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Complex Analysis
Mathematics 113. Analysis I: Complex Function Theory
Complex Analysis
Mathematics 113. Analysis I: Complex Function Theory
For the students of Math 113
Harvard University, Spring 2013
FOREWORD
Welcome to Mathematics 113: Complex Function Theory! These are the compiled lecturenotes for the class, taught by Professor Andrew Cotton-Clay at Harvard University duringthe spring of 2013. The course covers an introductory undergraduate-level sequence in com-plex analysis, starting from basics notions and working up to such results as the Riemannmapping theorem or the prime number theorem. We hope that these lecture notes will beuseful to you in the future, either as memories of the class or as a handy reference.
These notes may not be accurate, and should not replace lecture attendance.
Instructor: Andy Cotton-Clay, [email protected] assistants: Felix Wong, [email protected], Anirudha Balasubrama-nian, [email protected] websites: Professor Cotton-Clay’s website, http://math.harvard.edu/
~
acotton/
math113.html, Harvard course website http://courses.fas.harvard.edu/0405
Cover image: domain colored plot of the meromorphic function f(z) = (z�1)(z+1)2
(z+i)(z�i)2.
Source: K. Poelke and K. Polthier. Lifted Domain Coloring. Eurographics/ IEEE-VGTC Sympo-
sium on Visualization, (2009) 28:3. Credits: V.Y.Downloaded from www.flickr.com/photos/syntopia/6811008466/sizes/o/in/photostream/.
1
TABLE OF CONTENTS
The contents of this compilation are arranged by lecture, and generally have the topicscovered in a linear manner. Additional notes are provided in the appendices.
Foreword . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
2 Riemann Sphere, Complex-Di↵erentiability, and Convergence . . . . . 7
3 Power Series and Cauchy-Riemann Equations . . . . . . . . . . . . . . . 11
4 Defining Your Favorite Functions . . . . . . . . . . . . . . . . . . . . . . . 15
5 The Closed Curve Theorem and Cauchy’s Integral Formula . . . . . . 18
6 Applications of Cauchy’s Integral Formula, Liouville Theorem, MeanValue Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
7 Mean Value Theorem and Maximum Modulus Principle . . . . . . . . 28
8 Generalized Closed Curve Theorem and Morera’s Theorem . . . . . . 31
9 Morera’s Theorem, Singularities, and Laurent Expansions . . . . . . . 34
10 Meromorphic Functions and Residues . . . . . . . . . . . . . . . . . . . . 40
11 Winding Numbers and Cauchy’s Integral Theorem . . . . . . . . . . . . 42
12 The Argument Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
13 Integrals and Some Geometry . . . . . . . . . . . . . . . . . . . . . . . . . 49
14 Fourier Transform and Schwarz Reflection Principle . . . . . . . . . . . 52
2
15 Mobius Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
16 Automorphisms of D and H . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
17 Schwarz-Christo↵el and Infinite Products . . . . . . . . . . . . . . . . . . 62
18 Riemann Mapping Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 65
19 Riemann Mapping Theorem and Infinite Products . . . . . . . . . . . . 69
20 Analytic Continuation of Gamma and Zeta . . . . . . . . . . . . . . . . . 72
21 Zeta function and Prime Number Theorem . . . . . . . . . . . . . . . . . 78
22 Prime Number Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
23 Elliptic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86
24 Weierstrass’s Elliptic Function and an Overview of Elliptic Invariantsand Moduli Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
A Some Things to Remember for the Midterm . . . . . . . . . . . . . . . . 94
B On the Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97
C References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
Afterword . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106
3
CHAPTER 1
INTRODUCTION
Consider R but throw in i =p
�1 to get C = {a + bi : a, b 2 R}. We will be looking atfunctions f : C ! C which are complex-di↵erentiable. Complex-di↵erentiable, or holomor-phic, functions are quite a bit di↵erent from real-di↵erentiable functions.
We can think of the real world as rigid, and the complex world as flexible. Complex analysishas applications in topology and geometry (in particular, consider complex 4-manifolds),and of course, physics and everywhere else.
Anyway, write C = R2 with the identification (a, b) = a + bi. Addition and multiplica-tion by a 2 R is as for R. For multiplication in general, (a, b)(c, d) := (ac� bd, ad+ bc).
Claim. C is a field: in particular, it’s an additively commutative group, and multiplica-tively, C� {0} is a commutative group.
Proof. Should be immediate. ⌅
Claim. Every nonzero complex number z 2 C has a multiplicative inverse.Proof. In fact, z�1 := a
a
2+b
2 � ( b
a
2+b
2 )i works. ⌅
For z 2 C, z = a+ bi, we define the complex conjugate z := a� bi = a+ b(�i).
Claim. z + w = z + w, zw = zw for z, w 2 C.Proof. Write it out. ⌅
We also define the norm |z| :=p
a2 + b2 = zz. We can deduce a few properties of thenorm:
Claim. (multiplicative property) |z||w| = |zw| for z, w 2 C.Proof. It su�ces to work with squares. |z|2|w|2 = zzww = zwzw = |zw|2. ⌅
Algebraically, can we do better than the complex numbers (e.g. can we throw inp
1 + i,etc.)? The Fundamental Theorem of Algebra tells us we’re done as soon as we throw in i;
4
that is, we get all the solutions to polynomials.In particular, let p : C ! C, p(z) =
Pak
zk for ai
2 C. If the degree is > 0, then9z 2 C : p(z) = 0.
Consider the equation z2 � (1 + i) = 0. Let’s work this out explicitly for square roots.Given a+ bi, we want x+ iy s.t. (x+ iy)2 = a+ bi, which gives a system of equations:
(x2
� y2 = a
2xy = b=)
(a2 = (x2
� y2)2 = x4
� 2x2y2 + y4
(x2 + y2)2 = x4 + 2x2y2 + y4=) (x2
�y2)2+(2xy)2 = a2+b2
=) x2 + y2 =pa2 + b2 =)
(x2 = 1
2
(a+p
a2 + b2) � 0
y2 = 1
2
(�a+p
a2 + b2) � 0
=) x+ iy = ±
pa+
p
a2 + b2
2+ isgn(b)
p�a+
p
a2 + b2
2
!
We can take this big general formula and use it to solve our specific cases.
For z 2 C, z = a + bi, we define the real part Re(z) = a = z+z
2
, the imaginary partIm(z) = b = z�z
2i
.
The Geometry of the Complex Plane. We have a multiplicative norm |z|. View-ing z, w 2 C as vectors, addition is visualized as a parallelogram, and multiplication is bestseen in polar coordinates. Polar coordinates can be obtained by r = |z| and ✓ = Arg(z), theargument of z. So we get the pretty important identity
z = r cos ✓ + ir sin ✓ = r(cos ✓ + i sin ✓).
For multiplication, try z1
z2
, where z1
= r1
(cos ✓1
+ i sin ✓1
), z2
= r2
(cos ✓2
+ i sin ✓2
). Then
z1
z2
= r1
r2
(cos(✓1
+ ✓2
) + i sin(✓1
+ ✓2
)).
So |z1
z2
| = |z1
||z2
| and Arg(z1
z2
) = Arg(z1
) + Arg(z2
) mod 2⇡. Namely, multiplication byz = r(cos ✓ + i sin ✓) dilates the complex plane by r and rotates it counterclockwise by ✓.Another view of the complex number is that z = r(cos ✓ + i sin ✓) acts on C ⇠= R2 by
r
✓cos ✓ � sin ✓sin ✓ cos ✓
◆=
✓r cos ✓ �r sin ✓r sin ✓ r cos ✓
◆=
✓a �bb a
◆.
Of course, 1
z
= 1
r
(cos(�✓) + i sin(�✓)) = 1
r
(cos ✓ � i sin ✓) = 1
|z|2 z. So zn = rn(cos(n✓) +
i sin(n✓)).
Problem. Find all cube roots of 1.Solution. Want z3 = 1 = 1(cos 0 + i sin 0). So r3 = 1 =) r = 1 and 3✓ = 0 mod 2⇡ =)
✓ = 0, 2⇡
3
, 4⇡
3
. These give the cube roots of unity, and we can plot them on the unit circlein C. ⌅
Notation. ⇣n
= cos 2⇡
n
+ i sin 2⇡
n
, a primitive nth root of unity. The nth roots of unityare ⇣k
n
= cos 2⇡k
n
+ i sin 2⇡k
n
, k = 0, ..., n� 1.
5
The equation for nth roots of unity is zn � 1 = 0 = (z � 1)(zn�1 + zn�2 + ... + z + 1).This gives a simple formula for 1 + cos ✓ + cos 2✓ + ...+ cos(n� 1)✓. For z = cos ✓ + i sin ✓,1+z+ ...+zn�1 = (
Pn�1
k=0
cos k✓+ iP
sin k✓). Now multiply by z�1 = (cos ✓�1)+ i(sin ✓)to get
(cosn✓ � 1) + i sinn✓
(cos ✓ � 1) + i sin ✓=
(cosn✓ � 1)(cos ✓ � 1)� sinn✓ sin ✓
(cos ✓ � 1)2 + sin2 ✓.
Son�1X
k=0
cos k✓ =(cosn✓ � 1)(cos ✓ � 1)� i sinn✓ sin ✓
2� 2 cos ✓.
This isn’t the best formula, but we’re happy with it.
6
CHAPTER 2
RIEMANN SPHERE, COMPLEX-DIFFERENTIABILITY,AND CONVERGENCE
Stereographic projection. Consider a sphere S2 := {(u, v, w) : u2 + v2 + w2 = 1} in R3.We have a bijection ' : S2
� N ! C, where N is the north pole (point at 1), given bystereographic projection. Indeed, we define the Riemann sphere C := C [ {1}
⇠= S2.
Diagram 1: Stereographic projection. Source: Wikipedia
Given (u, v, w) 2 S2
�N and (x, y) 2 R2, the bijection is given by
' : (u, v, w) 7!
✓u
1� w,
v
1� w
◆, '�1 : (x, y) 7!
✓2x
x2 + y2 + 1,
2y
x2 + y2 + 1,x2 + y2 � 1
x2 + y2 + 1
◆.
Claim. Any circle on S2 maps to a circle or line in C under ' and vice-versa.Proof. I hope you paid attention in lecture. ⌅
Now consider the map (u, v, w) 7! (u,�v,�w), which swaps 0 2 C with S, the southpole, and 1 2 C with N , the north pole. What does it do to z = x + iy? We see thatz 7!
u�iv
1+w
.
7
Claim. This is 1
z
.Proof. Trivial. ⌅
In fact, 1
z
is just rotating the sphere around.
Claim. The inversion 1
z
sends (circles and lines) to (circle and lines).
If we have a function f : C ! C and we think it’s “nice” at 1, how do we quantifythat? We simply use the inversion 1
z
. In particular, we look at f( 1z
). Suppose f(1) = 1;then 1
f(
1z
)
sends 0 to 0.
For example, consider a polynomial of degree n, f(z) = c0
+ ...+ cn
zn and cn
6= 0. Take
1
f( 1z
)=
1
c0
+ ...+ cn
z�n
=zn
c0
zn + ...+ cn
,
which is well-defined in a neighborhood of 0 2 C and clearly, sends 0 7! 0.
