Let us define now the following two differential operators (sometimes called the

formal derivatives of f at the point z or the Wirtinger derivatives):

∂f

∂z=

1

2

(∂f

∂x− i

∂f

∂y

),

∂f

∂z=

1

2

(∂f

∂x+ i

∂f

∂y

).

(3.8)

These expressions are only symbolic derivatives with respect to z and z.

Remark 3.13 Let D ⊆ C be an open set, f : D → C, f = u+iv and z0 = (x0, y0) ∈D. Then, f is derivable at the point z0 if and only if u and v are R-differentiable

at (x0, y0) and∂f

∂z(z0) = 0. In other words, the Cauchy-Riemann equations are

equivalent to∂f

∂z(z0) = 0.

Also, if f is complex differentiable at z0, then, for any z0 ∈ C,

∂f

∂z(z0) = f

′

(z0).

The concept of C-differentiability at a given point is not enough for building an

interesting theory. Therefore, we shall be interested in dealing with functions which

are C-differentiable in a whole neighbourhood of a given point. A function f will be

called holomorphic at a point z0 ∈ C if it is C-differentiable in a neighbourhood of

z0 (this neighbourhood, usually, will remain unspecified). A function f will be said

to be holomorphic on a set D if it is C-differentiable in a neighbourhood of every

point in D. More precisely, we have the following definition.

Definition 3.14 Let D ⊆ C be an open nonempty set and f : D → C. The function

f is called holomorphic on D if it is C-differentiable at each point z0 ∈ D. The

50 COMPLEX ANALYSIS

function f is called holomorphic on an arbitrary set A ⊆ D if there exists an open set

D0, with A ⊆ D0 ⊆ D, such that f is holomorphic on D0. In particular, f : D → C

is holomorphic at a point z0 ∈ D if there exists r > 0 such that B(z0, r) ⊆ D and f

is holomorphic on B(z0, r).

We shall use the following notation:

H(D) = f : D → C | f is holomorphic on D.

Definition 3.15 Let f : C → C. If f is holomorphic on C, then f is said to be an

entire (integer) function.

Example 3.16 The function f : C → C, defined by f(z) = z3, is holomorphic on

C. So, it is an entire function.

Example 3.17 The function f : C → C, defined by f(z) = z, is not C-differentiable

anywhere in C, but it is R-differentiable everywhere on C.

Example 3.18 The function f(z) = 1/z is holomorphic on any open set in C that

does not contain the origin.

Example 3.19 Any polynomial function P = P (z) with complex coefficients is an

entire function.

Remark 3.20 If f is derivable at the point z0 and f′

(z0) 6= 0, then arg f′

(z0)

is the rotation angle of the tangent to a smooth path starting from z0 under the

transform f and |f ′

(z0)| is the linear deformation coefficient at this point (the linear

magnification ratio).

Remark 3.21 A function f is said to be holomorphic at infinity if the function

g(z) = f(1/z)

is holomorphic at z = 0. So, we can consider holomorphic functions in C∞.


Remark 3.22 Holomorphic functions are sometimes called regular functions. Since,

as we shall see later, holomorphic functions are analytic, the terms analytic, holo-

morphic or regular are often used interchangeably.

We shall consider now multi-valued functions. As we shall see, a multi-valued

function will be a map that assigns a finite or an infinite nonempty subset of C for

each argument in its domain of definition.

Definition 3.23 Let D be an open nonempty subset of C. We denote by P(C) the

set of all the parts (subsets) of C. A complex map F : D → P(C) is said to be a

multi-valued function on D.

Definition 3.24 Any continuous map f : D0 ⊆ D → C, with f(z) ∈ F (z), for

z ∈ D0, is called an uniform branch (or a determination) of the multi-valued function

F on D0. A point z0 ∈ C is called a critical point for the multi-valued function

F : D → P(C) if there exists no r > 0 such that on Ur(z0) there exists a uniform

branch of the function F .

Example 3.25 We have already seen that the argument function

Arg : C∗ → P(C)

is multi-valued, i.e.

Arg z = arg z + 2kπ | k ∈ Z ,

where we choose the principal argument arg z in (−π, π]. In fact, Arg : C∗ → P(R).

The argument function has an infinite number of branches. Unfortunately, if z

travels along a closed path winding around the branch point z = 0, the argument

changes by an integer multiple of 2π and we move from one branch to another.

Intuitively, if we want a branch to be a continuous single-valued function, we must

avoid working on a domain in which there are closed paths going around the origin.

Therefore, we can define a branch cut in the Argand plane, i.e. a line or a path in the

complex plane that will play the role of a barrier not to be crossed and preventing z

52 COMPLEX ANALYSIS

from making a complete loop around any branch point. In such a way, the function

will remain single-valued. For instance, a standard choice is to take the non-positive

real axis as a branch cut for the argument function (the origin is a branch point).

If we set C0 = C \ (−∞, 0], then arg : C0 → (−π, π) is a continuous map.

Moreover, if D is a domain in C0, any uniform branch f : D → C for Arg z is of the

following form:

f(z) = arg z + 2kπ,

with k ∈ Z fixed.

Remark 3.26 Let us notice that the point 0 ∈ C is a critical point for the multi-

valued function Arg.

Remark 3.27 Our choice to consider the principal branch of the argument function

defined by the restriction −π < arg z < π is not unique. Still, this is the most

common convention made in the literature for the argument and for many other

related functions, such as the logarithm function or the power function. Instead of

working with this particular branch cut obtained by removing from the complex plane

the non-positive real axis, one could consider, for instance, any radial cut from the

origin to infinity. When dealing with other functions, we need to consider one or

even more branch cuts to ensure single-valued definitions for different branches, the

choice of such cuts being sometimes a very difficult task.

We shall define now the multi-valued logarithm function as being the ”inverse”

of the exponential function. The exponential function is periodic, with complex

period 2πi. The complex exponential is not a one-to-one map. We can divide the

complex plane into horizontal strips of height 2π in such a way that in each strip

the exponential function is one-to-one. More precisely, for z = x+ iy ∈ C, we set

Sk = z = x+ iy ∈ C | x ∈ R, (2k − 1)π < y ≤ (2k + 1)π, k ∈ Z.

The sets Sk are called fundamental strips for the complex exponential function. It

follows that if z1 and z2 belong to the same strip Sk, then ez1 = ez2 implies that

z1 = z2.


From the periodicity of the complex exponential or, equivalently, from the fact

that the argument function is multi-valued, we shall see that the complex logarithm

is a multi-valued function. Also, since there is no solution for the equation exp(z) =

0, we shall not define the complex logarithm for the point 0. We shall use the

notation ln for the real logarithm and, respectively, log, for the complex logarithm

(there is no confusion because we work only with logarithms to base e).

Definition 3.28 The complex logarithm function Log : C∗ → P(C) is a multi-valued

map defined by

Log z = w ∈ C | ew = z .

Then,

Log z = ln |z|+ iArg z. (3.9)

The complex logarithm has an infinite number of branches. A standard choice is to

take the non-positive real axis as a branch cut for the logarithm function.

Let C0 = C \ (−∞, 0]. For z ∈ C0 and k ∈ Z, we set

logk(z) = ln |z|+ i (arg z + 2kπ),

where −π < arg z < π. The map

logk : C0 → C

is a uniform branch of the logarithm function. For k = 0, we get the principal branch

of the logarithm

log z = ln |z|+ i arg z. (3.10)

Moreover, if, for any k ∈ Z, we consider the so-called fundamental regions for the

exponential function

Tk = z = x+ iy | x ∈ R, (2k − 1)π < y < (2k + 1)π,

then the continuous map logk : C0 → Tk defined by

logk(z) = ln |z|+ i (arg z + 2kπ),

54 COMPLEX ANALYSIS

is a uniform holomorphic branch for the multi-valued logarithm function Log on C0

and for any z ∈ C0 and k ∈ Z, we have

(logk z)′

=1

z.

Example 3.29 For z = −5, we have

log (−5) = ln | − 5|+ i arg (−5) = ln 5 + i π,

Log (−5) = ln 5 + i π(2k + 1), k ∈ Z.

Example 3.30 For z = 1 + i, we have

log (1 + i) = ln√2 + i

π

4,

Log (1 + i) = ln√2 + i

(π4+ 2kπ

), k ∈ Z.

Let us note that

log (z w) 6= log z + logw, for any z, w ∈ C0.

As a matter of fact, there are complex numbers z and w for which log (z w) is not

even defined (for example, we can take z = w = i). For this particular example, z w

moved into the domain of a different holomorphic branch. Still,

Log (z w) = Log z + Logw, for any z, w 6= 0.

Remark 3.31 Let D be a simply connected domain with 1 ∈ D and 0 /∈ D. Then,

there exists in D a branch of the logarithm F (z), which is holomorphic in D and

for which eF (z) = z, for all z ∈ D. Each branch is an extension of the standard

logarithm defined for positive numbers.

Remark 3.32 Let us notice that we can make the logarithm function single-valued

in other regions of the complex plane by choosing a different branch for the argument

function. Typically, we can choose such a set as being the complement of a ray or


a path in the complex plane going from the origin (inclusive) to infinity in some

direction. In fact, such a line or a path which creates a domain of holomorphy for

a given function is called a branch cut and any point that lies on a branch cut is

called a branch point of that multi-valued function.

With the aid of the logarithm function, we can define complex powers of complex

numbers. Let α ∈ C. For any z 6= 0, we define the α-th power of z by

zα = eαLog z,

i.e.

zα = eα [ln |z|+i (arg z+2kπ)], k ∈ Z.

This is, in general, a multi-valued function. For each branch of the logarithm, we

obtain a branch of zα. Moreover, for any k ∈ Z, the map fk : C0 → C, defined by

fk(z) = eα logk(z),

is a holomorphic uniform branch of the map zα and

d

dz(zα) = αzα−1.

Example 3.33 Let us compute 1i. Since Log 1 = ln 1+2kπi, i.e. Log 1 = 2kπi, we

get

1i = e−2kπ, k ∈ Z.

If the exponent α is an integer, we get the standard concept of a power function.

If α is rational, i.e. α = m/n, with m and n > 0 integers with no common factors,

there are exactly n values of zα for each z ∈ C∗. So, we have n branches of the

power function zα. In particular, for α =1

n, with n ∈ N

∗, there are only n distinct

uniform branches of the multi-valued power function z1/n. In this case, we writen√z. If α is irrational or non-real, there are infinitely many branches of the complex

power function. If α > 0, the power function is also defined for z = 0. Finally, if

α = 0, the power function is defined as being the constant 1.

56 COMPLEX ANALYSIS

Exercise 3.34 Show that the function f : C → C, defined by f(z) = z2, is C-

derivable at any point z0 ∈ C.

Solution. For z0 ∈ C, we have

limz→z0

z2 − z20z − z0

= 2z0.

Therefore,

f ′(z0) = 2z0.

Exercise 3.35 Prove that the function f : C → C, defined by f(z) = z is not

complex differentiable everywhere on C.

Exercise 3.36 Prove that the function f : C → C, given by f(z) = (z)2 is differen-

tiable only at the point z = 0.

Exercise 3.37 Show that the exponential function exp : C → C defined by f(z) =

ex(cos y + i sin y), with z = x+ i y, is analytic in C.

Exercise 3.38 Find the points at which the function f : C → C, given by f(z) =

x2 + i y2 is complex differentiable.


z Im z is complex differentiable.


2xy − i (x+ y)2 is complex differentiable.

Exercise 3.41 Prove that if f(z) and f(z) are both holomorphic in the region

D ⊆ C, then f is constant in D.

Exercise 3.42 Let D ⊆ C be an open nonempty set and z0 ∈ D. If f, g : D → C

are derivable at z0, prove that:

1) f ± g is derivable at z0 and (f ± g)′

(z0) = f′

(z0)± g′

(z0);


2) fg is derivable at z0 and (fg)′

(z0) = f′

(z0) g(z0) + f(z0) g′

(z0);

3) if g(z0) 6= 0, then f/g is derivable at z0 and

(f

g

)′

(z0) =f

′

(z0) g(z0)− f(z0) g′

(z0)

g2(z0).

Exercise 3.43 Let D,E ⊆ C be open sets, z0 ∈ D and f : D → E, g : E → C. If f

is derivable at z0 and g is derivable at f(z0), show that g f is derivable at z0 and

we have

(g f)′(z0) = g′

(f(z0)) f′

(z0).

Exercise 3.44 Show that the function f : C → C, defined by f(z) = z, does not

satisfy the Cauchy-Riemann equations.

Solution. Indeed, since

u(x, y) = x, v(x, y) = −y,

it follows that∂u

∂x= 1,

while∂v

∂y= −1.

So, this function, despite the fact that it is continuous everywhere on C, it is R-

differentiable on C, is nowhere C-derivable.

Exercise 3.45 Show that the function f(z) = ez satisfies the Cauchy-Riemann

equations.

Solution. Indeed, since

ez = ex (cos y + i sin y),

it follows that

u(x, y) = ex cos y, v(x, y) = ex sin y

58 COMPLEX ANALYSIS

and∂u

∂x= ex cos y =

∂v

∂y;

∂u

∂y= −ex sin y = −∂v

∂x.