Di↵erentiation. Lets talk about something we all like, di↵erentiation. :-)
Definition. Given f(z) : C �
open
U ! C, z 2 U , we say f is complex-di↵erentiable(or holomorphic) at z if
f 0(z) := limh!0
f(z + h)� f(z)
h
exists for h 2 C.
Proposition. If f, g are holomorphic, then fg and f + g is as well, with (fg)0 = f 0g + fg0
and (f + g)0 = f 0 + g0.Proof. Also trivial. ⌅
Example. Consider complex polynomials in z, which are holomorphic: f(z) =P
ck
zk.Then f 0(z) exists and f 0(z) =
Pn�1(k + 1)c
k+1
zk.
Example. Let f(z) = z. By showing that the limits in the real and imaginary direc-tions do not agree, show that this is not holomorphic.
Before starting anything else, we would like to review real-di↵erentiable functions. Thereare “di↵erent severities” of being real di↵erentiable. Let f : R �
open
U ! R:
1. real di↵erentiable: f 0(x) exists 8x in domain
2. C1: f 0(x) exists and is continuous
3. Ck: f, f 0, ..., f (k) exists and are continuous
4. C1: (smooth) all derivatives exists and are continuous
5. real-analytic: f(x) =P1
ck
xk is a convergent power series
The punchline is that homolorphic functions are always C1 and analytic! (This is some-thing that should really be appreciated.) Indeed, we oftentimes interchange holomorphic,C1, and analytic in the complex world.
8
To show this, we will go back to convergence. Consider a sequence of functions {fk
(z)}defined on E ⇢ C that is containable in a compact set. We say that it is pointwise conver-gence if lim
k!1 fk
(z) exists 8z; the pointwise limit is defined as f(z).Suppose f
k
(z) is continuous. It’s uniformly convergent if 8✏, 9N : 8n � N : |fn
(z) �f(
z)| ✏, 8z 2 E.
Lemma. If {fk
(z)} are continuous and converge uniformly on E, then pointwise limitf(z) is continuous.
Proof. Use triangle inequality and a standard epsilon pushing argument. ⌅
Power series. Consider f(z) =P1
0
ck
zk. We need a condition on |ck
| for this to converge.
Definition. Given a sequence {ak
}
11
, we define the limit supremum
lim supk!1
ak
= limn!1
✓supk�n
ak
◆= lim
k!1ak
The polynomial condition is then to look at lim sup |ck
|
1/k.
Theorem. If lim|ck
|
1/k = L for L = 0, 0 < L < 1, or L = 1, then:
1. If L = 0, then f(z) =P
ck
zk converges for all z.
2. If 0 < L < 1, thenP
ck
zk converges for |z| < R = 1
L
, diverges for |z| > R, and R iscalled the radius of convergence. (We do not know for |z| = R.)
3. If L = 1,P
ck
zk diverges 8z 6= 0.
Proof. L = 0: 8✏ > 0, 9N : 8k > N, |ck
|
1/k < ✏. Let ✏ = 1
z
|z|. Then |ck
|
1/k < 1
2
|z|,
Re|ck
| < 1
2
k
1|z|k. We want |ck
|
1/k
|z| < 1
2
, so |ck
||z|k
1
2
k
soP
ck
zk converges. We can dothis with the M -test, by considering partial sums, etc.
0 < L < 1: lim|ck
|
1/k = L. Let |z| = R(1� 2�), then lim|z||ck
|
1/k = 1� 2� for R = 1
L
.
So 9N : |z||ck
|
1/k < 1� � 8n > N , andP
ck
zk converges because |ck
|zk < (1� �)k.Likewise, if we suppose |z| = R(1 + 2�). Then lim|c
k
|
1/k
|z| = 1 + 2�. So for infinitelymany k, |c
k
|
1/k
|z| > 1+� and |ck
zk| > 1. The last condition is proved in the same manner. ⌅
Example.P1
0
zk = 1
1�z
. Deduce this from the theorem above.
Example.P1
0
(k + 1)zk = 1
(1�z)
2 . We see this by taking ck
= 1 and showing that
lim(k + 1)1/k = 1.
Theorem. Suppose f(z) =P1
0
ck
zk converges for |z| < R. Then:
1.P1
1
kck
zk�1 converges for |z| < R.
2. f 0(z) exists and equalsP1
1
kck
zk�1.
Proof. (1) It su�ces to check lim|ck
|
1/k
1
R
=) lim(k + 1)1/l|ck
|
1/k
1
R
. Note that
lim(k + 1)1/k|ck
|
1/k = lim(k + 1)1/klim|ck
|
1/k.(2) R = 1: subtract and show that limits ! 0.
f(z + h)� f(z)
h�
Xkc
k
zk�1 =1
h
Xck
[(z + h)k � zk]�X
kck
zk�1
9
=X
ck
[1
h(z + h)k �
1
hzk � kzk�1].
Recall the binomial theorem,
(z + h)k =
✓k
l
◆hlzk�l,
which we can use to rewrite the expression above as
=1X
0
ck
[kX
l=1
✓k
l
◆zk�lhl�1
� kzk�1] =1X
0
kX
2
✓k
l
◆zk�lhl�1
= h
1X
0
kX
2
✓k
l
◆zk�lhl�2.
We also have ����X✓
k
l
◆zk�lhl�2
���� X✓
k
l
◆|z|k�1 = (|z|+ 1)k,
soP
ck
(|z|+ 1)k converges because R = 1 forP
ck
zk.R < 1: Let |z| = R� 2�, |h| < �; then |z + h| < R� �. Then
f(z + h)� f(z)
h�
Xkc
k
zk�1 = h
1X
0
ck
kX
2
✓k
l
◆zk�lhl�2.
Write ✓k
l
◆=
k(k � 1)...(k � l + 1)
l!.
For l � 2 we can bound�k
l
� k2
�k
l�2
�. So
�����X
2
✓k
l
◆zk�lhl�2
����� k2
|z|2(|z|+ |h|)k
and the equation above translates into
=h
|z|2
Xk2c
k
(|z|+ |h|)k,
which converges and ! 0 because of the h in the front. ⌅
10
CHAPTER 3
POWER SERIES AND CAUCHY-RIEMANN EQUATIONS
Power series. Last time we were dealing with power series, f(z) =P1
k=0
ck
zk. We definedthe radius of convergence R = 1
L
, where L = lim|ck
|
1/k and either L = 0, 0 < L < 1, orL = 1.
Inside the radius of convergence, f(z) is a convergent power series with some R. Whatare the consequences of this?
Corollary. Inside its radius of convergence, a power series is 1-di↵erentiable, with ex-pected derivatives as convergent power series.
Corollary. If R > 0, then ck
= f
(k)(0)
k!
.
Proof. Take f (k)(x). The corollary follows. ⌅
There are also several uniqueness properties:
Lemma. If f(z) =P
ck
zk is a convergent power series and f(zn
) = 0 for a sequence{z
n
}
1n=1
with zn
! 0, zn
6= 0, then ck
= 0 8k and Re(f(z)) ⌘ 0.
Proof. c0
= f(0) = limn!1 f(z
n
) = 0. Now form g1
(z) = f(z)
z
= c1
+ c2
z + c3
z2 + ...(exercise: show that this holds for some radius of convergence). Check c
1
= g1
(0) =
limn!1 g
1
(zn
) = limn!1
f(z)
z
= 0, and induct on gi
. ⌅
Proposition. If f(z) =P
ak
zk, g(z) =P
bk
zk and these agree on some set accumu-lating at 0, then a
k
= bk
8k, i.e. f(z) = g(z).Proof. Consider
P(a
k
� bk
)zk and apply the lemma. Show that lim|ak
|
1/k
L andlim|b
k
|
1/k
L =) lim|ak
� bk
|
1/k
L. ⌅
Note: To center the power series at w 2 C, considerP
ck
(z � w)k, which shifts the centerof the power series from 0 2 C to w and maintains the radius of convergence at R.
Complex-di↵erentiability. We refer to complex-di↵erentiability as either holomorphicor complex-analytic (some texts, e.g. Ahlfors, simply like to use analytic). This is a verynice property of functions that we will be exploring in the upcoming weeks.
11
Definition. f : C ! C is holomorphic at z 2 C if
limh!0
f(z + h)� f(z)
h
exists for h 2 C. Equivalently, 9 a number f 0(z) :
0 = limh!0
f(z + h)� f(z)
h� f 0(z).
The Cauchy-Riemann equations. By considering real and imaginary parts of holomor-phic functions, we get the celebrated Cauchy-Riemann equations.
Proposition. If f(z) : C ! C is holomorphic at z (alternatively, by R2
⇠= C, we canregard f(x, y) : R2
! R2), then @f
@x
and @f
@y
exist and
i@f
@x=@f
@y. (3.1)
Equivalently, for the decomposition f(x, y) = u(x, y) + iv(x, y) and
@f
@x=@u
@x+ i
@v
@x
@f
@y=@u
@y+ i
@v
@y,
what we mean is that, by comparing real and imaginary parts of the equation
i
✓@u
@x+ i
@v
@x
◆=@u
@y+ i
@v
@y,
we obtain the equalities (@u
@x
= @v
@y
�
@v
@x
= @u
@y
.(3.2)
Proof. This is a bit tedious to type up, so I hope you paid attention in lecture. The prooffollows from a simple computation of the partials; see any textbook. ⌅
The CR conditions (1) or (2), along with the condition that the partials are continuous,ascertains that f is itself holomorphic.
If the partials are not continuous, consider f(x, y) = xy(x+iy)
x
2+y
2 , z 6= 0, and 0 for z = 0.
Show that this is not di↵erentiable at 0 but @f
@x
(0, 0) = @f
@y
(0, 0) = 0.
Proposition. If f(x, y) = u(x, y)+iv(x, y) has continuous partial derivatives and i@f@x
= @f
@y
(satisfies CR equations), then f is holomorphic.Proof. Messy as well, but use the mean value theorem and write out the partials. ⌅
12
An equivalent (geometric) formulation of the CR equations. Consider the Jacobianof f : R2
! R2:
Jf
:=
@u
@x
@u
@y
@v
@x
@v
@y
!
and the rotation matrices ✓cos ✓ � sin ✓sin ✓ cos ✓
◆.
So multiplication by i is multiplication by the matrix (aside: see the relation to complexstructure) ✓
0 �11 0
◆.
The equivalent form is that Jf
I = IJf
; e.g. the Jacobian matrix commutes with rotationby ⇡/2. Check this:
@u
@x
@u
@y
@v
@x
@v
@y
!✓0 �11 0
◆=
@u
@y
�
@u
@x
@v
@y
�
@v
@x
!=
�
@v
@x
�
@v
@y
@u
@x
�
@u
@y
!=
✓0 �11 0
◆ @u
@x
@u
@y
@v
@x
@v
@y
!.
An algebraic interpretation. Consider complex polynomials, p(z) =P
n
k=0
ck
zk. Lookat complex-valued polynomials of real variables x, y:
p(x, y) =nX
m=0
X
k+l=m,
k,l�0
ck,l
xkyl (3.3)
with ck,l
= @p
@
k
x@
l
y
(0, 0), and take a “new basis” as z = x + iy and z = x � iy. Write thepolynomial in z, z :
p(z, z) =X
m=0
X
k+l=m,
k,l�0
ck,l
zkzl. (3.4)
We claim that there’s a 1-1 correspondence between polynomials given by (3) and (4).Indeed, @
@x
, @@y
act on these polynomials:
@
@z=
1
2
✓@
@x� i
@
@y
◆,
@
@z=
1
2
✓@
@x+ i
@
@y
◆
Claim. We have the equalities
@
@z(zkzl) = lzkzl�1,
@
@z(zkzl) = kzk�1zl.
Proof. We simply check:
@
@z(zkzl) =
1
2[kzk�1zl + lzkzl�1] +
i
2[kizk�1zl + (�i)lzkzl�1] = lzkzl�1,
13
and likewise for the second equality. ⌅
Claim. (equivalent form of CR equations) f(z) satisfies CR equations ()
@
@z
f = 0.Proof. A simple check. ⌅
The conclusion is that p(z, z) =PP
ck,l
zkzl is holomorphic i↵ ck,l
= 0 for l > 0 (e.g.no z0s). Alternatively, p(x, y) =
PPck,l
xkyl is holomorphic i↵ it can be written asp(z) =
Pck
zk.
Let’s give some basic hints that holomorphic functions behave a bit like power series:
Lemma. Suppose f(x, y) = u(x, y) + iv(x, y) is holomorphic on a disk of radius R andu(x, y) is constant. Then f is constant.
Proof. ux
= uy
= 0, so by CR equations, ux
= vy
, uy
= �vx
and vx
= vy
= 0. By themean value theorem, this shows that v is constant along horizontal and vertical lines in theplane. So v is constant throughout the disk. ⌅
Lemma. If f = u + iv is holomorphic on a disk in C and ||f ||2 is constant, then f isconstant.
Proof. We have u2 + v2 = c, so taking partials by x and y gives 2ux
u + 2vx
v = 0 and2u
y
u + 2vy
v = 0. Applying the CR equations, we get 2ux
v � 2vx
u = 0. Adding equationsgives 2u
x
(u2 + v2) = 0, and similarly 2vx
(u2 + v2) = 0. So either u2 + v2 = 0 (f = 0), or allpartials = 0.
In the second case, given any open ball, f is constant in that open ball since its partials= 0 in the ball. But the disk is connected, so f is constant everywhere.
14
CHAPTER 4
DEFINING YOUR FAVORITE FUNCTIONS
Today is about extending your favorite (real analytic) functions from R to C. When deter-mining their complex analogues, we could (1) ask for the same fundamental properties; (2)use their power series; (2’) extend such that the result is holomorphic (actually, this is thesame as 2). Happily, these all agree!
1. ex
We can consider (1) e0 = 1, d
dx
ex = ex, ... or (2) ex =P
x
n
n!
.
(1) Let’s look for f(z) with f(x) = ex and f(z1
+ z2
) = f(z1
)f(z2
) and holomorphic.Apparently, f(x+ iy) = exf(iy). Let f(iy) = A(y) + iB(y), so f(x+ iy) = exA(y) +exiB(y). Then use the Cauchy-Riemann equations to show that A(y) = cos y andB(y) = sin y by the Fundamental Theorem of ODEs (namely, any linear ordinary
di↵erential equation has an ez solution). So define ez = ex(cos y + i sin y), for z =
x+ iy.
(2) Let
f(z) =1X
n=0
zn
n!.
The radius of convergence is 1, since limn!1( 1
n!
)1/n = 0 (e.g. it’s an entire function).Also, f(z) is holomorphic on all of C. We can easily verify the properties that f 0(z) =f(z), f(0) = 1, f(z)f(w) = f(z + w), etc.
2. sin z, cos z
For f(z) = ez, look at f(iz) = 1+(iz)+ (iz)
2
2!
+... = (1� z
2
2!
+ z
4
4!
+...)+i(z� z
3
3!
+ z
5
5!
+...).Then we set f(iy) =: cos y + i sin y.
We can verify the usual properties that @
@z
cos z = � sin z, @@z
sin z = cos z, cos 0 = 1,sin 0 = 0, cos(�z) = cos z, sin(�z) = � sin z, eiz = cos z + i sin z, etc.
Likewise, we can express
cos z =eiz + e�iz
2, sin z =
eiz � e�iz
2i.
15
We can likewise check the addition formulas for cos(z1
+ z2
) and sin(z1
+ z2
).
3. log z
w = log z () z = ew. For w = a + bi, ew = ea(cos b + i sin b). Apparently, log z =log r+i✓+i(2⇡n), so the complex log is multi-valued. Write z = rei✓ = r(cos ✓+i sin ✓).
We define the principal branch of log z as follows. Consider C� R0
. log z has imag-inary part in (�⇡,⇡). Write log z = log |z|+ iArg(z) for Arg(z) 2 (�⇡,⇡); this is ourdefinition of log z for the principal branch, which has range �⇡ < Im(z) < ⇡.
Claim. log z is holomorphic given a branch cut, and @
@z
(log z) = 1/z.
Proof. Consider
limh!0
log(z + h)� log z
h.
Set w1
= log(z + h), w = log z. Then, noting that log is continuous given a branchcut, we see that
limh!0
log(z + h)� log z
h= lim
h!0
w1
� w
ew1� ew
=1
limw1!w
e
w1�e
w
w1�w
=1
ew=
1
z.
⌅Intuitively, log z looks like a helix above C� {0}; we call this the Riemann surface forlog z.
4. z↵,↵ 2 RDefine z↵ = e↵ log z = e↵(log z+2⇡ni), which is potentially multi-valued. e↵ log z is theprincipal branch; now examine e2⇡n↵i. If ↵ /2 Q, e2⇡n↵i takes infinitely many values.If ↵ = k
l
in reduced form, e2⇡nki/l takes l values (e.g. en⇡i = ±1 for ↵ = 1
2
).
For example, when consideringp
z, we still need a branch cut.p
rei✓ =p
rei✓/2,✓ 2 (�⇡,⇡).
Integration. Given f(z) holomorphic, can we find another function F (z) holomorphic withF 0(z) = f(z)? The answer is, generally, yes. We begin with integration along a contour,RC
f(z)dz for C a contour.
Definition. The image of �(t) : [a, b] ! C is a contour curve if:
1. � is continuously di↵erentiable except at finitely many points
2. When di↵erentiable, �0(t) 6= 0 except at finitely many points
For C a contour,RC
f(z)dz =Rb
a
f(�(t))�0(t)dt. Call �1
(t) ⇠ �2
(t) for �1
: [a, b] ! C and�2
: [c, d] ! C if 9�(t) : [c, d] ! [a, b] s.t. �2
(t) = �1
(�(t)).
Claim. This is well defined:
Zb
a
f(�1
(t))�01
(t)dt =
Zd
c
f(�2
(t))�02
(t)dt.
16
Lemma. (complex chain rule) Let g(t) = f(�(t)), f(z) holomorphic. Then g0(t) =�0(t)f 0(�(t)).
Proof. As usual, with limits or whatnot. ⌅
Example. Compute Z
S
1
zkdz
for k 2 Z, and S1 the unit circle in C oriented counterclockwise.Explanation. Let �(t) = eit = cos t+ i sin t. By chain rule, �0(t) = ieit. Then
Z
S
1
zkdz =
Z2⇡
0
eiktieitdt = i
Z2⇡
0
ei(k+1)tdt.
If k = �1, this is Z
S
1
1
zdz = i
Z2⇡
0
1dt = 2⇡i.
If k 6= 1, this isZ
2⇡
0
ei(k+1)tdt =ei(k+1)t
i(k + 1)
����2⇡
0
= 0.
So zk for k 2 Z, k 6= �1 has an antiderivative in C� {0}: z
k+1
k+1
. 1/z does not; we’d wantlog z, but it’s multivalued.
Proposition. If f(z) holomorphic with F (z) holomorphic and F 0(z) = f(z), then
Z
C
f(z)dz = F (end)� F (start),
for C a path from start to end.Proof. Use the complex chain rule. ⌅
Next time, we’ll consider functions that are defined on all of C (which we call entire func-tions), we’ll show 9F (z) : F 0(z) = f(z). Along the way, we’ll prove the closed curve theorem.
For motivation, we recall Green’s theorem:
Zb
a
(P,Q) · r0(t)dt =
Z Z
D
✓@Q
@x�
@P
@y
◆dxdy
for r(t) : [a, b] ! R2 a closed curve around a region D. This requires that Qx
, Py
becontinuous (or something like that). The theorem we’ll prove does not need this requirement.
17
CHAPTER 5
THE CLOSED CURVE THEOREM AND CAUCHY’SINTEGRAL FORMULA
Today we’ll be talking about the closed curve theorem andCauchy’s integral theorem.Recall from last lecture that, for a parameterization � of the contour C, we had
Z
C
f(z)dz =
Zf(�(t))�0(t)dt.
We also know that the path integral is determined by the antiderivative’s value at the curveendpoints:
Proposition. If F (z) is holomorphic, F 0(z) = f(z), then
Z
C
f(z)dz = F (final)� F (initial).
The closed curve theorem. We will first provide motivation for the closed curve theorem.Given f and �, consider splitting into real and imaginary parts:
f(z) = f(x, y) = u(x, y) + iv(x, y)
�(t) = a(t) + ib(t)Z
f(�(t))�0(t)dt =
Z(u+ iv)(a+ ib)dt =
Z[(ua� vb) + i(ub+ va)]dt
=
Z(u,�v) · (a, b)dt+ i
Z(v, u) · (a, b)df (5.1)
If C is the closed, counterclockwise boundary of D, then by Green’s theorem (which onlyapplies when all partials are continuous) we have that the above is equal to
Z Z
D
✓�
@v
@x�
@u
@y
◆dxdy + i
Z Z
D
✓@u
@x�
@v
@y
◆dxdy = 0,
18
where the last equality follows from the Cauchy-Riemann equations. We can also note that,in the language of vector fields, (1) is equivalent to
Z Z
D
curl(f)dA+
Z Z
D
div(f)dA,
where f is thought of as a vector field on R2 induced by f . The equivalent formulation ofthe CR equations would then be curl(f) = div(f) = 0 as a vector field on R2.
The conclusion above gives us a glimpse into the closed curve theorem, which, however,does not require the first partials to be continuous. Keep in mind that our objective is theresult that f being holomorphic on the interior of a disk implies that f has a convergentpower series expansion on that disk: we will do this using the closed curve theorem andfriends.
Claim. The closed curve theorem (stated below) is true for linear functions f(z) = a+ bz.Proof. Let F (z) = az + 1
2
bz2, so F 0(z) = f(z). So we haveRC
f(z)dz = F (final) �F (initial) = 0. ⌅
Theorem. (closed curve theorem) Suppose f(z) is holomorphic in an open disk D. Let Cbe any closed contour curve in D. Then
Z
C
f(z)dz = 0.
Remark. Our strategy is in three steps: (1) show this for any polygonal curve; (2) showthat f(z) has an antiderivative on D; (3) conclude the result.