Moreover, ez is complex derivable and, using (3.5), it follows immediately that its

complex derivative is ez.

Exercise 3.46 Prove that the function f : C → C, defined by

f(z) =

z5

|z|4 z 6= 0,

0, z = 0,

is continuous at z = 0, satisfies the Cauchy-Riemann equations at z = 0, but is not

C-derivable at z = 0.

Exercise 3.47 Prove that the function f : C → C, defined by f(z) =√

|xy|, wherez = x+ i y, is not differentiable at the origin.

Exercise 3.48 Prove that the function f : C → C, defined by f(z) =√

|xy|, wherez = x+ i y, satisfies the Cauchy-Riemann equations at the origin.

Exercise 3.49 Find the values of the complex numbers a, b and c for which the

function

f(z) = ax+ by + i(cx+ y)

is C-derivable on C.

Exercise 3.50 Let f be holomorphic on a domain D ⊆ C. Show that each of the

following conditions implies that f is constant on D:

(i) f ′ = 0 in D;

(ii) |f | is constant in D;

(iii) f is real-valued in D.

(iv) arg f is constant in D.

Are the above results still true if we replace D by any open subset of C?


Exercise 3.51 Let f(z) = z e−|z|2. Find the points at which f is derivable and

compute its derivative at these points.

Exercise 3.52 Let f be an entire function such that its real and imaginary parts

u and v satisfy the condition u(z) v(z) = 2, for all z. Prove that f is constant.

Exercise 3.53 Find all the complex numbers for which the function f(z) = (z − 2)i

is holomorphic.

Exercise 3.54 Find the solutions to the following equations:

a) ez = π i; b) sinh z = 0; c) cos z = 0.

3.2 Harmonic Functions

Definition 3.55 A real function u : D ⊆ R2 → R of class C2(D) is said to be

harmonic on the domain D if it satisfies Laplace’s equation ∆u = 0 in D, i.e., for

any (x, y) ∈ D,∂2u

∂x2+

∂2u

∂y2= 0.

If f = u + iv is holomorphic on a domain D, then f is infinitely differentiable on

D, and, hence, so are u and v. In particular, u, v ∈ C2(D). However, here we only

assume this in order to state the following result.

Proposition 3.56 Let f(z) = u(x, y)+iv(x, y) be holomorphic on a domain D ⊆ C.

If u, v ∈ C2(D), then u and v are both harmonic functions on D.

Example 3.57 The function u : R2 → R, defined by u(x, y) = x2 + y2, is not

harmonic and, so, it cannot be the real part of a holomorphic function.

Definition 3.58 Two harmonic functions u and v defined on a domain D ⊆ C and

related through Cauchy-Riemann relations are called harmonic conjugated.

Remark 3.59 If f = u+ iv is holomorphic, then u and v are harmonic conjugated.

60 COMPLEX ANALYSIS

Theorem 3.60 Let D ⊆ C be a simply connected domain and let u = u(x, y) be

a real-valued harmonic function on D. Then, there exists a real-valued function

v = v(x, y), defined up to an arbitrary constant, such that the complex function

f(z) = u(x, y) + iv(x, y)

is holomorphic on D. So, every harmonic function u(x, y) defined on a simply

connected domain is the real part of a complex-valued function f(z) = u(x, y) +

iv(x, y), defined for all z = x + iy ∈ D. In other words, for each function u =

u(x, y) that is harmonic in a simply connected domain D, there exists its harmonic

conjugated function v = v(x, y), defined up to an arbitrary constant term.

Exercise 3.61 Show that the function u : R2 → R, defined by u(x, y) = y3 − 3x2y,

is harmonic on R2.

Solution. It is not difficult to see that

∂2u

∂x2= −6y;

∂2u

∂y2= 6y.

So, ∆u = 0.

Exercise 3.62 Prove that the function u : R2 → R, defined by u(x, y) = ex cos y, is

harmonic on R2.

Solution. Since

∂2u

∂x2= ex cos y,

∂2u

∂y2= −ex cos y,

it follows immediately that

∆u = 0.

Exercise 3.63 Show that the function u : R2 → R, given by u(x, y) = e−x (x sin y − y cos y),

is harmonic on R2.


Solution. It is not difficult to check that

∂2u

∂x2= −e−x (2 sin y − x sin y + y cos y) ,

∂2u

∂y2= e−x (2 sin y − x sin y + y cos y) .

Therefore,

∆u = 0.

Exercise 3.64 Find an entire function f : C → C for which

Re f(z) = x2 − y2

and

f(0) = 3i.

Solution. We start by noticing that u : R2 → R, given by u(x, y) = x2 − y2, is

harmonic on R2. Therefore, there exists v(x, y) such that f(z) = u(x, y) + iv(x, y)

is holomorphic on C. Let (x0, y0) ∈ R2. Using Cauchy-Riemann equations, we get

v(x, y) =

(x,y)∫

(x0,y0)

−∂u

∂ydx+

∂u

∂xdy + C1, with C1 ∈ R,

along any path joining the points (x0, y0) and (x, y). Integrating along a particular

path consisting of two line segments parallel to the coordinate axes, we obtain

v(x, y) = 2xy + C, with C ∈ R.

So,

f(z) = x2 − y2 + i (2xy + C), with C ∈ R,

i.e.

f(z) = z2 + iC, with C ∈ R.

Imposing the condition f(0) = 3i, we get

f(z) = z2 + 3i.

62 COMPLEX ANALYSIS

Exercise 3.65 Let u(x, y) = a x2 + b x y + c y2, where a, b, c ∈ R. Prove that u is

harmonic if and only if a = −c.


4∂

∂z

∂

∂z= 4

∂

∂z

∂

∂z= ∆.

Exercise 3.67 Prove that the functions

a) u(x, y) = e−x sin y,

b) v(x, y) = sin x sinh y

are harmonic. Find the corresponding holomorphic function f = u+ iv in each case.

Exercise 3.68 Prove that the functions

a) u(x, y) = x2 − y2 + 2x− 4 y,

b) v(x, y) = cosh y sin x

are harmonic. Find the corresponding holomorphic function f = u+ iv in each case.

Exercise 3.69 Let v : R2 → R, defined by

v(x, y) = 3 + 4xy.

Find all the entire functions f : C → C such that Im f = v.

Exercise 3.70 Find the harmonic conjugates of the function v : R2 → R, defined

by

v(x, y) = x2 + 3xy − y2 + x+ 4y.

Exercise 3.71 Let u(x, y) = ln(x2 + y2

). Show that u is harmonic on C

∗.

Exercise 3.72 Prove that if f is holomorphic and nonzero in a region D ⊆ C, then

ln |f(x, y)| is harmonic in D.

Chapter 4

Complex Integration

4.1 The Complex Integral

Definition 4.1 A continuous function γ : [a, b] ⊆ R → C is called a path in C.

Remark 4.2 Sometimes, it is convenient to take γ : [0, 1] → C. Also, we can deal

with paths γ : [a, b] → C∞.

The points γ(a) and γ(b) are said to be the endpoints of the path γ; γ(a) is called

the initial point (or the start point) of the path γ and γ(b) is said to be its terminal

point.

Definition 4.3 We say that a path γ : [a, b] → C lies in a set D ⊆ C if γ(t) ∈ D,

for all t ∈ [a, b].

We denote the image of the map γ by γ or by γ([a, b]), but we shall often prefer

to refer to either the function or the image of the path with the same symbol γ.

Definition 4.4 A path γ : [a, b] → C is called closed if γ(a) = γ(b).

Definition 4.5 A path γ : [a, b] → C is called smooth if γ is differentiable, γ′

is

continuous and γ′

(t) 6= 0. A path γ is called piecewise smooth if it is continuous

and splits into a finite number of smooth pieces with no common interior points.

63

64 COMPLEX ANALYSIS

Definition 4.6 A path without self-intersection points is said to be simple. A simple

closed path is called a Jordan path or, sometimes, a loop.

Theorem 4.7 (Jordan Curve Theorem) Let γ be the image of a simple closed path

in C. The set C \ γ has exactly two connected components. One of these, known as

the interior, is bounded and the other one, known as the exterior, is unbounded.

So, any loop in the complex plane separates the plane into two domains having the

loop as common boundary. One domain, the interior, is bounded, while the other

one, the exterior, is unbounded.

Example 4.8 The line segment joining the points a and b in C, with a 6= b, is

defined as being:

γ : [0, 1] → C,

with

γ(t) = (1− t)a+ tb

or

γ(t) = a+ t(b− a).

Obviously, γ is smooth. Also, γ is a simple path.

Example 4.9 The circle of radius a, centered at the origin, is defined as being:

γ : [0, 2π] → C,

with

γ(t) = a eit,

for t ∈ [0, 2π]. It is easy to see that γ is a smooth closed path.

Example 4.10 The path γ : [−π, π] → C, given by γ(t) = cos t, is not a simple

path.

COMPLEX INTEGRATION 65

Definition 4.11 The path γ−1 : [a, b] → C, given by

γ−1(t) = γ(a+ b− t),

is called the inverse or the opposite path of γ.

If we take γ : [0, 1] → C, the inverse path γ−1 : [0, 1] → C is given by

γ−1(t) = γ(1 − t).

Example 4.12 A simple example of a smooth path is offered by the circle centered

at a point z0 ∈ C and having the radius r > 0, i.e. the set

C(z0, r) = z ∈ C : |z − z0| = r.

Its positive orientation (counterclockwise) is given by the following parametrization:

γ(t) = z0 + reit, t ∈ [0, 2π],

while its negative orientation (clockwise) is given by

γ(t) = z0 + re−it, t ∈ [0, 2π].

Remark 4.13 In what follows, if not otherwise mentioned, we shall work only with

simple positively oriented paths.

Definition 4.14 Let γ1, γ2 : [0, 1] → C be two paths such that γ1(1) = γ2(0). We

define their compound path (alternatively called the sum, the concatenation or the

composition of the paths γ1 and γ2) as being

γ : [0, 1] → C,

with

γ(t) = (γ1 ∨ γ2)(t) =

γ1(2t), 0 ≤ t ≤ 1

2,

γ2(2t− 1),1

2≤ t ≤ 1.

66 COMPLEX ANALYSIS

In a similar manner, if γ1 : [a, b] → C and γ2 : [c, d] → C are two paths such that

γ1(b) = γ2(c), then their sum can be defined as being

γ : [a, b+ (d− c)] → C,

with

γ(t) = (γ1 ∨ γ2)(t) =

γ1(t), t ∈ [a, b],

γ2(t− b+ c), t ∈ [b, b+ (d− c)].

Remark 4.15 Let us notice that γ ∨ γ−1 is a closed path.

Sometimes, a sequence of smooth paths γ1, γ2, . . . , γn such that the terminal point

of γk coincides with the initial point of γk+1, for 1 ≤ k ≤ n− 1, is called a contour.

A contour γ is said to be closed if its endpoints coincide. A closed contour is also

called, in some textbooks, a loop. Let us mention that some authors use the term

contour for a closed piecewise smooth path (what we call here a closed contour).

Definition 4.16 Let γ : [a, b] → C and γ : [α, β] → C be two smooth paths. We shall

say that γ and γ are equivalent (we denote γ ∼ γ) if there exists ϕ : [a, b] → [α, β]

bijective, of class C1, with ϕ′

(t) 6= 0, such that

γ = γ ϕ,

i.e., for any t ∈ [a, b],

γ(t) = γ(ϕ(t)).

Remark 4.17 The above relation is an equivalence relation on the set of all the

paths of class C1. Each corresponding equivalence class is called a curve.

We can define also an equivalence relation for oriented paths. Indeed, by asking that

ϕ′

(t) > 0, we get equivalent paths having the same orientation. Equivalent paths

have the same image and, moreover, equivalent oriented paths are traversed in the

same direction.


Definition 4.18 Let γ : [a, b] → C be a smooth path. The length of the path γ is

defined as being

l(γ) =

b∫

a

|γ′

(t)| dt.

This definition is independent of the parametrization of the path γ.

Remark 4.19 If γ is only a piecewise smooth path, then its length is the sum of

the lengths of its smooth parts.

Definition 4.20 Let D ⊆ C be a domain and γ1, γ2 : [0, 1] → D. A map ϕ :

[0, 1]× [0, 1] → D is called a continuous deformation of γ1 into γ2 if ϕ is continuous

and, for any t ∈ [0, 1],

ϕ(0, t) = γ1(t),

ϕ(1, t) = γ2(t).

Definition 4.21 Let D ⊆ C be a domain and γ1, γ2 : [0, 1] → D. The paths γ1

and γ2 are called homotopic in D (we shall denote this fact by γ1 ≈ γ2) if we can

continuously deform γ1 into γ2, without leaving D.

A closed path γ : [0, 1] → D is said to be homotopic to zero (we shall write γ ≈ 0)

in D if γ can be continuously deformed into a point, without leaving D.

The continuous map ϕ is also called a homotopy between the path γ1 and γ2. It is

not difficult to see that homotopy defines an equivalence relation.