Proof for triangle. Let T be a triangle contour. The idea is that f being holomorphicimplies that f is well-approximated by linear functions on small scales; see the claim above.
First, we subdivide the triangle. Suppose |
RT
f(z)dz| = C. Then 9Ti
with boundary �i
such that |
R�
i
f(z)dz| � c/4. Call this T (1) with boundary �(1). Keep subdividing to get
(dropping the parentheses) T 1, T 2, T 3, ... and �1,�2,�3, .... such that |
R�
k
f(z)dz| � c
4
k
.Note that \1
k=1
T k = {point}; call the point z0
. f(z) is holomorphic at z0
, so we have
limz!z0
����f(z)� f(z
0
)
z � z0
� f 0(z0
)
���� = 0.
Thusf(z) = f(z
0
) + f 0(z0
)(z � z0
) + ✏(z)(z � z0
)
with ✏(z) ! 0 as z ! z0
. Now consider
Z
�
k
f(z)dz =
Z
�
k
[f(z0
) + f 0(z0
)(z � z0
) + ✏(z)(z � z0
)]dz =
Z
�
k
✏(z)(z � z0
)dz.
Given ✏ > 0, 9N s.t. for k � N , ✏(z) < ✏ for z 2 T k. Let the perimeter of T be L and theperimeter of T k be L
2
k
. The max distance between two points in T k would then be
L
2
k
.Thus
c
4k
����Z
�
k
✏(z)(z � z0
)dz
���� L
2k✏L
2k=✏L2
4k,
19
where the first term after the equality is length of �k, the second is an upper bound on ✏(z),and the third is an upper bound on |z�z
0
|. We had c = |
R�
f(z)dz|, so c ✏L2 =) c = 0. ⌅
Corollary. (extension to polygons) If P is a polygon and f(z) is holomorphic on an openneighborhood of P with boundary �, then
RP
f(z)dz = 0.Proof. Divide P into triangles. ⌅
Theorem. (antiderivative theorem) If f(z) is holomorphic on an open disk D (or anopen polygonally simply connected region ⌦ ⇢ C), then 9F (z) holomorphic on D (or ⌦) s.t.F 0(z) = f(z).
Definition. An open region ⌦ ✓ C is polygonally connected (in topology, connected) if8z, w 2 ⌦, 9 piecewise linear curve in ⌦ connecting z, w. A region is simply polygonallyconnected (simply connected) if any closed polygonal curve is the boundary of a union ofpolygons in ⌦.
Proof of antiderivative theorem. Let p 2 D (or ⌦ ✓ C). Let F (z) =Rz
p
f(z)dz (theline integral along any polygonal path from p to z, which is well-defined by the result ontriangles above and the p.s.c. property of ⌦). We check that
F (z)� F (z0
)
z � z0
=1
z � z0
Zz
z0
f(z)dz.
Note ����F (z)� F (z
0
)
z � z0
� f(z0
)
���� =����
1
z � z0
Zz
z0
(f(z)� f(z0
))dz
����
because Zz
z0
f(z0
)dz = f(z0
)
Zz
z0
1dz = f(z0
)(z � z0
).
Then we have ����1
z � z0
Zz
z0
(f(z)� f(z0
))dz
���� 1
|z � z0
|
|z � z0
|✏,
where the middle term after the inequality is the length of our curve and ✏ is upper boundfor |f(z)� f(z
0
)| (by continuity). So the LHS ! 0 as ✏! 0. ⌅
Corollary. (closed curve theorem for regions) If f(z) is holomorphic on an open diskD (or p.c.r. region ⌦), then
RC
f(z)dz = 0 for C being any closed contour in D (or ⌦).Proof. Take F (z), the antiderivative of f(z), so that
RC
f(z)dz = F (final)�F (initial) =0. ⌅
Cauchy’s integral theorem. Given that f(z) is holomorphic, let’s look at
g(z) =
(f(z)�f(w)
z�w
z 6= w
f 0(w) z = w
for w 2 C.
Claim. g(z) satisfies the closed curve theorem.
20
Proof. g(z) is continuous because f 0(w) exists. The goal is to prove the above resultfor triangles for g(z); then it will follow 9G(z) : G0(z) = g(z), along with the closed curvetheorem for g(Z).
Consider w. If w is outside T , then g(z) holomorphic in an open neighborhood of T andthe old closed curve theorem applies.
If w 2 � = @T , then we cut o↵ small pieces {Ti
} soR�
i
f(z)dz = 0 (because they do notcontain w). The length of �i can be arbitrarily small at M , the upper bound on g(z) on T ,so it goes to 0 in the limit. If w is in the interior, we also use the same trick. ⌅
Cauchy integral formula. This is one of the central results of complex analysis. LetC be a circle surrounding w 2 C, and f(z) holomorphic on the closed boundary C of theopen disk D. Then we have the formula
f(w) =1
2⇡i
Zf(z)
z � wdz
Proof. Suppose C is a circle centered at w. The circle is parameterized by �(t) = w+Reit
for 0 t 2⇡. We know from the closed curve theorem that
0 =
Z
C
f(z)� f(w)
z � wdz,
because the integrand is g(z) defined on an interior satisfying the closed curve theorem.Now split this up:
Z
C
f(z)
z � wdz = f(w)
Z
C
1
z � wdz,
Z
C
1
z � wdz =
Z2⇡
0
iReitdt
Reit= 2⇡i
=) f(w) =1
2⇡i
Zf(z)
z � wdz.
⌅
Theorem. Suppose f(z) is holomorphic on the open disk Dk
(0) of radius R around 0 2 C.Then there exists a convergent power series
P1k=0
ck
zk with lim|ck
|
1/k
1
R
such thatf(z) =
P1k=0
ck
zk.Proof. From Cauchy’s integral formula, we have
f(z) =1
2⇡i
Zf(w)
w � zdw.
Let |z| < R < R, and write
1
w � z=
1
w
1
1� z
w
=1
w
✓1 +
z
w+
z2
w2
+ ...
◆
|
z
w
| = |z|˜
R
< 1 so this converges uniformly to 1
w�z
for w 2 C. Then
f(z) =1
2⇡i
Z
˜
C
f(w)(1
w+
z
w2
+z2
w3
+ ...)dz
21
=1
2⇡i
Z
C
f(w)
wdw +
✓1
2⇡i
Z
C
f(w)
w2
dw
◆z +
✓1
2⇡i
Z
C
f(w)
w3
dw
◆z2 + ...,
i.e.
Ck
=1
2⇡i
Z
˜
C
f(w)
wk + 1dw.
We check this as follows:
|Ck
|
1
2⇡2⇡RM
1
Rk+1
= M/Rk,
where the terms of the right of the inequality come as follows: 2⇡R is the length of R and
M is an upper bound on f in D˜
R
(0). So lim|Ck
|
1/k
limM
1/k
˜
R
= 1
˜
R
, and
f (k)(0)
k!=
1
2⇡i
Z
C
f(w)
wk+1
dw
for C containing 0. ⌅
Thus we have the result that f holomorphic =) f analytic =) f infinitely di↵erentiable.
22
CHAPTER 6
APPLICATIONS OF CAUCHY’S INTEGRAL FORMULA,LIOUVILLE THEOREM, MEAN VALUE THEOREM
Applications of Cauchy’s integral formula. Recall that we have the Cauchy integralformula:
f(z) =1
2⇡i
Z
C
f(w)
w � zdw,
for C a curve containing z. For f(w) =P
ck
wk, note that we can write
f(0) =1
2⇡i
Z
C
f(w)
wdw =
1
2⇡i
Z
C
1X
k=1
ck
wk�1dw =1
2⇡i
Z
C
c0
wdw = c
0
.
Let’s use this formula to compute some stu↵. The general method is to decompose a closedcontour (over which the integral is zero) into a sum of directed edges, then evaluating theintegral over the other edges to give us results.
Claim. Z 1
0
sinx
xdx =
⇡
2.
Proof. Let CR
denote the contour of an upper half-circle of radius R centered at theorigin. For r < R, we can use Cauchy’s integral theorem to write (noting that the integrand
is equivalent to the imaginary part of e
ix
x
:
0 =
ZR
r
eix
xdx+
Z
C
R
eiz
zdz +
Z �r
�R
eix
x�
Z
C
r
eiz
zdz.
Now parameterize Cr
with z = rei✓ for ✓ 2 [0,⇡] and likewise with z = Rei✓ to see
0 =
ZR
r
eix
xdx+
Z⇡
0
eiRe
i✓
Rei✓iRei✓d✓ +
Z �r
�R
eix
xdx�
Z⇡
0
eirei✓
rei✓irei✓d✓.
=
ZR
r
eix
xdx+
Z⇡
0
ieiRe
i✓
d✓ +
Z �r
�R
eix
xdx�
Z⇡
0
ieirei✓
d✓.
23
But note that the second integral ! 0 as R ! 1 and the fourth integral ! ⇡i as r ! 0(show this yourself), so we’re done as sin x
x
(the imaginary part of the remaining two inte-grals) is symmetric over the y-axis and its integral from 0 to 1 would then be equal to theimaginary part of our answer divided by 2. Namely, we have
R10
sin x
x
dx = ⇡
2
as desired. ⌅
Claim (added by Felix, for more practice)
Z 1
0
sin(x2)dx =
Z 1
0
cos(x2)dx =
p
2⇡
4.
Proof. We first show thatR1�1 e�x
2
dx =p
⇡ as follows:
✓Z 1
�1e�x
2
dx
◆2
=
Z 1
�1
Z 1
�1e�(y
2+z
2)dydz =
Z2⇡
0
Z 1
0
re�r
2
drd✓ = ⇡,
with y = r cos ✓ and z = r sin ✓.Using what we showed above, we integrate f(z) = e�z
2
on the contour defined byC = {Reit : 0 < t < ⇡/4}. Observe that we can bound the integral by
0
����Z
C
f(z)dz
���� Z⇡/4
0
e�R
2(cos
2(t)�sin
2(t))Rdt =
Z⇡/4
0
e�R
2cos(2t)Rdt,
but also (from brute force bounding) we know that RR⇡/4
0
e�R
2cos(2t)dt ! 0 as R ! 1
(⇤). This is good, because we can invoke Cauchy’s theorem to see thatRC
0 f(z) for C 0 ={reit : 0 < r < R, 0 < t < ⇡/4} = �
1
+ �2
+ �3
(where �1
is the path from 0 to R, �2
is theeighths-circle, and �
3
is the line from Rei⇡/4 back to 0), a closed boundary, is zero. LettingR ! 1, we saw from above that
R�2
f(z) = 0, so what we have left is (I’ll drop the f(z)’s
to make writing easier)
Z
�1
+
Z
�3
=
Z 1
0
e�z
2
+
Z
�3
=p
⇡/2 +
Z
�3
= 0 )
p
⇡/2 = �
Z
�3
.
So to finish, we evaluate �
R�3:
�
Z
�3
=
ZR
0
e�(e
i⇡/4t)
2
ei⇡/4dt = ei⇡/4Z
R
0
e�it
2
= ei⇡/4Z
R
0
cos(t2)� i sin(t2)dt.