Definition 4.22 Let γ : [a, b] → C be a smooth path and f : γ([a, b]) → C a

continuous function. The complex integral of the function f along the path γ is

defined as follows:∫

γ

f(z) dz =

b∫

a

f(γ(t)) γ′

(t) dt. (4.1)

Remark 4.23 If γ is a piecewise smooth path, then the integral of f along γ is just

the sum of the integrals of f along the smooth parts of γ.

68 COMPLEX ANALYSIS

Proposition 4.24 1. (Linearity) If the functions f and g are continuous on a

piecewise smooth path γ, then, for any complex numbers α and β, we have

∫

γ

(αf + βg)(z) dz = α

∫

γ

f(z) dz + β

∫

γ

g(z) dz.

2. (Additivity) Let γ1 and γ2 be two piecewise smooth paths and f a continuous

function on their union γ1 ∨ γ2. Then,

∫

γ1∨γ2

f(z) dz =

∫

γ1

f(z) dz +

∫

γ2

f(z) dz.

3. (Invariance) The complex integral is invariant under a re-parametrization of

the integration path.

4. (Orientation) Let γ be a piecewise smooth path. If f is continuous on γ, then

∫

γ−1

f(z) dz = −∫

γ

f(z) dz.

5. The modulus of the integral of a complex function f along a path γ is bounded

by the integral of its modulus with respect to the arc length, i.e.

∣∣∣∣∣∣

∫

γ

f(z) dz

∣∣∣∣∣∣≤∫

γ

|f(z)| ds.

If, in addition, we suppose that there exists a constant M ≥ 0 such that |f(z)| ≤ M

on γ, then ∣∣∣∣∣∣

∫

γ

f(z) dz

∣∣∣∣∣∣≤ Ml(γ).

Definition 4.25 Let D be a nonempty open subset of C. A function F : D → C is

called a primitive or an anti-derivative of f : D → C if it is holomorphic on D and

F′

(z) = f(z), for any z ∈ D.


Remark 4.26 A primitive of a continuous function f on a domain D, if it exists,

is defined up to an arbitrary constant. So, any two primitives of f on D, if they

exist, differ by a constant.

Theorem 4.27 (The Fundamental Theorem of Calculus for Path Integrals) Let D

be a domain in C and f : D → C a continuous function which possesses an anti-

derivative F : D → C. If γ : [a, b] → D is a smooth path contained in D, then

∫

γ

f(z) dz = F (γ(b)) − F (γ(a)). (4.2)

Corollary 4.28 If γ is a closed path in a domain D ⊆ C and f is continuous and

has an anti-derivative in D, then

∫

γ

f(z) dz = 0.

Remark 4.29 Using this result, we can prove that the function f(z) = 1/z does not

have a primitive in the open set C \ 0. Indeed, if we take γ as being the positively

oriented unit circle |z| = 1, given by γ(t) = eit, with t ∈ [0, 2π], we get

∫

γ

f(z) dz =

2π∫

0

i eit

eitdt = 2πi 6= 0.

Exercise 4.30 Find a parametrization for the circle C(1 + i, 1), oriented counter-

clockwise.

Exercise 4.31 Find a parametrization for the line segment from 1− i to 3 i.

Exercise 4.32 Find a parametrization for the he rectangle with vertices at the

points 1± 3 i, oriented counter-clockwise.

Exercise 4.33 Show that any two positively oriented circles C(z0, r) and C(w0, R)

are homotopic in C.

70 COMPLEX ANALYSIS

Exercise 4.34 Show that the positively oriented circles C(0, 2) and C(2, 4) are not

homotopic in C \ 0.

Exercise 4.35 Show that in a simply connected domain, any two paths with com-

mon endpoints are homotopic and any closed path is homotopic to zero.

Exercise 4.36 Compute the integral of the function f : C → C given by f(z) = z2

over the line segment from 0 to 1 + i.

Solution. A parametrization of this path is γ : [0, 1] → C, γ(t) = t+ i t. Therefore,

I =

1∫

0

(1− i t)2 (1 + i) dt =2

3(1− i) .

Exercise 4.37 Compute the integral

I =

∫

γ

z dz,

where γ is the line segment [−i, i].

Solution. The path γ can be parameterized as γ : [0, 1] → C, with γ(t) = (1 −t)(−i) + ti, i.e.

γ(t) = (2t− 1)i, with t ∈ [0, 1].

Obviously, γ is a smooth path and f(z) = z is continuous along γ. Then, we have

I =

1∫

0

[−(2t− 1) i 2i] dt =

1∫

0

(4t− 2) dt = 0.

Exercise 4.38 Evaluate the integral

I =

∫

γ

z 2 dz,

where γ is the circle |z − 1| = 1.


Solution. The path γ can be represented as γ : [0, 2π] → C, with γ(t) = 1 + eit.

Obviously, γ is a smooth path and f(z) = z 2 is continuous along γ. Then, we get

I =

2π∫

0

(1 + e−2it + 2e−it

)i eit dt =

2π∫

0

(i eit + ie−it + 2i

)dt = 4πi.


In =

∫

γ

(z − a)n dz,

where n ∈ Z and γ is the circle |z − a| = r, with a ∈ C and r > 0.

Solution. The path γ can be represented as γ : [0, 2π] → C, with γ(t) = a+ reit. So,

γ is a smooth path and f(z) = (z − a)n is continuous along γ. We have

I =

2π∫

0

rn eint ri eitdt =

2π∫

0

rn+1 i ei(n+1)t dt.

For n 6= −1, we obtain

In = 0,

while for n = −1, we get

I−1 = 2πi.

Exercise 4.40 Evaluate the following integral:

I =

∫

γ

z2 dz,

where γ is the quarter of the unit circle lying in the first quadrant (traversed coun-

terclockwise).

Solution. We can parametrize the path γ as being γ(t) = eit, with t ∈[0,

π

2

]. Using

this parametrization, we get:

I =

π/2∫

0

e2it i eit dt = −1

3(1 + i).

72 COMPLEX ANALYSIS

Exercise 4.41 Let |a| < r < |b|. Show that

∫

γ

1

(z − a)(z − b)dz =

2πi

a− b,

where γ is the positively oriented circle |z| = r.


I =

∫

γ

(z2 + 3 z) dz,

where γ is the path connecting the origin to 1 + 2i along a straight line.


I =

∫

γ

(z3 + 1) dz,

along the path γ : [0, π] → C, defined by

γ(t) = eit.

Solution. Since the function

F (z) =z4

4+ z

is a primitive of the continuous function f(z) = z3 +1 in C and γ is a smooth path,

it follows that

I = F (γ(π)) − F (γ(0)),

i.e.

I = −2.


4.2 Cauchy’s Theorem. Cauchy’s Integral Formula. Ap-

plications

Theorem 4.44 (Cauchy) Let f be a holomorphic function in a simply connected

domain D ⊆ C and γ a closed path in D. Then,∫

γ

f(z) dz = 0. (4.3)

Theorem 4.45 (Path Independence Theorem) Let D ⊆ C be a simply connected do-

main. If f is holomorphic in D and γ1 and γ2 are piecewise regular paths connecting

two given points z1 and z2 in D, then∫

γ1

f(z) dz =

∫

γ2

f(z) dz.

Theorem 4.46 (Cauchy) Let D be an open nonempty set in C. If f is holomorphic

in D and γ is a piecewise smooth closed path that is homotopic to zero in D, then∫

γ

f(z) dz = 0.

Theorem 4.47 (Deformation Theorem) Let D be an open nonempty set in C. If

f is holomorphic in D and γ1 and γ2 are closed piecewise regular paths that are

homotopic in D, then ∫

γ1

f(z) dz =

∫

γ2

f(z) dz.

Theorem 4.48 Any function f which is holomorphic in a simply connected domain

D possesses an anti-derivative in this domain.

If we consider the function f = 1/z, which is holomorphic in the annulus U =

z ∈ C | 0 < |z| < 2, we can see immediately that, in Theorem 4.44, the assumption

that D is simply connected is essential: the global existence theorem of an anti-

derivative does not hold in general for multiply connected domains. The same

example shows that the integral of a holomorphic function over a closed path in a

multiply connected domain might not vanish.

74 COMPLEX ANALYSIS

Proposition 4.49 Let D be a domain in C and f : D → C be a continuous function.

Then, if ∫

γ

f(z) dz = 0

for any closed path γ in D, f admits an anti-derivative on D.

Remark 4.50 (Morera) a) If f is a continuous complex function on a domain D

and

∫

γ

f(z) dz = 0 for every closed contour γ in D, then f is holomorphic on D.

b) If a complex function f possesses an anti-derivative on a domain D, then f

is holomorphic on D.

To summarize, if D is a domain in C and f is a continuous function on D, then

the following three statements are equivalent:

a) f has an antiderivative in D;

b) The integral

∫

γ

f(z) dz vanishes for all piecewise smooth closed paths γ

contained in D;

c) The integral

∫

γ

f(z) dz is path-independent.

Remark 4.51 (Extended Cauchy-Goursat Theorem) Let Γ and γ1, . . . , γn be simple

closed positively oriented contours such that all the paths γi lie in the interior of Γ.

Assume that the interiors of γi and γj are disjoint sets, for all i 6= j. If f is

holomorphic on a domain containing the paths Γ and γi, for i = 1, n, and the points

between them, then ∫

Γ

f(z) dz =

n∑

i=1

∫

γi

f(z) dz.

Theorem 4.52 (Cauchy’s Integral Formula) Let f be a holomorphic function in

a simply connected domain D and γ a positively oriented simple closed path in D.


Then, for any point z in the interior of the domain bounded by γ, we have

f(z) =1

2πi

∫

γ

f(ξ)

ξ − zdξ. (4.4)

Let us notice that (4.4) is a kind of a ”holographic” formula, stating that a

holomorphic function in a simply connected domain is determined by its behaviour

on the boundary of the domain. As a consequence, it follows that if two holomorphic

functions coincide on the boundary of a simply connected domain, they coincide

everywhere in the domain.

Remark 4.53 Formula (4.4) gives a representation of a holomorphic function in

the closure of a domain in terms of an integral over the boundary of the domain. In

other words, if f is holomorphic in the closure of a domain Ω that is bounded by a

finite number of continuous paths, then, at any point z ∈ Ω, f can be represented as

f(z) =1

2πi

∫

∂Ω

f(ξ)

ξ − zdξ,

where ∂Ω is the positively oriented boundary of Ω. If the point z lies outside Ω, the

functionf(ξ)

ξ − zis holomorphic in Ω and we have

1

2πi

∫

∂Ω

f(ξ)

ξ − zdξ = 0.

Remark 4.54 It is not difficult to see that we can extend Cauchy’s formula (5.4)

for the case in which the closed contour γ is the oriented boundary of a multiple

connected domain D. For instance, if ∂D = γ1 ∨ γ2, then

f(z) =1

2πi

∫

γ1

f(ξ)

ξ − zdξ − 1

2πi

∫

γ2

f(ξ)

ξ − zdξ, z ∈ D.

Proposition 4.55 If f is holomorphic in a simply connected domain D ⊆ C and γ

is a positively oriented simple closed path in D, then f is indefinitely differentiable

76 COMPLEX ANALYSIS

in D and, for any z ∈ D \ γ and n ∈ N, we have

f (n)(z) =n!

2πi

∫

γ

f(ξ)

(ξ − z)n+1dξ. (4.5)

Definition 4.56 Let γ be a piecewise smooth path in C. If z0 ∈ C \ γ, we can

define the so-called index of γ with respect to z0 as being

n(γ, z0) =1

2πi

∫

γ

dξ

ξ − z0.

It can be proven that if γ is a closed path in C, its index with respect to a point

z0 ∈ C \ γ is an integer. This integer is also known as the winding number of the

closed path γ about the point z0. Intuitively, the index gives the number of times

a closed contour γ winds counterclockwise around a point z0. The sign of the index

is determined by the orientation we choose on the path.

It is not difficult to see that the index n(γ, z0) has the following properties:

1) if γ1 and γ2 are homotopic paths in C \ z0, then n(γ1, z0) = n(γ2, z0);

2) n(γ−1, z0) = −n(γ, z0);

3) if γ = γ1 ∨ γ2, then n(γ, z0) = n(γ1, z0) + n(γ2, z0).

Theorem 4.57 Let f be a holomorphic function on an open set D ⊆ C and γ a

closed contour which is homotopic to zero in D. Then, f is indefinitely derivable on

D and, for any z ∈ D \ γ and n ∈ N, we have

n(γ, z) f (n)(z) =n!

2πi

∫

γ

f(ξ)

(ξ − z)n+1dξ. (4.6)

Theorem 4.58 (Generalized Cauchy’s Theorem) Let D be a domain in C and let

f be holomorphic on D. Let γ1, . . . , γn be n closed paths in D. If

n(γ1, z) + · · ·+ n(γn, z) = 0,


for all z /∈ D, then ∫

γ1

f(z) dz + · · ·+∫

γn

f(z) dz = 0.