Thus we see that
p
⇡/2 = ei⇡/4Z 1
0
cos(t2)�i sin(t2)dt )
Z 1
0
cos(t2)�i sin(t2)dt =
p
⇡/2p
2/2 +p
2/2i=
p
2⇡
4�
p
2⇡
4i.
Equating real and imaginary parts, we see thatR10
sin(x2)dx =R10
cos(x2)dx =p2⇡
4
, asdesired.
(⇤) Brute force bounding: For more exercise, we’ll explicitly show that
R
Z⇡/4
0
e�R
2cos(2t)dt ! 0
24
as R ! 1 here. We have
R
Z⇡/4
0
e�R
2cos(2t)dt = R
Z⇡/4
0
e�R
2sin(2t)dt = R
Z⇡/6
0
e�R
2sin(2t)dt+R
Z⇡/4
⇡/6
e�R
2sin(2t)dt.
Then, since we can bound sinx � (1/2)x for 0 x 1/3 ( d
dx
sinx � 1/2 for x 1/3) andsinx 1/2 for ⇡/6 ⇡/4, we see that
R
Z⇡/6
0
e�R
2sin(2t)dt+R
Z⇡/4
⇡/6
e�R
2sin(2t)dt
Z⇡/6
0
Re�R
2tdt+
Z⇡/4
⇡/6
Re�R
2/2dt
=⇡
12Re�R
2/2 +
1
R(1� e
�⇡R
2
6 ) ! 0,
as R ! 1. Integrals are fun stu↵. ⌅
Now let’s turn to other theorems we can prove using the Cauchy integral formula.
Theorem. (Liouville) A bounded entire function f(z) is constant.Proof. Suppose |f(z)| M for all z. Consider f(a), f(b) for |a|, |b| < R. We would like
to show that f(a) = f(b). Let C0
(R) be the circle of radois R about 0. Then
|f(a)� f(b)| =1
2⇡
�����
Z
C0(R)
f(z)
z � adz �
Z
C0(R)
f(z)
z � bdz
�����
andf(z)
z � a�
f(z)
z � b=
(b� a)f(z)
(z � a)(z � b).
For |b� a| 2R, we can also get
1
2⇡
�����
Z
C0(R)
f(z)
z � adz �
Z
C0(R)
f(z)
z � bdz
����� =1
2⇡
�����
Z
C0(R)
f(z)(b� a)
(z � a)(z � b)dz
����� 1
2⇡
2⇡RM |b� a|1
4
R2
=4M |b� a|
R! 0
as R ! 1. ⌅
Let’s look at polynomials P (z) =P
n
k=0
ck
zk, cn
6= 0. What can we say about them?
Claim. limz!1 P (z) = 1 in the sense that 8M, 9r : 8|z| > R, |P (z)| > M .
Proof. Should be obvious. ⌅
Theorem. (Fundamental Theorem of Algebra) Given a nonconstant polynomial P (z) =Pn
k=0
ck
zk, cn
6= 0, n � 1, 9w 2 C : p(w) = 0.Proof. Suppose not. Then 1
P (z)
=: f(z) is a bounded entire function. Choose R suchthat
1
R>
n�1X
k=0
|ck
|
|cn
|
25
and R > 1 so that for |z| > R we have
1
P (z)�
11
2
R|cn
|
Note that, in obtaining this inequality, we noted that the highest power of any polynomialdominates in the limit |z| ! 1. So f(z) is bounded by C for |z| > R, and it’s similarlybounded on any compact region D
0
(R); thus f(z) is constant by Liouville’s theorem. ⌅
Theorem. (Generalization of Liouville’s Theorem) If |f(z)| A + B|z|N and f(z) isentire, then f(z) is a polynomial of degree N .
Proof. By induction. The base case is Liouville’s theorem. Now define
g(z) =
(f(z)�f(↵)
z�↵ z 6= ↵
f 0(↵) z = ↵
for any ↵. Notice that g(z) is holomorphic, since we saw 9G(z) that is holomorphic withG0(z) = g(z), and we showed that holomorphic implies C1 so g(z) is also holomorphic.Notice further that |g(z)| C +D|z|N�1.
Addendum by Felix: We can see the last statement more clearly as follows. If to thecontrary |g(z)| > D|z|n�1, then taking limits as |z| ! 1 we see that the growth rate
(growth order � N) would not match with f(z)�f(↵)
z�a
or f 0(↵) (growth order N � 1).(Also, as said in lecture, we can see the last statement as follows: look inside a large
ball, then outside the large ball. Inside, it’s bounded by some constant; outside, because ofthe induction hypothesis, it has the form D|z|N�1.)
So g(z) is a polynomial of degree N � 1 by induction and f(z) = g(z)(z � ↵) is apolynomial of degree N . This completes the inductive step. ⌅
Uniqueness theorem. It turns out that holomorphic functions are determined up totheir values in a certain region; this is the content of the uniqueness theorem.
Definition. A region ⌦ in C is a polygonally connected open set.
Proposition. (uniqueness theorem) If f and g are holomorphic on a region D ✓ C withf(z) = g(z) on a collection of points accumulating somewhere, then f(z) = g(z) on D.
Proof. Let A = {z 2 D : 9 infinitely many points w in any disk containing z s.t.f(w) = g(w)}. Let B = D\A. We claim that both A and B are open.
To show that B is open, note that if z 2 B, then 9 a disk containing z with only finitelymany such w’s. Let � < lim
such w,w 6=z
|z|. Then every point in the open disk of radius � isin B.
To show that A is open, suppose z 2 A. In D, 9 a disk of radius r containing z. Onthat disk, both f and g being holomorphic implies that both f and g have convergent powerseries expansions centered at z of radius of convergence � r. So by uniqueness for powerseries, f = g on this disk, i.e. every point in this disk in in A.
So we have that A and B are both open, A [ B = D, and A \ B = ;. By elementarytopology, one must be the empty set. ⌅
It turns out that polynomials are the only entire functions that go to infinity with z inthe limit. Geometrically, this means that if we consider the Riemannian sphere C with
26
f : C ! C entire, then f : 1 7! a for a finite implies that f(z) is bounded, i.e. a constant.If f : 1 7! 1, then f(z) is a polynomial.
Proposition. If f(z) is entire and f(z) ! 1 as z ! 1, then f(z) is a polynomial.Proof. The zeroes of f all lie in a disk of some radius, call it R. There are only finitely
many zeroes, else they accumulate somewhere and then f = 0. Now consider
g(z) =f(z)
(z � z1
)m1(z � z2
)m2 ...(z � zk
)mk
.
Note that g(z) has no zeroes and is entire. lim f(z) = 1 =) |g(z)| � 1
|z|m , wherem =P
mi
.
So 1
g(z)
is entire and bounded, and by Liouville’s theorem it must be a constant. Thus f(z)is a polynomial. ⌅
We take the time to state one last important result, the mean value theorem, withoutproof:
Theorem. (mean value) If f(z) is holomorphic on D↵
(R), then 8r : 0 < r < R, wehave
f(↵) =1
2⇡
Z2⇡
0
f(↵+ rei✓)d✓.
Corollary. This is true as well for u = Re(f), v = Im(f) in place of f .Proof. Use the integral formula. ⌅
27
CHAPTER 7
MEAN VALUE THEOREM AND MAXIMUM MODULUSPRINCIPLE
Today, we’ll be talking about the maximum modulus principle, the mean value theorem, andsome topological concerns. Recall from last time that we had the mean value theorem:
Theorem. (mean value) For f holomorphic in DR
(z), and r < R,
f(z) =1
2⇡
Z2⇡
0
f(z + rei✓)d✓.
We also have the maximum modulus principle, which is one of the most useful theorems incomplex analysis:
Theorem. (maximum modulus) Let f be holomorphic in DR
(z). If |f | takes a localmaximum value at z in the interior, then f is constant.
Proof. Use the mean value theorem. Alternatively, an equivalent formulation (Ahlfors)is as follows: if f is holomorphic and nonconstant in D
R
(z), then its absolute value |f | hasno maximum in D
R
(z).To show this, suppose that w = f(z) is any value in D
R
(z), so that the neighborhoodD✏
(w) is contained in the image of DR
(z). In this neighborhood, there are points of modulus> |w| so |f(z)| is not the maximum of |f |. ⌅
Claim. If |f | is constant and holomorphic, then f is constant on a disk (or region).Proof. Use the CR equations. ⌅
Corollary. (minimum modulus) If f is holomorphic and achieves a minimum of |f(z)|in a disk, then f(z) = 0 there.
Proof. If f(z) 6= 0 at a minumum then 1/f is holomorphic in a neighborhood of z. Thenuse the maximum modulus theorem.
Likewise, we can also use power series to prove this. ⌅
We went over another proof of the maximum modulus theorem in class using power se-ries; it may be nice to go over this proof as well (which is, again, rather long); see page 87
28
in the textbook.
Corollary. If f is holomorphic in DR
(z), then on Dr
(z), r < R, the maximum valueof |f(z)| on D
r
(z) is achieved on the boundary.Proof. Consult the textbook. ⌅
The following corollary is more general than the above:
Corollary. If f is continuous on a closed bounded set E and holomorphic in its interiorE � @E, then the maximum of |f | is attained on @E.
Proof. (Ahlfors) Since E is compact, |f(z)| has a maximum on E. Suppose that it isattained at z
0
; if z0
2 @E, we’re done. Else if z0
is an interior point, then |f(z0
)| is alsothe maximum of |f(z)| in a disk |z � z
0
| < � contained in E. But this is not possible unlessf(z) is constant in the component of the interior of E which contains z
0
. It follows bycontinuity that |f(z)| is equal to its maximum on the whole boundary of that component.This boundary is not empty and is contained in the boundary of E, so the maximum isalways attained at a boundary point. ⌅
Proposition (“anti-calculus”) If |f(z)| achieves its max value on Dr
(z) at w, then f 0(w) 6=0.
Proof. Sketched in class; see page 88 of the textbook. ⌅
We will use the maximum modulus principle to prove the following theorem:
Theorem. (open mapping) The map of an open set under a non-constant holomorphicmap f : U ! C, U ✓ C, is open. Note that f continuous at z means that 8✏ > 0, 9� > 0 :f(B
�
(z)) ✓ B✏
(f(z)).Sketch of another proof. If f 0(z) 6= 0, then
Jf
=
✓ux
uy
vx
vy
◆
is nonsingular. Then det Jf
= ux
vy
� uy
vx
= u2
x
+ v2x
= |f 0(z)|2 6= 0. The inverse functiontheorem gives the result.
If f 0(z) = 0, then consider f(B�
(0)) = Bf(�)
(0). Write f(z) = zk(ck
+ck+1
z+ ...). When
f has a zero of order k at 0, i.e. f(0) = f 0(0) = ... = f (k�1)(0) = 0, it turns out thatf(z) = [g(z)]k, g(z) = Az + ... i.e. g0(0) 6= 0 (but we cannot prove this as of yet). ⇤
Our proof. See page 93 of the textbook.