Proposition 4.59 (Cauchy’s Inequalities) Let z0 ∈ C, r ∈ R+ and f holomorphic

on an open set G ⊇ B(z0, r). If

M = sup|f(z)| | z ∈ B(z0, r),

then, for any n ∈ N, we have

∣∣∣f (n)(z0)∣∣∣ ≤ n!

M

rn. (4.7)

Theorem 4.60 (Liouville) If f is holomorphic on C and there exists a positive

number M such that |f(z)| ≤ M for any z ∈ C, then f is constant on C.

Theorem 4.61 (The Fundamental Theorem of Algebra) Any polynomial

P (z) = a0 + a1z + · · ·+ anzn,

with ai ∈ C, an 6= 0 and n ≥ 1, has at least one root in C.

Exercise 4.62 Show that if γ is a simple closed path in C and a is a point not lying

on γ, then∫

γ

1

z − adz =

2πi, a is inside γ,

0, a is outside γ.

Solution. Note that f(z) = 1/(z − a) is holomorphic everywhere, except at z = a,

where it has, as we shall see later, a simple pole. If a lies outside γ, then, fromCauchy’s theorem, the integral is zero. If a is inside γ, then the integral is equal to

the integral along a suitable chosen circle centered at z = a, which can be computed

directly by using the parametrization γ(t) = a+ reit, with t ∈ [0, 2π] and r > 0. As

we saw in a previous example, the value of the integral in this case is 2πi.

78 COMPLEX ANALYSIS

Exercise 4.63 Prove that the integral

∫

γ

sin z dz = 0, for any Jordan curve γ in

the complex plane.

Solution. Since the function f(z) = sin z is entire and γ is a Jordan path, using

Cauchy’s theorem, it follows immediately that our integral is equal to zero.


∫

γ

1

(z2 + 4)(z2 + 9)dz = 0,

where γ is the positively oriented circle |z| = 1.

Solution. The zeros of the polynomial (z2+4)(z2+9) are ±2i and, respectively, ±3i

and they lie outside the disk D = B(0, 1). Therefore, the function

f(z) =1

(z2 + 4)(z2 + 9)

is holomorphic on D and, since the path γ is smooth and closed, it follows that the

integral is zero.

Exercise 4.65 Show that if f is holomorphic in a domain D and its derivative is

zero on D, then f is constant on D.

Exercise 4.66 Prove that if f is holomorphic in the annulus

U = z ∈ C | r < |z − a| < R,

then the integral

I =

∫

|z−a|<ρ

f(z) dz

has the same value for all ρ with r < ρ < R.


Exercise 4.67 Compute the value of the integral

I =

∫

γ

z + 1

z2(z − 1)dz,

where γ is the circle |z − 2| =√2, traversed counterclockwise.

Solution. We note thatz + 1

z2(z − 1)=

f(z)

z − 1,

where

f(z) =z + 1

z2.

The function f is holomorphic in the disk B(2,√2), its only singularity, at z = 0,

lying outside the closed contour γ. Therefore, using Cauchy’s integral formula (...),

we get

I = 2πi f(1) = 4πi.


I =

∫

γ

z − 1

z2(z − 2)dz,

where γ is the circle |z| = 1, traveled counterclockwise.

Solution. We remark thatz − 1

z2(z − 2)=

f(z)

z2,

where

f(z) =z − 1

z − 2.

The function f is holomorphic in the disk B(0, 1). Using Cauchy’s integral formula

(5.5), we get

I = 2πif′

(0) = −πi

2.

80 COMPLEX ANALYSIS


I =

∫

γ

sinh z

(z − πi)4dz,

where γ is the positively oriented circle |z − 2i| = 3.

Solution. Since

f(z) = sinh z

is holomorphic on the disk B(2i, 3), using Cauchy’s integral formula (5.5), we get

I =2πi

3!f

′′′

(πi) = −πi

3.


I =

∫

γ

ez

z2 − 2zdz,

where γ is the positively oriented circle C(0, 4).

Solution. Since

I =1

2

∫

γ

ez

z − 2dz − 1

2

∫

γ

ez

zdz,

it follows immediately that

I =1

22π ie2 − 1

22π ie0 = π i

(e2 − 1

).

Exercise 4.71 Find the length of the following paths:

a) the positively oriented circle C(1 + 2 i, 1);

b) the line segment from −1 + 2 i to 3 i;

c) the rectangle with vertices ±2± 3 i.


Exercise 4.72 Evaluate the integral I =

∫

γe4z dz for each of the following paths:

a) γ is the line segment from 2 to 3 i;

b) γ is the positively oriented circle C(0, 2).

Exercise 4.73 Show that ∫

γz3 ez

2dz = 0

for any closed piecewise smooth path γ.

Exercise 4.74 Compute the following integrals:

a) I =

∫

C(0,2)

ez

z (z − 7)dz;

b) I =

∫

C(0,8)

ez

z (z − 7)dz.

Exercise 4.75 Show that ∫

C(0,3)

1

z2 + 1dz = 0.

Chapter 5

Taylor and Laurent Series

5.1 Sequences and Series of Functions. Power Series

Definition 5.1 Let us consider a sequence of complex functions (fn)n≥0, defined on

a nonempty set D ⊆ C. The sequence (fn)n≥0 is said to be pointwise convergent at

the point z0 ∈ D if the number sequence (fn(z0))n≥0 is convergent.

Definition 5.2 The sequence (fn)n≥0 is said to be pointwise convergent on the set

D if (fn)n≥0 is convergent at any point z ∈ D. In this case, we can define a function

f : D → C, by putting, for z ∈ D,

f(z) = limn→∞

fn(z).

The function f is called the limit of (fn)n≥0 on D. We shall use the notation

fn → f

to denote the fact that f is the limit of the sequence (fn)n≥0.

Definition 5.3 The sequence (fn)n≥0 is said to be uniformly convergent on D to f

if for any ε > 0 there exists Nε ∈ N such that, for any z ∈ D,

|fn(z)− f(z)| < ε,

83

84 COMPLEX ANALYSIS

for any n ≥ Nε. We shall write

fnu−→ f

to denote the fact that the sequence (fn)n≥0 is uniformly convergent to f .

Remark 5.4 Of course, uniform convergence implies pointwise convergence.

Proposition 5.5 Let fn : D ⊆ C → C, with n ≥ 0, be a pointwise convergent

sequence and let f(z) = limn→∞

fn(z). If

an = supz∈D

|fn(z)− f(z)|

and

limn→∞

an = 0,

then (fn)n≥0 is uniformly convergent to f on D.

Proposition 5.6 If the sequence (fn)n is uniformly convergent to f on D and all

the maps fn are continuous at z0 ∈ D, then f is continuous at z0.

Remark 5.7 Let us notice that if the sequence (fn)n is uniformly convergent to f

on D and all the maps fn are continuous at z0 ∈ D, then

limz→z0

limn→∞

fn(z) = limn→∞

limz→z0

fn(z).

Proposition 5.8 Let (fn)n≥0 be a sequence of continuous functions defined on a

domain D containing a contour γ. If (fn)n≥0 converges uniformly to f on D, then

limn→∞

∫

γ

fn(z) dz =

∫

γ

limn→∞

fn(z) dz =

∫

γ

f(z) dz.

Definition 5.9 Let fn : D → C, with n ≥ 0, and Sn : D → C given by

Sn(z) =n∑

i=0

fi(z). (5.1)

The pair ((fn)n≥0, (Sn)n≥0) is said to be a series of complex functions (the series

associated to the sequence (fn)n≥0).

TAYLOR AND LAURENT SERIES 85

We shall denote such a series by∑

n≥0

fn. Sn is called the nth partial sum of the series

∑

n≥0

fn and fn is said to be its general term.

Definition 5.10 The series∑

n≥0

fn is said to be convergent at z ∈ D if the sequence

(Sn)n≥0 is convergent at z ∈ D. The limit S(z) of (Sn(z))n≥0 is called the sum of

the convergent series∑

n≥0

fn at the point z. We shall use the following symbol to

denote the sum of a convergent series:

S =

∞∑

n=0

fn.

Definition 5.11 A series which is not convergent is called divergent.

Definition 5.12 A series∑

n≥0

fn is called absolutely convergent at z0 ∈ D if the

number series∑

n≥0

|fn(z0)| is convergent.

Definition 5.13 The series∑

n≥0

fn is called uniformly convergent on the set D if

(Sn)n≥0 is uniformly convergent on D.

Theorem 5.14 (Weierstrass Test) Let fn : D ⊆ C → C, with n ≥ 0. If there exists

an ≥ 0 such that

|fn(z)| ≤ an,

for any n ≥ 0 and for any z ∈ D and if the series∑

n≥0

an is convergent, then∑

n≥0

fn

is absolutely and uniformly convergent on D.

Theorem 5.15 Let fn : D → C, with n ≥ 0. If the series∑

n≥0

fn is uniformly

convergent on D and fn are continuous at z0 ∈ D, then the sum of the series∑

n≥0

fn

is continuous at z0 ∈ D.

86 COMPLEX ANALYSIS

Proposition 5.16 Let∑

n≥0

fn be a series of continuous functions which converges

uniformly to f on a domain D. If γ is a contour lying entirely in D, then

∫

γ

∞∑

n=0

fn(z) dz =

∞∑

n=0

∫

γ

fn(z) dz.

Proposition 5.17 a) Let (fn)n≥0 be a sequence of holomorphic functions on a do-

main D ⊆ C. If the sequence (fn)n≥0 converges uniformly to f on D, then f is

holomorphic on D.

b) If (fn)n≥0 is a sequence of holomorphic functions that converges uniformly to

a function f on every compact subset of a domain D, then f is holomorphic on D.

Moreover, the sequence of derivatives (f(k)n )n≥0 converges uniformly to f (k) on every

compact subset of D, for any k ≥ 1.

Proposition 5.18 a) Let (fn)n≥0 be a sequence of holomorphic functions on a do-

main D. If the series∑

n≥0

fn converges to f uniformly on D, then f is holomorphic

on D.

b) Let (fn)n≥0 be a sequence of holomorphic functions on a domain D. If∑

n≥0

fn

converges to f uniformly on every compact subset of D, then for any k ≥ 1 and for

all z ∈ D, we have∞∑

n=0

f (k)n (z) = f (k)(z),

i.e. the series∑

n≥0

fn may be differentiated term-by-term. Moreover, for any k ≥ 1,

the above convergence is uniform on any compact subset of D.

Definition 5.19 Let z0 ∈ C and an ∈ C, with n ≥ 0. A series of functions for

which

fn(z) = an(z − z0)n, for n ≥ 0,

is called a power series on C. We shall denote such a power series by

∑

n≥0

an(z − z0)n. (5.2)


Let us notice that the above series is convergent for z = z0.

Theorem 5.20 (Cauchy-Hadamard) Let R ≥ 0 be defined as follows:

R =1

limn→∞

n√

|an|,

with the convention that R = 0 if limn→∞

n√|an| = ∞ and R = ∞ if lim

n→∞

n√

|an| = 0.

Then:

1) if R > 0, the power series (5.2) is absolutely convergent for all the points z

with |z − z0| < R and divergent for any z with |z − z0| > R;

2) if R > 0, the power series (5.2) is uniformly convergent over any compact

disk |z − z0| ≤ r, with 0 < r < R;

3) if R = 0, the series (5.2) is convergent only for z = z0.

Remark 5.21 The non-negative quantity R is called the radius of convergence for

the series (5.2).

The set

Dc = z ∈ C | |z − z0| < Ris called the disk or the domain of convergence for the series (5.2). On the boundary

of this disk, i.e. at the points |z − z0| = R, the power series (5.2) may be either

convergent or divergent. Therefore, we have to examine separately what happens

at these points. Alternatively, we can apply other theorems (see [7], [9] and [10]) to

deal with this case (for instance, we can use Abel’s theorem).

Remark 5.22 If there exists limn→∞

∣∣∣∣anan+1

∣∣∣∣, then

R = limn→∞

∣∣∣∣anan+1

∣∣∣∣ .

Moreover, if there exists limn→∞

∣∣∣∣anan+1

∣∣∣∣, then

R = limn→∞

∣∣∣∣anan+1

∣∣∣∣ .

88 COMPLEX ANALYSIS

Remark 5.23 A power series about the point z0 = 0, i.e. a series of the form

∑

n≥0

anzn,

is called a Maclaurin series.

Remark 5.24 If the power series (5.2), having the radius of convergence R, is

uniformly convergent over |z − z0| < r, with 0 < r < R, then its sum defines a

continuous function at any point z with |z− z0| < R. Moreover, in |z− z0| < R, the

sum of the series (5.2) is a holomorphic function and its derivative can be obtained by

termwise differentiation. The series∑

n≥1

n an(z − z0)n−1, obtained by differentiating

(6.2) with respect to z term by term, has the same radius of convergence as the series

(5.2).

Remark 5.25 A power series is infinitely complex differentiable in its disk of con-

vergence and all its derivatives are power series that can be obtained by termwise

differentiation.

Using power series, we can rigorously define the most commonly used elementary

functions of a complex argument.

I. A polynomial functions P = P (z) can be defined as being a power series with

a finite number of nonzero coefficients:

P (z) = a0 + a1z + · · ·+ anzn,

with ai ∈ C. The radius of convergence of this power series is ∞ and the polynomial

function is indefinitely derivable on C.

II. Rational functions are complex functions of the form

R(z) =P (z)

Q(z),

where P and Q are polynomial functions in z and Q is not the zero polynomial. The

domain of R is the set of all the points z for which the denominator Q(z) is not zero.


III. The exponential function exp : C → C is defined by

exp (z) =∑

n≥0

zn

n!.

This function is also denoted by ez and its radius of convergence is ∞.

We have already seen that this function is an entire one and its derivative is

d

dz(ez) = ez.

We shall give now other properties of the exponential function.

a) ez+w = ez ew, for any z, w ∈ C;

b) (ez)′

= ez, for any z ∈ C;

c) eiz = cos z + i sin z, for any z ∈ C (Euler’s formula);

d) The exponential function is periodic, with complex period 2πi, i.e.

ez+2kπi = ez , for any z ∈ C, k ∈ Z.

e) The complex exponential is not a one-to-one map. As already mentioned, we

can divide the complex plane into horizontal strips of height 2π in such a way that in

each strip the exponential function is one-to-one. More precisely, for z = x+ iy ∈ C,

we set

Sk = z = x+ iy ∈ C | x ∈ R, (2k − 1)π < y ≤ (2k + 1)π, k ∈ Z.

The sets Sk are called fundamental strips for the complex exponential function. It

follows that if z1 and z2 belong to the same strip Sk, then ez1 = ez2 implies that

z1 = z2.

IV. The sine function sin : C → C is defined by

sin z =∑

n≥0

(−1)nz2n+1

(2n + 1)!.

90 COMPLEX ANALYSIS

Its radius of convergence is ∞.

V. The cosine function cos : C → C, defined by

cos z =∑

n≥0

(−1)nz2n

(2n)!,

has the radius of convergence R = ∞.

VI. The hyperbolic sine function sinh : C → C, defined by

sinh z =∑

n≥0

z2n+1

(2n + 1)!.


VII. The hyperbolic cosine function cosh : C → C, defined by

cosh z =∑

n≥0

z2n

(2n)!.


Properties

Let z, w ∈ C. It is not difficult to see that the following properties hold true:

1. sin z =eiz − e−iz

2i, cos z =

eiz + e−iz

2;

2. sinh z =ez − e−z

2, cosh z =

ez + e−z

2;

3. sin2 z + cos2 z = 1, cosh2 z − sinh2 z = 1;

4. (sin z)′

= cos z, (cos z)′

= − sin z;

5. (sinh z)′

= cosh z, (cosh z)′

= sinh z;


6. sin (−z) = − sin z, cos (−z) = cos z;

7. sinh (−z) = − sinh z, cosh (−z) = cosh z.

8. sin (z + w) = sin z cosw + cos z sinw;

9. cos (z +w) = cos z cosw − sin z sinw;

10. sinh (z + w) = sinh z coshw + cosh z sinhw;

11. cosh (z + w) = cosh z coshw + sinh z sinhw.

Exercise 5.26 Solve the equation cos z = 0.

Exercise 5.27 Solve the equation sin z = 5.

Exercise 5.28 Solve the equation sin z = 1 + i.

Example 5.29 Let D = z ∈ C | |z| < 1 and fn : D → C, with n ≥ 1, defined by

fn(z) = zn.

Then, for any z ∈ D,

limn→∞

fn(z) = 0,

i.e. (fn)n is convergent on D.

Exercise 5.30 Let fn(z) = zn, with n ≥ 1, and f(z) = 0. Prove that:

(a) (fn)n≥1 is pointwise convergent to f on the open disk B(0, 1);

(b) (fn)n≥1 converges uniformly to f on any closed disk B(0, ρ), with 0 < ρ < 1;

(c) (fn)n≥1 is not uniformly convergent to f on the closed disk B(0, 1).

Exercise 5.31 Compute the radius and the set of convergence for the series∑

n≥0

n! zn.

92 COMPLEX ANALYSIS

Solution. It is easy to see that the radius of convergence for this series is equal to

zero. Therefore, the series is convergent only for z = 0, i.e. Dc = 0.

Exercise 5.32 Find the radius of convergence for the series∑

n≥0

zn

n!.

Solution. The radius of convergence for the above series is equal to ∞. So, the

domain of convergence is Dc = C.

Exercise 5.33 Prove that the series∑

n≥0

zn

n2is absolutely convergent on |z| < 1.

Exercise 5.34 Find the radius of convergence for the series

∑

n≥1

nnzn

n!.

Exercise 5.35 Compute the radius of convergence for the series

∑

n≥1

2−n2zn.

Exercise 5.36 Determine the radius of convergence for the series

∑

n≥2

zn

lnn.

Exercise 5.37 Find the radius of convergence for the series

∑

n≥1

(1 +

1

n

)n2

zn.

Exercise 5.38 Compute the radius of convergence for the series

∑

n≥1

zn2

n2.


Exercise 5.39 Find the radius of convergence of the hypergeometric series

F (α, β, γ; z) = 1 +∞∑

n=1

α (α+ 1) · · · (α+ n− 1)β (β + 1) · · · (β + n− 1)

n! γ (γ + 1) · · · (γ + n− 1)zn,

for α, β ∈ C and γ 6= 0,−1, . . . .

Exercise 5.40 Prove that ez = 1 if and only if z = 2kπi, with k ∈ Z.

Exercise 5.41 Prove that ez1 = ez2 if and only if z1 = z2 + 2kπi, with k ∈ Z.


ez = ez, for any z ∈ C.


sin (2z) = 2 sin z cos z; cos (2z) = cos2 z − sin2 z;

sinh (2z) = 2 sinh z cosh z.


| sin z|2 = sin2 x+ sinh2 y; | cos z|2 = cos2 x+ sinh2 y;

| sinh z|2 = sinh2 x+ sin2 y; | cosh z|2 = sinh2 x+ cos2 y.


cos (iz) = cosh z; sin (iz) = i sinh z;

sinh (iz) = i sin z; cosh (iz) = cos z.

5.2 Taylor Series

Theorem 5.46 (Taylor’s Theorem) Let z0 ∈ C and r > 0. If f : B(z0, r) → C

is holomorphic on B(z0, r), then there exists a unique power series∞∑

n=0

an(z − z0)n

having the radius of convergence R ≥ r such that, for any z ∈ B(z0, r),

f(z) =

∞∑

n=0

an(z − z0)n. (5.3)

94 COMPLEX ANALYSIS

The coefficients an in (5.3), called the Taylor coefficients, are given by

an =f (n)(z0)

n!=

1

2πi

∫

γ

f(ξ)

(ξ − z0)n+1dξ, (5.4)

where γ = ∂B(z0, ρ), with 0 < ρ < r.

Corollary 5.47 If a complex function f is holomorphic on a domain D, then, for

any n ≥ 0, its n-th derivative f (n) exists and is holomorphic on D.

Remark 5.48 If z0 = 0, the Taylor series (5.3) becomes

f(z) = f(0) +f

′

(0)

1!z +

f′′

(0)

2!z2 + · · · . (5.5)

This series is said to be the Maclaurin series of f .

Definition 5.49 Let D be an open nonempty set in C. We say that the complex

function f : D → C is expandable in a Taylor series about the point z0 ∈ D if there

exists r > 0 such that B(z0, r) ⊆ D and there exists a power series∞∑

n=0

an(z − z0)n

convergent in B(z0, r) such that, for any z ∈ B(z0, r), we have:

f(z) =∞∑

n=0

an(z − z0)n. (5.6)

Definition 5.50 A function f : D → C is called analytic on the open set D if it is

expandable in a convergent Taylor series about every point z0 in D.

Remark 5.51 This definition characterizes the class of the functions which can be

locally approximated by convergent power series.

So, a function f defined on an open set D is said to be analytic (or to possess a power

series expansion) at a point z0 ∈ D if there exists a power series∞∑

n=0

an(z − z0)n,

with a positive radius of convergence, such that

f(z) =

∞∑

n=0

an(z − z0)n,

for any z in a neighbourhood of z0.


Remark 5.52 Let us note that any power series with a positive radius of conver-

gence defines an analytic function on the interior of its region of convergence.

Elementary operations with analytic functions can be performed. If f and g are

analytic at a point z0, then f ± g, c f (where c is a constant) and fg are analytic

functions at z0, too. Also, if h is analytic at f(z0), then (h f)(z) is analytic at z0.

Theorem 5.53 A function f : D → C is holomorphic on the open set D if and only

if f is analytic on D.

So, the terms holomorphic and analytic can be used in an interchangeable manner.

Definition 5.54 Let f : D ⊆ C be holomorphic on a domain D. A point z0 ∈ D

is called a zero of the function f if f(z0) = 0. If there exists n ∈ N∗ such that

f(z0) = f′

(z0) = · · · = f (n−1)(z0) = 0 and f (n)(z0) 6= 0, then z0 is called a zero of

order n for f .

Example 5.55 For the function f(z) = sin z − z, the point z0 = 0 is a third order

zero. Indeed, it is not difficult to see that f(0) = f′

(0) = f′′

(0) = 0 and f′′′

(0) 6= 0.

Remark 5.56 If f is holomorphic on a domain D ⊆ C and z0 is a zero of order

n for f , then there exists a function g, holomorphic on D, such that g(z0) 6= 0 and

f(z) = (z − z0)n g(z), for all z ∈ D.

Remark 5.57 It is not difficult to see that if f is holomorphic on a domain D ⊆ C,

has a zero at the point z0 ∈ D and does not vanish identically in D, then there exists

a neighbourhood U ∈ V(z0), U ⊆ D, a non-vanishing holomorphic function g on U

and a unique n ∈ N∗ such that

f(z) = (z − z0)n g(z), for any z ∈ U.

Proposition 5.58 Let f and g be holomorphic on a domain D ⊆ C. Then, f ≡ g

if and only if there exists z0 ∈ D such that f (n)(z0) = g(n)(z0), for any n ∈ N.

96 COMPLEX ANALYSIS

Remark 5.59 If f and g are two holomorphic functions on a domain D ⊆ C and

f(z) = g(z) for all z in a nonempty open subset of D, then f(z) = g(z), for any

z ∈ D.

Theorem 5.60 (The Maximum Modulus Principle) If f : D → C is holomorphic

on a domain D ⊆ C and there exists z0 ∈ D such that

|f(z)| ≤ |f(z0)|,

for any z ∈ D, then f is constant on D.

Remark 5.61 If f is holomorphic on a bounded domain D and continuous on D,

then

maxz∈D

|f(z)| = maxz∈∂D

|f(z)| .

Theorem 5.62 (The Analytic Continuation Principle) Let f : D → C be holomor-

phic on a domain D. Assume that there exist a domain D ⊇ D and a function f ,

holomorphic on D and such that f |D= f . Then, f is unique. The function f is

said to be the analytic continuation of f into f .

So, if f and f are analytic on the domains D and, respectively, D, with D ⊆ D and

if they agree on D, then f is said to be an analytic continuation of f into the domain

D. The above theorem states that there is only one such analytic continuation, i.e.

f is uniquely determined by f .

Exercise 5.63 Prove that if f : D → C is holomorphic on a domain D ⊆ C and

there exists a point z0 ∈ D such that f (n)(z0) = 0, for any n ∈ N, then f(z) = 0, for

any z ∈ D.

Exercise 5.64 Prove that if f : D → C is holomorphic on a domain D ⊆ C and the

set z ∈ D | f(z) = 0 has an accumulation point contained in D, then f(z) = 0,

for any z ∈ D.


1

1− z= 1 + z + z2 + · · · , for |z| < 1.



1

1 + z= 1− z + z2 − · · · , for |z| < 1.


1

(1− z)2= 1 + 2z + 3z2 + · · · , for |z| < 1

and1

(1 + z)2= 1− 2z + 3z2 − · · · , for |z| < 1.

Exercise 5.68 Let f : C \ 0, 1 → C, defined by

f(z) =1

z(z − 1).

Expand f in a Taylor series about the point z = 0.

Exercise 5.69 Find the Maclaurin series for the error function erf (z), defined by

erf (z) =2√π

z∫

0

e−t2 dt.

Exercise 5.70 Find a power series representation about the origin for the following

functions:

a) f(z) = cos(z2);

b) f(z) = z3 sin(z2);

c) f(z) = (sin z)2.

Exercise 5.71 Find the Taylor series expansion of the function f : C \ ±i → C

given by

f(z) =1

1 + z2

about the point z = 0 and compute its radius of convergence.

98 COMPLEX ANALYSIS

Solution. We have

f(z) =1

1− (−z2)=

∞∑

k=0

(−z2

)k=

∞∑

k=0

(−1)k z2k.

So, the radius of convergence of the above power series is R = 1.

....

Solution Manual -desktop .......

5.3 Laurent Series

As we saw, Taylor series are suited for representing holomorphic functions on a disk

centered at a given point z0. Now, we shall consider more general power series, con-

taining both positive and negative powers of (z−z0), and characterizing holomorphic

functions in annuli, i.e. in sets of the form

D = z ∈ C | r < |z − z0| < R,

with z0 ∈ C and 0 ≤ r < R ≤ +∞. Using such a generalized expansion, we shall be

able to describe the behavior of a function having a singularity at the point z0.