Topology toward general closed curve theorem. We would like to take the timeto revisit some topological concerns about the complex plane. In particular, recall that wedefined the notions of polygonally connected and polygonally simply connected :
Definition. A set U ✓ C is polygonally connected if 8p, q 2 U , there exists any piece-wise linear path from p to q.
A polygonally connected set is connected, but a connected set may not be polygonallyconnected (what is an example?). For open sets, they are equivalent:
29
Proposition. If U ✓ C is open, then U is connected i↵ it’s polygonally connected.
Definition. A set U ✓ C is polygonally simply connected (or, in the topological sense,simply connected) if for any closed polygonal curve is the boundary of a union of polygonsin U .
Definition. A region ⌦ in C is an open connected (or polygonally connected) subsetof C.
If E(U) denotes the space of directed edges (for purpose of integrating over) in the re-gion U , then by composing elements in E(U) to form a closed contour � we know fromCauchy’s integral theorem that
R�
f(z)dz = 0 for f holomorphic.The notion of connectedness goes into topology, involving such topics as homotopy and
homology. For those of you that are familiar with topology, a simply connected set hastrivial fundamantal group (e.g. every loop is homotopic to a constant). The Cauchy integralformula also has a more general, homology version involving winding numbers.
30
CHAPTER 8
GENERALIZED CLOSED CURVE THEOREM ANDMORERA’S THEOREM
Today we’ll talk about the general closed curve theorem and Morera’s theorem.
Theorem. (general closed curve theorem) If D is a region (open, polygonally connectedsubset) in C and C�D is connected, then the closed curve theorem holds, i.e.
Z
C
f(z)dz = 0
for f holomorphic on D and C a contour curve in D.
Definition. A set S in C ⇠= S2 is connected if @A,B open in C with A [ B � S andA \B 6= ;.
The method for proving the general closed curve theorem is:1. C�D is connected implies D p.s.c.2. D p.s.c. implies closed curve theorem holds (see textbook for pictures).
To show this, we’ll need to go back to last time: first let U be a region in C. Thendefine
• T (U) := {
Pn
i=1
ai
Ti
}, for Ti
a triangle in U , ai
2 Z, and any n.
• E(U) := {
Pn
i=1
ai
Ei
}, for Ei
an edge in U (recall that an edge means a pair of points{P,Q} with P 6= Q).
• P (U) := {
Pn
i=1
ai
Pi
}, for Pi
a point in U .
• @T := E1
+ E2
+ E3
for T a triangle.
• @E = Q� P ; @ : E(U) ! P (U) and @ : T (U) ! E(U).
• e 2 E(U) is closed if @e = 0, e.g. e = E1
+ ...+ En
is closed i↵ ±Ei
form loops.
Definition. e1
is homologous to e2
, denoted e1
⇠ e2
, if 9t : @t = e1
� e2
. This is anequivalence relation.
31
Definition. U is p.s.c. if e is closed, e.g. e is null-homologous, which means e ⇠ 0.
Proposition. If U is convex in C, then U is p.s.c.Proof. Prove this yourself, by dividing the set into triangles. ⌅
Proposition. If U is p.s.c., then the closed curve theorem holds.Proof. Note that the antiderivative theorem holds. Let e be a (not necessarily closed)
piecewise linear path from P to Q, @e = Q� P . ThenZ
e
f(z)dz =X
ai
Z
E
i
f(z)dz.
Define F (z) =Rz
z0f(z)dz. This makes sense because
Z
e1
f(z)dz =
Z
e2
f(z)dz.
if e1
⇠ e2
, since then Z
e1�e2
f(z)dz =X
bj
Z
�
f(z)dz
where � is the boundary of the triangle and the integral is zero. So the antiderivativetheorem implies the closed curve theorem as before:
Z
C
f(z) = f(final)� f(initial) = 0,
as desired. ⌅
We’ll introduce Morera’s theorem since we have some time. First some topology:
Proposition. If U is an open subset in C, U is polygonally connected i↵ U is connected.
Claim. Let D be a region in C ⇢ C. If C�D is connected, then D is p.s.c.
Theorem. (Morera) If f(z) is continuous on a region D andR�
f(z)dz = 0 for � a triangle,then f(z) is holomorphic.
Proof. See textbook. ⌅
Morera’s theorem is quite useful. Some of the uses are as follows:
1. Showing limn!1 f
n
= f is holomorphic.
2. Showing that some functions defined by sums or integrals are holomorphic, e.g. theRiemann zeta function and the gamma function:
⇣(z) :=1X
n=1
1
nz
�(z) :=
Z 1
0
e�ttz�1dt
32
3. Showing functions that are continuous and holomorphic on all but some set (e.g. somepoint) are holomorphic.
Proposition. Suppose {fn
} is holomorphic in an open set D, and fn
! f uniformly oncompact subsets of D. Then f = lim
n!1 fn
is holomorphic.Proof. It su�ces to work in B
✏
(↵) ⇢ D. B✏/2
(↵) is compact, so by assumption fn
! fis uniform here. Note that f
n
! f uniformly implies f is continuous. AlsoZ
�
f(z)dz =
Z
�
limn!1
fn
(z)dz = limn!1
Z
�
fn
(z)dz = 0
(Uniform convergence implies that we can swapRand lim, as we all know from elementary
analysis.) By Morera’s theorem, f is holomorphic. ⌅
The Riemann zeta function. The zeta function ⇣(z) is defined as
⇣(z) =1X
n=1
1
nz
,
for Re(z) > 1. We claim that ⇣(z) is convergent for Re(z) > 1, and indeed, is uniformlyconvergent for Re(z) > R > 1. (Hence by Morera’s theorem ⇣(z) is holomorphic.)
Proof of claim. First note that 1
n
z
= e�(logn)z. We first show uniform convergence. Let
fN
(z) :=NX
n=1
1
nz
=NX
n=1
e�(logn)z.
We boundP1
n=N+1
|e�(logn)z
| for the whole region Re(z) � R > 1:
|e�(logn)z
| = |e� logneRe(z)
| R|e� logn
|
So we’re left withP1
n=N+1
1
n
R
, R > 1. The integral test enures convergence, telling us thatthe tail end is small. ⌅
33
CHAPTER 9
MORERA’S THEOREM, SINGULARITIES, AND LAURENTEXPANSIONS
Let’s finish the topology stu↵, talk about Morera’s theorem, and then move on to singulariesand Laurent expansions.
First, what does it mean to fix an open set in C = S2?
Definition. A neighborhood of 1 is given by {1} [ {z : |z| > R}. Then S ⇢ C isopen if:
(1) S \ C is open in C(2) If 1 2 S, then 9R : {z : |z| > R} ⇢ S.
Lemma. If P is a closed polygonal path which doesn’t cross itself (i.e. P is a piecewise linearmap froman interval that is injective except at the endpoints), then 9A,B open in C s.t. A[B = C�Pand A \B = ;.
Proof. Note 9R : P \ {|z| > R} = ; for z 2 C� P . Take L, a line segment from z to apoint in S = {|z| > R}, and count the number of times P crosses L to R and the numberof times P crosses R to L. This makes sense because P \ L is a finite union of edges of Pand points. Let z 2 A if the count is odd; else z 2 B.
Exercise. Show that z 2 A or B is well-defined independent of the choice of L. Hint:use wedges. Conclude that nearby points are in the same set. ⌅
Integration over other regions. We use C � R0
to denote the complex plane with
a ray on the real axis missing. Note that C � (C � R0
) = {1} [ R0
is connected.Proof: any interval is connected. Thus we see that C�R0
is polygonally simply connected(PSC), and we can integrate along any closed curve in this region by applying the closedcurve theorem.
How about C⇤ = C�{0} ? Applying the closed curve theorem for the di↵erence contour
34
C2
� C1
, we see that Z
C2�C1
f(z)dz = 0
for f(z) holomorphic on C⇤. An image of C2
(the outside counterclockwise contour) and C1
(the inside counterclockwise contour) is as follows:
Diagram 1: C1
and C2
in C⇤.
Now let’s pick o↵ with Morera’s theorem, which as you recall states
Theorem. (Morera) If f is continuous on an open set D andR�
f(z)dz = 0 for � be-ing the boundary of a triangle in D, then f(z) is holomorphic.
Corollary. If fn
(z) is holomorphic on D open and fn
! f uniformly on compact sub-sets of D, then f(z) is holomorphic on D.
Proof. Proved last time. ⌅
The Riemann zeta function. The Riemann zeta function is a particular form of aDirichlet series or L-function given by
⇣(z) =1X
n=1
1
nz
.
We claim that ⇣(z) is holomorphic on Re(z) > 1, and indeed we showed this last time. ⇤The gamma function. The gamma function, which is useful in probability and numbertheory, is given by
�(z) =
Z 1
0
e�ttz�1dt.
We claim that �(z) is holomorphic for Re(z) > 0.Remark. �(z) is a natural extension of the usual factorial function in that �(n+1) = n!.
Show this yourself using integration by parts and induction.Proof of the claim. We show that
R10
e�ttzdt is holomorphic in Re(z) > �1. To thatextent we work in �1 < R
1
Re(z) R2
; note that any compact subset of {z : Re(z) > �1}is contained in one of these sets. Define
fn
(z) :=
Z1
1/n
e�ttzdt+
Zn
1
e�ttzdt,
which we will show converges uniformly. For t � 1, we can bound
|e�ttz| e�ttRe(z)
e�ttR2
becausetz = ez log t = e(Re(z)) log teiIm(z) log z,
35
the last term of which has norm 1. Then
|fn
(z)/2� f(z)/2|
Z 1
n
e�ttR2dt,
which converges for n = 1. So 8✏ > 0, 9N : 8n > N ,
Z 1
n
e�ttR2dt < ✏/2.
For t 1, |e�ttz| = e�ttRe(z)
e�ttR1 and
�����
Z1
1/n
e�ttzdt�
Z1
0
e�ttzdt
����� =
�����
Z1/n
0
e�ttzdt
����� Z
1/n
0
e�ttR1dt
for R1
> �1. Let s = 1/t and ds = �
1
t
2 dt, so that we getR1n
e�1/ss�2�R1ds. We need�2� R
1
< �1, i.e. �1 < R1
. Thus the sequence converge uniformly, so 8✏ > 0, 9N : 8n >N,R1n
e�1/ss�2�R1ds < ✏/2. ⇤
Corollary. If f is continuous on D and f is holomorphic on D except at ↵ 2 D, thenf is holomorphic on D.
Proof. Let � = @T . Then Z
�
f(z)dz = 0
if 0 /2 T . If ↵ 2 T , we split the contour so that ↵ is located at the corners of the triangle.If ↵ is at a corner, then
0 = liml!1
Z
�
l
f(z)dz =
Z
�
f(z)dz
by continuity of f . ⌅
Singularities of functions. (Newman and Bak §9) Singularities of functions will beimportant when trying to access their behaviors around certain points so that we can definethings like Laurent series or perform residue calculus. Needless to say, an understanding ofthe di↵erent types of singularities and how they pertain to classes of functions is importantfor complex anaysis in general.
Definition. Given ↵ 2 C, a deleted neighborhood D of ↵ is a neighborhood �{↵}, e.g.{z : 0 < |z � ↵| < ✏}. f is said to have an isolated singularity at ↵ if f is defined andholomorphic on a deleted neighborhood of ↵.Types of singularities. For D a deleted neighborhood of ↵, we have the following types ofsingularities:
1. Removable singularity : 9g(z) on D [ {↵} holomorphic with g(z) = f(z) on D.
2. Pole of order k: 9A(z), B(z) holomorphic on D [ {↵} s.t. A(↵) 6= 0, B(↵) = 0,
f(z) = A(z)
B(z)
and B(z) has a zero of order k at ↵. Recall that we can expand B(z) into
a power series as B(z) = ck
(z � ↵)k + ck+1
(z � ↵)k+1.
3. Essential singularity: Neither of the above.
36
The following proposition allows us to recognize if something has a removable singularity.
Proposition. Iflimz!↵
f(z)(z � ↵)
exists and is equal to 0, then f(z) has a removable singularity at ↵.Proof. Define h(z) as
h(z) =
(f(z)(z � ↵) z 6= ↵
0 z = ↵.
This is continuous, and holomorphic except at ↵. So h(z) is holomorphic in a neighborhood
of ↵ by the corollary to Morera’s theorem. Moreover, we see that h(↵) = 0, so g(z) = h(z)
z�↵is holomorphic and equal to f(z) away from ↵. (If h(↵) = 0 and h is holomorphic, thenh(z)
z�↵ is holomorphic; one can prove this easily using power series.) ⌅
Likewise, the following proposition allows us to recognize a pole of f .
Proposition. Suppose f(z) is holomorphic in a deleted neighborhood of ↵ and 9n :
limz!↵
f(z)(z � ↵)n = 0.
Then letting k = 1 be the least such n, f(z) has a pole of order k at ↵.Proof. Again, let
h(z) =
(f(z)(z � ↵)k+1 z 6= ↵
0 z = ↵.
h(z) is continuous on a neighborhood and holomorphic on a deleted neighborhood, so it’sholomorphic on a neighborhood of ↵ again by the corollary to Morera’s theorem. h(↵) = 0
so g(z) = h(z)
z�↵ is holomorphic in D[{↵} and g(z) = f(z)(z�↵)k on D. Then f(z) = g(z)
(z�↵)k .
(Note limz!↵
g(z) 6= 0 by assumption; also notice it exists). ⌅
For essential singularities, we’ve shown that it must be the case that limz!↵
f(z)(z � ↵)n
does not exist for all n. Notice as well the following:
• If f is bounded in a deleted neighborhood, then f has a removable singularity.
• If f < C1
1
|z|N + C2
1
|z|N�1 + ...+ C as z ! 0, then f has a pole of order N .
We have a theorem for essential singularities, which tells us that the image of any deletedneighborhood of an essential singularity under a holomorphic function is necessarily densein the complex plane:
Theorem. (Casorati-Weierstrass) If f has an essential singularity at ↵ and D is anydeleted neighborhood of ↵, then f(D) = {f(z) : z 2 D} is dense in C (i.e for any ✏ > 0,w 2 C, B
✏
(w) \ f(D) 6= ;).Proof. Suppose not. Then 9B
✏
(w) with B✏
(w)\f(D) = ;. This means |f(z)�w| > ✏, or1
|f(z)�w| <1
✏
. So 1
f(z)�w
is defined on D, a deleted neighborhood of ↵, and bounded there.
Thus g(z) = 1
f(z)�w
is holomorphic on a neighborhood of ↵.
Now consider 1
g(z)
= f(z) � w, so that f(z) = wg(z)+1
g(z)
is a ratio of two holomorphic
functions; we see that the singularity must be removable, or f(z) must have a pole.
37
Exercise left to the reader: prove the converse of the theorem. ⌅
Example. Consider f(z) = e1/z defined on C⇤, which has an essential singularity at 0 2 C.We claim that f(B
✏
(0)� {0}) is C⇤.To see this, note that f(z) is the composition of 1
z
and then ez. Under 1
z
, B✏
(0) � {0}gets inverted to fill C outside the ball. ez is 2⇡i-periodic, so we can draw horizontal linesin the complex plane at ⇡ + 2⇡n, n 2 Z. Then the interior of any strip created by thesehorizontal lines gets mapped under ez to C�R�0
. Our claim is that there are strips outsideB
0
(1/✏) so that f(B✏
(0)� {0}) = C⇤. ⇤
Diagram 2: The mappings 1
z
: B✏
(0)� {0} ! C� (B✏
(0)� {0}) andez : {z : ⇡ Im(z) 3⇡} ! C� R�0
.
Laurent expansions. Simply put, Laurent expansions are Taylor series of the formf(z) =
P1k=�1 c
k
zk.
Theorem. If f(z) is holomorphic on the annulus A(R1
, R2
) := {z : R1
< |z| < R2
},then f(z) =
P1k=�1 c
k
zk converges on all of A(R1
, R2
). (Taking R1
= 0 and R2
= 1 isalso fine.)
We sayP1
k=�1 ak
= L ifP1
k=0
ak
exists,P�1
�1 ak
=P1
l=1
a�l
exists, and their sum isL.
Proposition. If1
limk!1|c
k
|
1/k
� R2
and limk!1
|c�k
|
1/k
R1
,
then f(z) =P1
k=�1 ck
zk converges and is holomorphic on A(R1
, R2
).Proof. Write
f(z) =1X
k=0
ck
zk +1X
k=1
c�k
✓1
z
◆k
.
The first sum is convergent for |z|
1
lim
k!1|ck
|1/k . The second sum is convergent for
|
1
z
1
lim
k!1|c�k
|1/k =) |z| � limk!1|c�k
|
1/k. The first sum is clearly holomorphic. The
second sum is holomorphic as a function of 1/z; e.g. g(z) =P1
k=1
c�k
wk is holomorphicfor |w| 1
R1. The second sum is a composition of 1
z
and g, so by the chain rule it’s alsoholomorphic. So f is holomorphic. ⌅
38
Example. The following Laurent series expansion converges on A(0,1):
e1/z = 1 +1
z+
1
2z2+
1
6z3+ ...
⇤Example. Near z = 0 (pole of order 1), we have
1
z(1 + z)2=
1
z
1
(1 + z)2=
1
z(1� 2z + 3z2 � 4z3 + ...).
This is because @
@z
( 1
1+z
) = �
1
(1+z)
2 and 1
1+z
= 1 � z + z2 � z3 + .... We can do the samenear z = �1. ⇤
39
CHAPTER 10
MEROMORPHIC FUNCTIONS AND RESIDUES
Last time, we introduced the Laurent series for f(z) as f(z) =P1
�1 ck
(z � ↵)k. This isdefined on A
↵
(R1
, R2
), the annular region centered at ↵.
Theorem. If f(z) is holomorphic on A↵
(R1
, R2
), then f(z) =P1
k=�1 ck
(z � ↵)k andthe series converges on A
↵
(R1
, R2
).Proof. Do this yourself by considering the function
g(w) =
(f(w)�f(z)
w�z
w 6= z
f 0(w) w = z
and using the closed curve theorem. ⌅
Remark. Laurent series are unique to every function. We define the principal part off(z) =
Pnear ↵ as the sum
P�1
k=�1 ck
(z � ↵)k and observe the following:(1) ↵ is a removable singularity i↵ c
k
= 0 for k < 0.(2) ↵ is a pole of order n i↵ c
k
= 0 for k < �n, c�n
6= 0.(3) ↵ is an essential singularity i↵ c
k
6= 0 for infinitely many negative k (why?).
Here’s an idea: at a pole, we can make the function f(z) still be holomorphic if we ex-pand its range to C, so that f(z) = 1 there and holomorphically so.
Proposition. If f(z) has a pole of order k at ↵, then 1
f(z)
is holomorphic near ↵ andhas a pole of order k.
Proof. Write f(z) = 1
(z�↵)k g(z). Then g(z) holomorphic near ↵, g(z) 6= 0, so 1
f(z)
=(z�↵)kg(z)
, g(↵) 6= 0. This is holomorphic on a neighborhood at ↵ and has zero of order k. ⌅
Definition. f is meromorphic on D if f(z) is holomorphic on D except at isolated singu-larities at which f has poles.
The proposition above can then be restated as: a function f being meromorphic in Dreally means that f : D ! C is holomorphic.
40
Theorem. Let f be meromorphic in C and at infinity, and suppose that limz!1 f(z) = 1
for some w 2 C. (The limit means 8M, 9R : |f(z)| > M when |z| > R.) Then f(z) is arational function.
Theorem. If f : C ! C is holomorphic, then f is a rational function.
Residues of functions. By introducing residues, we seek to deduce the residue theo-rem as a means for computing integrals more e↵ciently. Given a holomorphic function f(z)in a deleted neighborhood of ↵, by the Laurent expansion we have f(z) =
P1�1 c
k
(z�↵)k.We define the residue of f(z) at the point ↵, Res(f ;↵), as
Res(f ;↵) = c�1
.
From Cauchy’s integral formula, we notice that
1
2⇡i
Z
C
R
(↵)
f(w)dw = c�1
.
Also recall the expression for ck
:
ck
=1
2⇡i
Z
C
R
(0)
f(w)
wk+1
dw.
We see from this that the “limsup stu↵” shows that Laurent series are uniformly convergenton A(r
1
, r2
) by comparison with geometric series.On a ball of radius ✏, we can compute the residue of f at ↵ by using the formula:
Res(f ;↵) = c�1
=1
2⇡i
Z
C
✏
(0)
f(w � ↵)dw =1
2⇡i
Z
C
✏
(↵)
f(w)dw.
There are a few ways to compute residues in general:(0) Compute the integral
RC
✏
(↵)
f(w)dw.
(1) Compute the Laurent series for f centered at ↵.(2) If f has a simple pole, then
c�1
= limz!↵
(z � ↵)f(z).
Also, if f(z) = A(z)/B(z) for A a nonzero function, and B having a simple zero, then wehave:
Lemma. Res(f ;↵) = A(↵)
B
0(↵)
.
We finish by working out an example. ConsiderZ 1
�1
1
1 + x2
dx.
We can compute this by first taking a semicircle contour of radius R, for which there is onepole contained inside (↵ = i). If we compute the residue at this pole, we can use the residuetheorem to get the value of the integral (this is left to the reader as an exercise).
41
CHAPTER 11
WINDING NUMBERS AND CAUCHY’S INTEGRALTHEOREM
Today, we’ll cover residues, winding numbers, Cauchy’s residue theorem, and possibly theargument principle. This is in the textbook, §10.1, §10.2.
Last time, we defined resudies. For f holomorphic in a deleted neighborhood of ↵, wehave
Res(f ;↵) =1
2⇡i
Z
C
✏
(↵)
f(z)dz = c�1
,
where c�1
appears in the Laurent expansion for f around ↵, f =P1
�1 cn
zn. Note that fis holomorphic in C
✏
(↵) except for the point at ↵.
Example. Res( 1
1+z
2 ; i)Method 1: Find the Laurent expansion about i, i.e. in terms of z � i.
1
1 + z2=
1
(z + i)(z � i)=
�i
2
1
z � 1+
1
4�
1
8(z � i) + ....