Definition 5.72 Let z0 ∈ C and an ∈ C. A series of the form

∞∑

n=−∞

an(z − z0)n (5.7)

is called a Laurent series.

So, a Laurent series is a series of functions of the form

∞∑

n=−∞

an(z− z0)n = · · ·+ a−n

(z − z0)n+ · · · a−1

z − z0+a0+ · · ·+an(z− z0)

n+ · · · . (5.8)

In fact, a Laurent series about a given point z0 is a sum of two independent power

series, one consisting of positive powers of (z − z0) and the other one of negative

powers of (z − z0):∞∑

n=0

an(z − z0)n +

∞∑

n=1

a−n(z − z0)−n.


Still, we shall often use the short notation (5.8), despite the fact that, conceptually,

a Laurent series is the sum of two distinct power series.

Definition 5.73 The series

∞∑

n=1

a−n(z − z0)−n is called the principal part or the

singular part of the Laurent series (5.8). The series∞∑

n=0

an(z− z0)n is said to be the

Taylor part or the regular part of the series (5.8).

Remark 5.74 Any power series is a Laurent series, with an = 0, for n < 0.

Definition 5.75 The Laurent series∞∑

n=−∞

an(z − z0)n converges on a set D ⊆

C\z0 if both its principal part and its Taylor part are convergent on D. Moreover,

if we denote by Π(z) and T (z) the sums of the principal part and, respectively, the

Taylor part of the series (5.8), then the sum

S(z) =

∞∑

n=−∞

an(z − z0)n

of the Laurent series (5.8) is defined as being

S(z) = Π(z) + T (z), ∀z ∈ D ⊆ C \ z0.

Theorem 5.76 (Laurent) For the Laurent series (6.8), let

r = limn→∞

n√

|a−n| (5.9)

and

R =1

limn→∞

n√

|an|, (5.10)

with the convention that R = 0 if limn→∞

n√|an| = ∞ and R = ∞ if lim

n→∞

n√

|an| = 0.

a) If r < R, the Laurent series (5.8) is absolutely and uniformly convergent over

compact sets in the annulus U(z0; r,R) and divergent in C \ U(z0; r,R).

100 COMPLEX ANALYSIS

b) If R < r, the Laurent series (5.8) is divergent on C.

c) If R = r, we may have points of convergence only on the circle ∂B(z0, r).

It is not difficult to see that the sum

S(z) =∞∑

n=−∞

an(z − z0)n

is a holomorphic function in the annulus U(z0; r,R).

Theorem 5.77 If f is holomorphic in the annulus D = U(z0; r,R), with 0 ≤ r <

R ≤ ∞, then f can be expanded uniquely as

f(z) =

∞∑

n=−∞

an(z − z0)n, for any z ∈ D, (5.11)

with the coefficients an given, for n ∈ Z, by

an =1

2πi

∫

γ

f(ξ)

(ξ − z0)n+1dξ. (5.12)

Here, γ = ∂B(z0, ρ), with r < ρ < R.

Remark 5.78 The Laurent expansion of a function f which is holomorphic on a

whole disk D centered at a point z0 has no principal part and its regular part coincides

with the power expansion of f on the disk D.

Definition 5.79 Let f be a single-valued holomorphic function defined on an open

set D ⊆ C. A point z0 ∈ C is called an isolated singular point for the function f if

z0 /∈ D, but there exists a punctured neighbourhood of z0 contained in D, i.e there

exists R > 0 such that U(z0;R) ⊆ D.

So, a function f has an isolated singularity at a point z0 if f is differentiable in some

punctured disk centered at z0 and f is not defined at z0.


Example 5.80 For the function f : C \ 0, ±4i → C, defined by

f(z) =z + 3

z3(z2 + 16),

the points z = 0 and z = ±4i are isolated singularities.

Remark 5.81 Let us notice that if z0 is an isolated singularity for f , then z0 belongs

to the boundary of D and D∪z0 is an open set. Moreover, if D is a domain, then

D ∪ z0 is a domain, too.

Definition 5.82 Let f : D ⊆ C → C be a holomorphic function on an open set D

and z0 ∈ C be an isolated singular point for f .

a) The point z0 is said to be a removable singularity if f is holomorphically ex-

tendable to D = D ∪ z0.

b) The point z0 is called a pole if there exists limz→z0

f(z) = ∞.

c) The point z0 is said to be an essential singularity if the limit limz→z0

f(z) does not

exist.

Remark 5.83 The points at which f is holomorphic, together with those at which

f has a removable singularity, are said to be regular points for f .

Remark 5.84 The point z0 is a removable singular point for f if and only if there

exists limz→z0

f(z) and it is finite. We shall often use the same notation f for the

extension of f to D = D ∪ z0, by putting f(z0) = limz→z0

f(z).

Remark 5.85 Let D ⊆ C be an open subset of the complex plane, f ∈ H(D) and

z0 an isolated singular point of f . There are several ways to decide if an isolated

singularity is a removable one. More precisely, it is not difficult to see that the

following statements are equivalent:

(i) f is holomorphically extendable to z0;

(ii) f is continuously extendable to z0;


(iii) there exists a deleted neighbourhood of z0 on which f is bounded;

(iv) limz→z0

(z − z0) f(z) = 0.

Remark 5.86 Let f ∈ H(D) and z0 an isolated singular point for f . The point z0

is a pole of order n for f if z0 is a removable singularity for1

fand a zero of order

n for the holomorphic extension to z0 of this function.

Remark 5.87 Let us notice that if z0 is a pole for f , then there exist a unique

n ∈ N∗ and a unique function g ∈ H(D ∪ z0) such that g(z0) 6= 0 and

f(z) = (z − z0)−n g(z), for any z ∈ D.

In fact, f has a pole of order n at z0 if n is the smallest positive integer for which

(z − z0)n f(z) is holomorphic at z0.

An important tool for analyzing the behavior of a holomorphic function f about an

isolated singularity z0 is offered by its Laurent series around z0. Let f ∈ H(D), z0

an isolated singularity for f and R > 0 such that U(z0; 0, R) ⊆ D. Then, in this

annulus, we have

f(z) =∞∑

n=−∞

an(z − z0)n, (5.13)

with

an =1

2πi

∫

γ

f(ξ)

(ξ − z0)n+1dξ, γ = ∂B(z0, r), 0 < r < R. (5.14)

Theorem 5.88 Let f ∈ H(D) and z0 ∈ C be an isolated singular point for f .

a) The point z0 is a removable singularity for f if and only if the principal part

of the Laurent series of f about z0 in a punctured disk centered at z0 is identically

zero.

b) The point z0 is a pole for f if and only if the principal part of the Laurent

series of f about z0 in a punctured disk centered at z0 consists only of a finite number


of terms, i.e. there exists a unique n ∈ N∗ such that, in a punctured disk centered at

z0, we have

f(z) =a−n

(z − z0)n+ · · ·+ a−1

z − z0+ a0 + a1(z − z0) + · · · , (5.15)

with a−n 6= 0. The integer n, called the order or the multiplicity of the pole, describes

the rate at which the function grows near z0.

c) The point z0 is an essential singularity if and only if the principal part of

the Laurent series of f about z0 in a punctured disk centered at z0 has an infinite

number of terms.

Remark 5.89 Let us notice that there are functions that have non-isolated singular

points. For example, the function

f(z) =1

sin1

z

has singular points at z = 0 and, respectively, at zk =1

kπ, for k = ±1,±2, . . . . The

points zk are simple poles that accumulate in z = 0. Thus, z = 0 is a non-isolated

singular point for f (it is an accumulation point of poles).

Definition 5.90 Let D be an open nonempty subset of C. A function f is called

meromorphic on D if there exists a set E ⊆ D such that f ∈ H(D \ E) and E

contains only removable singularities or poles for f .

If we denote by M(D) the set of all the meromorphic functions on D, it follows that

H(D) ⊆ M(D).

The above definition holds true also for D ⊆ C∞.

Definition 5.91 Let f ∈ H(D). If there exists r > 0 such that z ∈ C | |z| > r ⊆D, i.e. f is holomorphic on the domain r < |z| < ∞, then the point at infinity

z = ∞ is said to be an isolated singular point for f .


In order to study the behavior of the function f at ∞, we shall consider the function

g(ξ) = f

(1

ξ

),

which is holomorphic on U

(0;

1

r

). In other words, ξ = 0 is an isolated singularity

for g.

We shall say that z = ∞ is a removable singularity, a pole of order n or, respec-

tively, an essential isolated singular point for f if ξ = 0 is a removable singularity, a

pole of order n or, respectively, an essential isolated singular point for g.

Example 5.92 Let us analyze the behavior of the function

f(z) =1

z(z + 1)

in a neighbourhood of the point at infinity. Let

z =1

ξ

and

g(ξ) = f

(1

ξ

).

Expanding g for small |ξ|, we get

g(ξ) = ξ2 − ξ3 + ξ4 − · · · .

Therefore, the point z = ∞ is a regular point for the function f .

Exercise 5.93 Expand in a Laurent series about the point z0 = −1 the function

f(z) =1

1− z2

in the annulus D = z ∈ C | 0 < |z + 1| < 2.


Solution. The function f is holomorphic in the annulus D. Thus, on D, it can be

expanded in a convergent Laurent series:

f(z) =

∞∑

n=−∞

an(z + 1)n.

We have

f(z) =1

1− z2=

1

2(z + 1)

(1− z + 1

2

) .

Therefore,

f(z) =1

2(z + 1)

∑

n≥0

(z + 1)n

2n=

1

2(z + 1)+

1

4+

1

8(z + 1) + · · · .

This expansion is valid provided that z ∈ D.

Exercise 5.94 Let D = z ∈ C | 0 < |z − i| < ∞. Expand the holomorphic

function f : D → C, defined by

f(z) =e2z

(z − i)3,

in the annulus D, in a convergent Laurent series about the point z = i.

Solution. We have

e2z

(z − i)3=

e2i

(z − i)3e2(z−i)

=e2i

(z − i)3

[1 + 2(z − i) +

(2(z − i))2

2!+ · · ·

].

So,

f(z) =e2i

(z − i)3+

2e2i

(z − i)2+

2e2i

(z − i)+

4e2i

3+

2e2i

3(z − i) + · · · .

Exercise 5.95 Expand in a Laurent series about the point z0 = 1 the function

f(z) =1

1− z2

in the annulus D = z ∈ C | 2 < |z − 1| < ∞.


Exercise 5.96 Develop in a Laurent series in the punctured disk D = z ∈ C | 0 <

|z| < 1 the function

f(z) =1

z(z − 1).

Exercise 5.97 Find the Laurent series about the point z0 = 1 for the function

f(z) = sin1

z − 1

in the annulus D = z ∈ C | 0 < |z − 1| < ∞.

Exercise 5.98 Let z0 ∈ C, an ∈ C and 0 ≤ r < R ≤ ∞. Prove that if the Laurent

series

∞∑

n=−∞

an(z − z0)n is convergent in the annulus D = U(z0; r,R) and its sum is

zero in D, then an = 0, for any n ∈ Z.

Exercise 5.99 Prove that, in the annulus D = z ∈ C | 3 < |z| < ∞, the function

f(z) =1

z4 + 9z2

can be expanded in the following Laurent series:

f(z) =1

z4

∞∑

n=0

(−1)n(

9

z2

)n

.

Exercise 5.100 Let z0 ∈ C, an, bn ∈ C and 0 ≤ r < R ≤ ∞. Prove that if the

Laurent series

∞∑

n=−∞

an(z− z0)n and

∞∑

n=−∞

bn(z− z0)n are convergent in the annulus

D = U(z0; r,R) and their sums are equal in D, then an = bn, for any n ∈ Z.

Exercise 5.101 Let f : C \ 0 → C be defined by

f(z) =sin z

z.

Show that the point z0 = 0 is a removable singularity for f .


Solution Due to the fact that

limz→0

f(z) = 1,

we can extend f to C by defining the function f at origin as being f(0) = 1.

Exercise 5.102 Let f : C \ 1 → C be defined by

f(z) =z

z − 1.

Prove that the point z0 = 1 is a pole for f .

Solution. We have

limz→1

f(z) = ∞.

Hence, the point z0 = 1 is a pole for f .

Exercise 5.103 Let us consider f : C \ 0 → C, defined by

f(z) = e1z2 .

Show that z0 = 0 is an essential singularity for the function f .

Solution. Since the limit limz→0

f(z) doesn’t exist, the point z0 = 0 is an essential

singularity for the function f .

Exercise 5.104 Show that the function f : C \ 0 → C, defined by

f(z) =1− cos z

z5,

has a third order pole at z0 = 0.

Solution. In the annulus D = z ∈ C | 0 < |z| < ∞, the function f is holomorphic

and can be expanded in the following Laurent series:

f(z) =1

2! z3− 1

4! z+

z

6!− · · · ,

which implies that z0 = 0 is a pole of order three for f .


Exercise 5.105 Prove that the function f : C \ 0 → C, defined by

f(z) = sin1

z,

has an essential singularity at z0 = 0.

Solution. In the annulus D = z ∈ C | 0 < |z| < ∞, the function f is holomorphic

and can be expanded in the following Laurent series:

f(z) =1

z− 1

3! z3+

1

5! z5− · · · ,

which implies that z0 = 0 is an essential singularity for f .

Exercise 5.106 Find the zeros and their multiplicities of the following functions:

a) f(z) =(1 + z2

)4;

b) f(z) = z4 cos z;

c) f(z) = sin2 z.

Exercise 5.107 Determine the power series of f(z) = ez centered at the point

z = −2.

Exercise 5.108 Find the poles of the following functions and determine their or-

ders:

a) f(z) =z

(z2 + 1)4 (z − 2)2;

b) f(z) =sin z

z6;

c) f(z) =z

ez − 1.


5.4 Residues

We shall discuss now the concept of residue of a single-valued holomorphic function

f at an isolated singular point z0. Let us recall that if we consider f ∈ H(D), z0 an

isolated singularity for f and R > 0 such that U(z0; 0, R) ⊆ D, then, in this annulus,

the function f can be expanded in a Laurent series of the form

f(z) =

∞∑

n=−∞

an(z − z0)n, (5.16)

with

an =1

2πi

∫

γ

f(ξ)

(ξ − z0)n+1dξ, γ = ∂B(z0, r), 0 < r < R. (5.17)

Definition 5.109 Let z0 ∈ C and f be a regular function in the punctured disk

D = z ∈ C | 0 < |z − z0| < R. The coefficient a−1 of the Laurent series (6.19) of

f about the point z0 in D is called the residue of the function f at the point z0 and

is denoted by Res (f, z0).

So,

Res (f, z0) =1

2πi

∫

γ

f(z) dz, (5.18)

where γ = ∂B(z0, r), with 0 < r < R. Rearranging this formula, we get∫

γ

f(z) dz = 2πiRes (f, z0) (5.19)

and, hence, the value of the residue of f at the point z0 is instrumental for evaluating

complex integrals. In fact, formula (6.22) holds true for any simple closed contour

in D having the index n(γ, z0) = 1. Let us notice that the only term in the Laurent

series expansion for f that has a nonzero contribution to the integral in (6.22) is

a−1 and this justifies the use or the name residue for this coefficient.

Example 5.110 If D = z ∈ C | 0 < |z| < 1 and f(z) =1

z3(1− z), then

f(z) =1

z3+

1

z2+

1

z+ 1 + z + z2 + · · · .


In this case, the residue of the function f at the point z0 = 0 is a−1 = 1.

Example 5.111 If D = z ∈ C | 0 < |z| < 1 and f(z) =ez

z3, then

f(z) =1

z3+

1

z2+

1

2! z+

1

3!+

z

4!+ · · ·

So, the residue of the function f at the point z0 = 0 is a−1 =1

2.

Example 5.112 If D = z ∈ C | 0 < |z| < +∞ and f(z) = e1/z

, then

f(z) = · · ·+ 1

3! z3+

1

2! z2+

1

z+ 1

The residue of the function f at the point z0 = 0 is a−1 = 1.

Let us see now how we can compute the residue of a given function f at an

isolated singular point z0. As we saw, such a point can be a removable singularity,

a pole or an essential singularity.

I. If z0 is a regular point for f , then

Res (f, z0) = 0. (5.20)

II. If z0 is a first order pole for f , then

Res (f, z0) = limz→z0

(z − z0) f(z). (5.21)

In particular, if

f(z) =g(z)

h(z),

with g, h ∈ H(D), g(z0) 6= 0, h(z0) = 0, h′

(z0) 6= 0, then

Res (f, z0) =g(z0)

h′(z0). (5.22)

Indeed, since in a punctured neighbourhood of z0,

f(z) =a−1

(z − z0)+ a0 + a1(z − z0) + · · · ,


it follows that

(z − z0) f(z) = a−1 + a0(z − z0) + a1(z − z0)2 + · · · ,

which gives immediately (6.24).

If f(z) =g(z)

h(z), then

limz→z0

(z − z0)f(z) = limz→z0

g(z)

h(z) − h(z0)

z − z0

=g(z0)

h′(z0).

III. If z0 is a pole of order m, then

Res (f, z0) = limz→z0

1

(m− 1)!

dm−1

dzm−1[(z − z0)

mf(z)] . (5.23)

Indeed, since in a punctured neighbourhood of z0,

f(z) =a−m

(z − z0)m+ · · · a−1

z − z0+ a0 + a1(z − z0) + · · · ,

it follows that

[(z − z0)mf(z)](m−1) = (m− 1)! a−1 +m! a0(z − z0) + · · · ,

which implies that

limz→z0

[(z − z0)mf(z)](m−1) = (m− 1)! a−1.

IV. If z0 is an essential singularity, then

Res (f, z0) = a−1.

So, in this case, the residue of f can be computed only as being the coefficient a−1

in the Laurent series of f about z0.


Example 5.113 For the function f : C \ 0, 1 → C, defined by

f(z) =1

z2(1− z),

the point z = 0 is a second order pole and the point z = 1 is a simple pole. Then,

it is not difficult to see that

Res (f, 0) = 1, Res (f, 1) = −1.

Example 5.114 Let f : C \ ±i → C, defined by

f(z) =z

z2 + 1.

Then,

Res (f, i) =1

2, Res(f,−i) =

1

2.

Example 5.115 For the function f : C \ (2k + 1)πi | k ∈ Z → C, defined by

f(z) =1

ez + 1,

the residue at the point z0 = πi, which is a simple pole, can be computed using

formula (6.25). We get

Res (f, πi) = −1.

Example 5.116 Let f : C \ 1 → C, defined by

f(z) =1

z − 1e

1

z − 1.

The point z0 = 1 is an essential singularity for f . Indeed, in the annulus 0 <

|z − 1| < ∞, the function f can be expanded in a Laurent series of the form

f(z) =1

z − 1+

1

(z − 1)2+

1

2! (z − 1)3+ · · · .

Thus, we obtain

Res (f, 1) = 1.


Exercise 5.117 Locate and classify the singularities of the following functions:

a) f : C \ 0, 2 → C, f(z) =z3 + 1

z4(z − 2);

b) g : C∗ → C, g(z) = z2e1/z ;

c) h : C∗ → C, h(z) =sin z − z

z2.

Exercise 5.118 Find and classify the singularities of the function f : C\1, 4 → C,

given by

f(z) =z ez

(z − 1)2 (z − 4)3.

Exercise 5.119 Compute the residue of the function f : C∗ → C, defined by

f(z) = z3 cos1

z,

at the point z = 0.

Exercise 5.120 Evaluate the residue of the function f : C \ i → C, defined by

f(z) = z sin

(1

iz + 1

),

at the point z = i.

Exercise 5.121 Compute the residue at the point z = 0 of the function

f(z) =sin z

z3.

Solution. The function f has a double pole at z = 0 and

Res (f, 0) = 0.

Exercise 5.122 Compute the residue at the point z = 0 of the functions:

a) f(z) =cos z

z4;

b) f(z) =ez − 1

sin2 z.


Let us introduce now the notion of residue at the point at infinity.

Definition 5.123 If z = ∞ is an isolated singular point for f , i.e. f is holomorphic

in a punctured neighbourhood D = U(0; r,+∞) of ∞, then f can be expanded in D

in a Laurent series

f(z) =

∞∑

n=−∞

cnzn in r < |z| < ∞. (5.24)

The residue of the function f at ∞ is defined as being the coefficient −c−1 in the

expansion (6.27), i.e.

Res (f,∞) = −c−1.

Remark 5.124 Let us notice that if we compute, for z = ∞, the integral of f along

a closed path lying in the punctured domain of holomorphy and having the index 1

with respect to z = ∞, we get

1

2πi

∫

γ

f(z) dz = −c−1, (5.25)

where γ = γ−11 , γ1 = ∂B(0, R), for r < R, and this justifies the above definition.

Example 5.125 Let f : C \ ±i → C, defined by

f(z) =z

z2 + 1.

For |z| > 1, we can expand f in the following Laurent series:

f(z) =1

z− 1

z3+

1

z5− · · ·

So,

Res (f,∞) = −c−1 = −1.

Chapter 6

The Residue Theorem.

Applications

6.1 The Residue Theorem

Theorem 6.1 (Cauchy’s Residue Theorem) Let γ be a positively oriented simple

closed path in a simply connected domain D. We assume that f is analytic every-

where on D, except at a finite number of isolated singularities z1, . . . , zn lying in the

interior of γ. Then,

∫

γ

f(z) dz = 2πi

n∑

k=1

Res (f, zk). (6.1)

Theorem 6.2 If f is holomorphic on an open set D \ zii∈I , where zi are isolated

singularities for f , then for any compact set K ⊂ D with γ = ∂K being a piecewise

smooth path which doesn’t contain singular points for f , there exists a finite number

of points zi ∈ K and

∫

γ

f(z) dz = 2πi∑

zi∈K

Res (f, zi). (6.2)

Theorem 6.3 Let f : D ⊆ C → C be holomorphic on the open set D and let S be

the set of the isolated singular points of f . If γ is a closed contour in D which is

115


homotopic to zero in D ∪S, then the set z ∈ S | n(γ, z)Res(f, z) 6= 0 is finite and

∫

γ

f(z) dz = 2πi∑

z∈S

n(γ, z)Res (f, z). (6.3)

Remark 6.4 In what follows, in all the applications of Cauchy’s residue theorem,

we shall work only with paths γ having the winding numbers around all the involved

singularities equal to one or zero.

Proposition 6.5 Let f be holomorphic in D = C\z1, z2, . . . , zn and let us denote

zn+1 = ∞. Then,n+1∑

k=1

Res (f, zk) = 0.

Example 6.6 Let us look at the integral

I =

∫

γ

z + 1

z (z2 + 9)2dz,

where γ = ∂B(0, 7).

The function f has isolated singularities at the points 0 and ±3i and they belong to

the domain bounded by the contour γ. So, one way to compute the integral I is to

write

I = 2πi [Res (f, 0) + Res (f, 3i) + Res (f,−3i)] ,

which gives

I = 0.

On the other hand, we can evaluate the residue of the function f at z = ∞ and, then,

using Proposition 7.5, we can easily compute the value of the integral I. Indeed, for

|z| > 3, we have

f(z) =z + 1

z (z2 + 9)2=

z + 1

z5

1

1 +9

z2

2

,

i.e.

f(z) =

(1

z4+

1

z5

) (1− 9

z2+

81

z4− · · ·

)2

,

THE RESIDUE THEOREM. APPLICATIONS 117

which implies that

Res (f,∞) = 0

and, hence,

I = 0.


I =

∫

|z|=2

4

(z2 + 1)(z − 3)2dz.

Solution. The function f has isolated singularities at the points z = ±i and z = 3.

Since only the singular points ±i, which are simple poles, belong to the domain

bounded by the closed contour γ, it follows that

I = 2πi [Res (f,−i) + Res (f, i)] =12

25πi.


I =

∫

γ

sin z

z6dz,

where γ = ∂B(0, 3).

Solution. The function f can be expanded, in the annulus 0 < |z| < ∞, in a Laurent

series about the point z = 0 of the form

f(z) =1

z6

(z − z3

3!+

z5

5!− · · ·

)=

1

z5− 1

3! z3+

1

5! z− · · · .

So, z = 0 is a pole of order five and we obtain

I = 2πiRes (f, 0) =2πi

5!.


I =

∫

γ

z + 1

z (z2 + 4)2dz,

where γ = ∂B(0, 5).


Solution. Since the singular points 0 and ±2i belong to the domain bounded by the

contour γ, we get

I = 2πi [Res (f, 0) + Res (f, 2i) + Res (f,−2i)] ,

i.e.

I = 0.


I =

∫

γ

1

ez + 1dz,

where γ = ∂B(2i, 2).

Solution. Since z0 = πi is a first order pole for f which belongs to B(2i, 2) and all

the other singular points of f lie outside the domain bounded by γ, we obtain

I = 2πiRes (f, πi),

i.e.

I = −2πi.


I =

∫

γ

z2 + 1

ln z − πidz,

where γ is the positively oriented circle |z + 1| = 1

2and the logarithm is considered

on its principal branch.

Solution. We notice that z0 = −1 is a first order pole for f . Therefore,

I = 2πiRes (f,−1),

i.e.

I = −4πi.


6.2 Evaluation of Real Integrals

Cauchy’s residue theorem proves to be a powerful tool for computing a wide range

of definite integrals. As we shall see, using this theorem, we shall be able to evaluate

some trigonometric integrals of the form

I =

2π∫

0

R (sin t, cos t) dt, (6.4)

where R is a suitable real rational function, or important types of improper real

integrals of the form

I =

∞∫

−∞

f(x) dx. (6.5)

I. Trigonometric Integrals

Let us consider trigonometric integrals of the form

I =

2π∫

0

R (sin t, cos t) dt,

where R is a real rational function of its arguments.

Theorem 6.12 Let R = R(u, v) be a real rational function of its arguments, which

has no poles on the circle u2 + v2 = 1. Then,

2π∫

0

R (sin t, cos t) dt = 2πi∑

|z|<1

Res (g, z), (6.6)

where

g(z) =1

izR

(z − z−1

2i,z + z−1

2

). (6.7)

The notation∑

|z|<1

Res (g, z) means the sum of the residues of g at all the

poles lying inside the disk |z| < 1.


Exercise 6.13 Compute

I =

π∫

0

1

2− cos tdt.

Solution. We notice that the integral is defined only over [0, π]. Since cos (2π− t) =

cos t, we get2π∫

π

1

2− cos tdt =

π∫

0

1

2− cos tdt,

which implies that

I =1

2

2π∫

0

1

2− cos tdt.

Therefore,

I =1

2

∫

γ

1

2− 1

2

(z +

1

z

) dz

iz= −1

2

∫

γ

1

i

2

z2 − 4z + 1dz.

The function

g(z) = −1

i

1

z2 − 4z + 1

has simple poles at z = 2 ±√3. Hence, since only the pole z = 2 −

√3 lies inside

the unit disk and

Res (g, 2 −√3) =

1

2i√3,

we get

I =π√3.

Remark 6.14 In a similar manner, we can treat integrals of the form

In =

2π∫

0

cosntR (sin t, cos t) dt

or

Jn =

2π∫

0

sinntR (sin t, cos t) dt,


for n ∈ N. In this case, we consider the integral In+ iJn and, using the same change

of variables and de Moivre’s formula, we can easily compute the above integrals.

Exercise 6.15 For a > 1, show that

2π∫

0

1

a+ cos tdt =

2π√a2 − 1

.


2π∫

0

1

1 + a cos tdt =

2π√1− a2

, for 0 < |a| < 1.


π∫

0

1

(a+ cos t)2dt =

aπ

(a2 − 1)3/2, for a > 1.

Exercise 6.18 For n ∈ N, prove that

2π∫

0

(1 + 2 cos t)n cosnt dt = 2π.

II. Improper Integrals

If the improper integral

I =

∞∫

−∞

f(x) dx (6.8)

is convergent, the main idea for computing its value by using Cauchy’s residue

theorem will be to extend f to the complex plane and to choose a suitable family of

contours of integration γr such that

limr→∞

∫

γr

f(z) dz =

∞∫

−∞

f(x) dx.

In order to deal with integrals of the form (6.8), we shall make use of a famous

lemma due to Camille Jordan.


Lemma 6.19 (Jordan’s Lemma) Let D = z ∈ C | Im z ≥ 0 and f : D → C be a

continuous function. If, for r > 0, the path γr : [0, π] → C is defined by

γr(t) = r eit

and

limz→∞

f(z) = 0, (6.9)

then

limr→∞

∫

γr

f(z) eiz dz = 0. (6.10)

Moreover, if f is an analytic function in the upper-half complex plane such that f

tends to zero, uniformly on any upper semi-circle γr centered at the origin and with

radius r → ∞, then, for a > 0,

limr→∞

∫

γr

eiazf(z) dz = 0.

Let us remark that if a < 0, a similar result holds for the semicircle γ−r lying in the

lower half-plane.

Lemma 6.20 Let f be a continuous function on the closed sector S0[θ1, θ2] and

γr(t) = r ei[θ1+t(θ2−θ1)], with t ∈ [0, 1].

(i) If limz→∞

z f(z) = 0, then limr→∞

∫

γr

f(z) dz = 0.

(ii) If limz→0

z f(z) = 0, then limr→0

∫

γr

f(z) dz = 0.

(iii) If θ1 = 0, θ2 = π/p and limr→∞

z1−p f(z) = 0, then

limr→∞

∫

γr

f(z) eizpdz = 0.


(iv) If f is holomorphic on the sector S0[θ1, θ2] \ 0 and z = 0 is a simple pole for

f , then

limr→0

∫

γr

f(z) dz = (θ2 − θ1) iRes(f, 0).

Remark 6.21 The Jordan’s lemma is obtained from (iii) if we take p = 1.

II.1. Improper integrals of rational functions

We intent to evaluate now integrals of the form

I =

∞∫

−∞

R(x) dx,

with R a real rational function. We shall consider here only integrals for which

the integrand is a rational function with no real poles and such that the degree

of the denominator is at least two units higher than the degree of the numerator.

Therefore, the integrals will be convergent.

Theorem 6.22 Let R = P/Q be a real rational function, with P and Q polynomials

of degree n and, respectively, m. We suppose that Q has no zeros on the real axis

and lim|z|→∞

z R(z) = 0, i.e. n ≤ m− 2. Then, the integral

∞∫

−∞

R(x) dx is convergent

and∞∫

−∞

R(x) dx = 2πi∑

Im z>0

Res (R, z). (6.11)

Let us notice that in (6.11) we consider the sum of the residues of the function

R = R(z) at all its poles that are situated in the upper half-plane.


I =

∞∫

0

2x2

(x2 + 1)(x2 + 4)dx.


x

y

O− r r

γ+r

Figure 6.1: The closed contour γ

Solution. Obviously,

I =1

2

∞∫

−∞

2x2

(x2 + 1)(x2 + 4)dx.

Let R : C \ ±i,±2i → C, defined by

R(z) =2z2

(z2 + 1)(z2 + 4).

The function R has simple poles at the points ±i and ±2i. We define the closed

contour γ = γ+r ∨[−r, r], where r > 2 and γ+r is the semicircle going counterclockwise

from (r, 0) to (−r, 0) (see Figure 6.1). The function R satisfies the conditions of

Theorem 6.22. Its singular points in the upper half-plane are the simple poles z1 = i

and z2 = 2i. Thus,

I = πi [Res (R, z1) + Res (R, z2)] = πi

(i

3− 2i

3

)=

π

3.


∞∫

0

1

(x2 + a2)2dx =

π

4a3, whenever a > 0.



∞∫

−∞

1

x4 + a4dx =

π√2a3

, for a > 0.


I =

∞∫

−∞

x2

(x2 + 1)2(x2 + 9)dx.

II.2. Improper integrals of products of rational and trigonometric func-

tions

We shall consider now integrals of the following form:

I =

∞∫

−∞

R(x) eiax dx; I =

∞∫

−∞

R(x) cos (ax) dx;

I =

∞∫

−∞

R(x) sin (ax) dx.

In order to deal with such integrals, we state the following result.

Theorem 6.27 Let R = P/Q be a rational real function, with P and Q polynomials

of degree n and, respectively, m. We assume that Q has no zeros on the real axis

and lim|z|→∞

R(z) = 0, i.e. n ≤ m− 1. Then, for a > 0,

∞∫

−∞

R(x) eiaxdx is convergent

and∞∫

−∞

R(x) eiax dx = 2πi∑

Im z>0

Res (g, z), (6.12)

where

g(z) = R(z) eiaz .



I1 =

∞∫

−∞

x− 1

x2 − 2x+ 5cos 5x dx.

Solution. We consider also the integral

I2 =

∞∫

−∞

x− 1

x2 − 2x+ 5sin 5x dx.

We compute

I = I1 + iI2,

i.e.

I =

∞∫

−∞

x− 1

x2 − 2x+ 5ei5x dx.

The rational function

R(z) =z − 1

z2 − 2z + 5

is holomorphic on C\1±2i and has simple poles at z = 1±2i. Since the polynomial

z2 − 2z+5 has no roots on the real axis and limz→∞

R(z) = 0, using Theorem 6.27, we

get∞∫

−∞

R(x) ei5x dx = 2πi∑

Im z>0

Res (g, z), (6.13)

where

g(z) = R(z) ei5z .

Since in the upper half-plane lies only the simple pole z = 1 + 2i, computing the

residue of g at this pole, we get

I = πi(e−10 cos 5 + ie−10 sin 5

),

i.e.

I1 = Re I = −πe−10 sin 5

and

I2 = Im I = πe−10 cos 5.



I =

∞∫

0

x sinαx

1 + x2dx =

π

2e−α, for α > 0.


∞∫

−∞

cos x

x2 + a2dx = π

e−a

a, for a > 0.


∞∫

−∞

x sinx

x2 + a2dx = π e−a, for a > 0.

Exercise 6.32 Evaluate the following improper integrals:

a) I =

∞∫

0

x sin 2x

(x2 + 4)(x2 + 1)2dx.

b) I =

∞∫

−∞

x+ 2

x2 + 2x+ 5sin 3x dx.

III. The Poisson Integral

Using Cauchy’s residue theorem, we can prove that

∞∫

0

sinx

xdx =

π

2.

Consider the function f : C∗ → C, defined by

f(z) =eiz

z,


which is holomorphic on C∗. This function has a simple pole at z = 0 and an

essential singularity at z = ∞. Let 0 < r < R and the paths γR : [0, π] → C, defined

by γR(t) = Reit, and, respectively, γr : [0, π] → C, given by γr(t) = r ei(π−t) (see

Figure 6.2). Let

γ = γR ∨ [−R,−r] ∨ γr ∨ [r,R].

x

y

O−R R− r r

γR

γr

Figure 6.2: The contour of integration for the Poisson integral

So, using this indented semicircle and Cauchy’s residue theorem, it follows that

∫

γR

eiz

zdz +

−r∫

−R

eix

xdx+

∫

γr

eiz

zdz +

R∫

r

eix

xdx = 0. (6.14)

Since

eix = cos x+ i sinx,

it follows that

−r∫

−R

eix

xdx+

R∫

r

eix

xdx =

R∫

r

eix − e−ix

xdx = 2i

R∫

r

sinx

xdx. (6.15)


Using Jordan’s lemma, we get

limR→∞

∫

γR

eiz

zdz = 0. (6.16)

Also,

limr→0

∫

γr

eiz

zdz = −iπ. (6.17)

Hence, for r → 0 and R → ∞, from (7.20)-(7.23), we obtain

∞∫

0

sinx

xdx =

π

2.

IV. Other Improper Integrals

Using the same techniques, we can compute other important types of real integrals.


∞∫

−∞

eax

1 + exdx =

π

sinπa, for 0 < a < 1.

Solution. Let us consider the function

f(z) =eaz

1 + ez,

which is holomorphic on C \ (2k + 1)πi | k ∈ Z and a contour consisting of a

rectangle in the upper half-plane with one side lying on the real axis and another

one parallel with the real axis and situated at the level of the line Im z = 2π (see

Figure 6.3). More precisely, if R ∈ (0,∞), we take

ΓR = [−R,R] ∨ [R,R + 2πi] ∨ [R+ 2πi,−R + 2πi] ∨ [−R+ 2πi,−R].


x

y

O−R R

−R+ 2π i R+ 2π i

π i

Figure 6.3: The contour ΓR

The only singularity of f in this rectangle is z0 = πi, which proves to be a simple

pole. Then,

Res (f, z0) = −eaπi

and, using the residue theorem, we get

∫

ΓR

f(z) dz = −2πi eaπi.

Let

IR =

R∫

−R

f(x) dx,

which tends, when R → ∞, to

I =

∞∫

−∞

eax

1 + ex.

Let us evaluate now the integrals of f over each side of the rectangle. It is not

difficult to see that the integrals over the vertical sides tend to zero as R → ∞, while

the integral on the top side of the contour is equal to −e2πiaIR.

Therefore, when R tends to infinity, we get

I − e2aπiI = −2πi eaπi,


i.e.

I =π

sinπa.

In a similar manner, we can compute the Fresnel’s integrals, occurring in the theory

of diffraction. We have

∞∫

0

cos x2 dx =

∞∫

0

sin x2 dx =1

2

√π

2.


∞∫

−∞

x sinx

x2 + 1dx = πe−1.


∞∫

0

√x lnx

x2 + a2dx =

π√2a

(ln a+

π

2

), for a > 0.


∞∫

0

x lnx

(x2 + 1)3dx = −1

8.

Exercise 6.37 Evaluate the following improper integrals:

a) I =

∞∫

0

sin ax

x(x2 + b2)2dx, for a, b > 0.

b) I =

∞∫

0

√x lnx

(1− x)2dx.


Remark 6.38 Let f be a continuous real function. An improper integral of the

form (6.8) is convergent if there exists and it is finite

limR → ∞r → −∞

R∫

r

f(x) dx.

In this case,∞∫

−∞

f(x) dx = limR → ∞r → −∞

R∫

r

f(x) dx.

If there exists

limR→∞

R∫

−R

f(x) dx

and it is finite, the integral (6.8) is said to be convergent in the sense of Cauchy

principal value. In this case, the limit is called the Cauchy principal value of the

integral I. We use the notation

p.v.

∞∫

−∞

f(x) dx = limR→∞

R∫

−R

f(x) dx.

If the integral (6.8) is convergent, then it is convergent also in the Cauchy sense,

the converse implication being, in general, false. In the case in which the integral

(6.8) is not convergent and exists only as an improper Lebesgue integral, we have

to consider its Cauchy principal value, i.e. the value we must assign to such an

improper integral is the principal value integral.

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