So c�1
= �i
2
.
Method 2: If f(z) has a simple pole at ↵, i.e. if f(z) = A(z)
B(z)
for A(↵) 6= 0, B(↵) =
0, B0(↵) 6= 0, then Res(f ;↵) = A(↵)
B
0(↵)
. Set Res( 1
1+z
2 ; i) =1
2i
= �i/2. ⇤
An application of this is as follows. Consider the integral
Z 1
�1
1
1 + x2
dx = limR!1
arctan(R)� limR!1
arctan(�R) = ⇡/2� (�⇡/2) = ⇡.
We can also do this by contour integration over the contours C1
, C2
, C3
, where C1
= [�R,R],C
2
is the CCW semicircle connecting R to �R, and C3
is the CCW circle omitting the pointat i. Then Z
C1+C2�C3
f(z)dz = 0
42
by the closed curve theorem, and by the estimation lemma we have����Z
C2
1
1 + z2dz
���� ⇡R ·
1
R2
=⇡
R! 0
as R ! 1. Also,Z
C3
1
1 + z2dz = 2⇡iRes(
1
1 + z2; i) = 2⇡i(�i/2) = ⇡.
So
limR!1
ZR
�R
1
1 + x2
dx = ⇡
after substituting for the values of the integrals over C1
, C2
, C3
.
Winding numbers. These intuitive give the number of times a path loops around apoint, and are also dealt with in topology.
Definition. Let � : [0, 1] ! C be a contour curve with �(t) 6= ↵ 8t. Then the wind-ing number of � about ↵, denoted n(�,↵), is the value
1
2⇡i
Z
�
1
z � ↵dz.
Theorem. n(�,↵) 2 Z.Proof. We have Z
�
1
z � ↵dz =
Z1
0
�0(t)
�(t)� ↵dt.
Note that the integrand on the RHS is d
dt
log(�(t)�↵), with a caveat that log is a multivaluedfunction. log z = log |z|+ iArg(z) modulo 2⇡i. We’ll show that
Z1
0
�0(t)
�(t)� ↵dt = 2⇡ik
for some k. This k = n(�,↵). Let F (s) =Rs
1
�
0(t)
�(t)�↵dt (secretly, log(�(s)�↵)� log(�(1)�↵).
Then F 0(s) = �
0(s)
�(s)�↵ . We claim that
F (s) =�(s)� ↵
�(0)� ↵.
To see this, let G(s) = (�(s) � ↵)e�F (s). Then G0(s) = ��0(s)e�F (s) + �0(s)e�F (s) = 0,so G(s) is a constant. Check that at s = 0, we get G(0) = (�(0) � ↵)e�F (0), so indeed
eF (0) = �(0)�↵�(t)�↵ .
Now eF (1) = �(1)�↵�(t)�↵ . F (1) =
R1
0
�
0(t)
�(t)�↵dt and �(1) = �(0), so eF (1) = 1. So F (1) = 2⇡ikfor k 2 Z. ⌅
Corollary. (version of Jordan curve theorem) Let � : [0, 1] ! C be a closed curve. ThenC� image(�) is disconnected.
43
Let’s call a simple closed curve one that doesn’t intersect itself. For us, if � has n(�,↵) = 0or 1, then we call � “winding simple.” The Jordan curve theorem shows that any closedcurve has what we call an interior and exterior. For the interior region B, we would haven = 1. For the exterior region A, we would have n = 0.
Cauchy’s residue theorem. It turns out that we can evaluate any contour integralof a holomorphic function by considering its residues and winding numbers.
Theorem. Let f be holomorphic on a region D with zero first homology (i.e. PSC, i.e.closed curve theorem holds for holomorphic functions on D) except for isolated singularitiesat ↵
1
, ...,↵n
2 D. Let � be any contour curve in D. Then
Z
�
f(z)dz = 2⇡inX
k=1
n(�,↵k
)Res(f ;↵k
).
Proof. The gist is to repeatedly subtract the principal parts of f at each ↵k
. This proof isa bit long to type up, so it is left as an exercise to the reader. ⌅
44
CHAPTER 12
THE ARGUMENT PRINCIPLE
Recall from last time that we had Cauchy’s residue theorem:
Theorem. Let D be a region with C�D connected (i.e. closed curve theorem applies), i.e.simply connected (all loops contract to a point in D). Let f be holomorphic on D exceptat ↵
1
, ...,↵n
, and let � be a contour curve in D missing these points. Then
Z
�
f(z)dz = 2⇡inX
k=1
Res(f ;↵k
)n(�,↵k
).
The argument principle. This is essentially relating the number of zeroes and poles of fto (# of times f winds around 0). Let � be a regular contour curve, e.g. if 8↵ 2 C� im(�),either n(�,↵) = 0 or n(�,↵) = 1. Intuitively, this means that the points are either “insideof �” or outside; the former is given by the set {z 2 C� im(�) : n(�, z) = 1} and the latteris given by the set {z 2 C� im(�) : n(�, z) = 0}.
Theorem. If f is holomorphic in a simply connected region D, and � a regular curvein D, then the following are equal:
(1) (# of zeroes of f with multiplicity) � (# of poles of f with multiplicity inside �)(2) The winding number around zero of f(�(t)), i.e. n(f � �, 0)
(3) 1
2⇡i
R�
f
0(z)
f(z)
dz
Remark. This assumes that for f |�
6= 0, there are no poles. The multiplicity is also calledthe order.
Proof. Let’s do (2) () (3). For � : [0, 1] ! C, we have
n(f � �; 0) =1
2⇡i
Z
f��
1
zdz =
1
2⇡i
Z1
0
�0(t)f 0(�(t))
f(�(t))dt =
1
2⇡i
Z
�
f 0(z)
f(z)dz.
The integrand is actually @
@z
(log f), which keeps track of how Arg(f) changes and goesaround �.
(3) () (1). To do this we check the residues of f 0/f . The only singularities of f 0/foccur at the zeroes or poles of f . Near a zero or pole ↵ of f , we have f(z) = (z � ↵)ng(z)
45
for g(z) holomorphic near ↵, g(↵) 6= 0. Also,
f 0(z) = n(z � ↵)n�1g(z) + (z � ↵)ng0(z),
sof 0(z)
f(z)=
n
z � ↵+
g0(z)
g(z),
where we note that the residue at ↵ of n
z�↵ is n, and g(z) 6= 0 implies that the term on theright is holomorphic and is the residue at ↵. Hence
1
2⇡i
Z
�
f 0
fdz =
X
↵ pole of f
outside �
n(�,↵) · order(f,↵)�X
↵ pole of f
inside �
n(�,↵) · order+poles(f,↵)
And n(�,↵) ! 1 in all these sums. ⌅
Moreover, we can say that the “zeroes of order n look like zn”; they wrap around 0 ntimes CCW, and “the poles of order n look like 1
z
n
”; they wrap around 0 n times CW.The idea of computing the number of zeroes in a curve by computing an integral is quite
nice. Most of the time, however, we just use the following corollary (also dropping the ideaof poles for now, since they aren’t really used in it):
Corollary. (Rouche Theorem) (Assume f has no zeroes on �.) Let f, g be holomorphic inthe unit disk (in �), and ||g(z)|| < ||f(z)|| on the unit circle (on �). Then # of zeroes of fin the unit disk = # of zeroes of f + g in the unit disk, for � regular.Examples. The # of zeroes of 3z8(+iz6 + 1) in the unit disk is the same as 3z8, which has8 zeroes. iz8 + 3z6 + 1 has 6 zeroes in the unit circle. iz8 + z6 + 3 has no zeroes. ⇤
Proof of Rouche’s Theorem. First note that for any functions A and B,
(AB)0
AB=
A0
A+
B0
B
Now write
(f + g) = f
✓1 +
g
f
◆,
and note |
g
f
| < 1 on �. The # of zeroes of f + g in � is given by
1
2⇡i
Z
�
(f + g)0
f + gdz =
1
2⇡i
Z
�
f 0
fdz +
1
2⇡i
Z
�
(1 + g
f
)0
(1 + g
f
)dz,
but this is just
= (# of zeroes with multiplicity of f in �) + n
✓1 +
g � �
f � �; 0
◆.
Note that 1 + g��f�� is contained inside B
1
(1) ✓ {Re(z) > 0}, the upper half-plane. So thewinding number is zero, and the above is
=1
2⇡i
Z
⇢
1
zdz
46
where ⇢ is the curve traced out by the fact that 1 + g��f�� < 1, which has an antiderivative
given by the principal branch of log z in Re(z) > 0. ⌅
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Generalized Cauchy’s integral formula. Let � be regular, and z be inside �. Then
f (k)(z) =k!
2⇡i
Z
�
f(w)
(w � z)k+1
dw.
Proof. The residue of f(w)
(w�z)
k+1 at zero is
1
w � z
✓f(w)
(w � z)k+1
◆= (w � z)kf(w) =
f (k)(z)
k!.
⌅
Corollary. Suppose fn
is holomorphic and fn
! f uniformly on compact sets (henceholomorphic by Morera’s theorem). Then f 0
n
! f 0 uniformly as well.Proof. We have
f 0n
(z)� f 0(z) =1
2⇡i
Z
�
fn
(w)� f(w)
(w � z)2dw
for � = Cr
(↵), z 2 Dr
(↵). On Dr/2
(↵), we have
|f 0k
(z)� f 0(z)| =1
2⇡
����Z
�
fn
(w)� f(w)
(w � z)2dw
���� 1
2⇡· 2⇡r ·
4
r2· maxC
r
(↵)
|fn
(w)� f(w)|,
where the 4
r
2 comes as an upper bound on 1
|w�z|2 because |w � z| � r/2 =) 1
|w�z|2
4
r
2 .Now choose n large enough so that
|fn
(w)� f(w)| <✏r
4
on Cr
(↵). Then |fn
(z)� f(z)| < ✏ for z 2 Dr/2
(↵), n > N .Now cover the compact set in euqation by disks of half the radius. Because the set is
compact, finitely many su�ice. Take the largest N↵
among these. ⌅
Theorem. (Hurwitz) Suppose fn
! f holomorphic in D, and uniformly so on compactsets. Suppose that none of the f
n
’s are zero in D. Then either f = 0 on D or f has nozeroes.
Proof. Suppose f 6= 0, and let f(↵) = 0 for some ↵ 2 D. ↵ is not the limit of zeroes of f(other than 0), because then f = 0 by the uniqueness theorem. Hence there is some closeddisk D
✏
(↵), ✏ > 0, with no zeroes of f in D✏
(↵) other than ↵. We have f 0n
! f 0 uniformly,and there are no zeroes on C
✏
(↵) of f ; none of fn
are zero, so 1
f
n
!
1
f
uniformly on C✏
(↵)
(show this yourself). Thusf 0n
fn
!
f 0
f
uniformly on C✏
(↵). It follows that
limn!1
1
2⇡i
Z
C
✏(↵)
f 0n
fn
=1
2⇡i
Z
C
✏
(↵)
f 0
f.
47
Solim
n!1# of zeroes of f
n
in D✏
(↵) = # of zeroes of f in D✏
(↵) = 1,
a contradiction, so we can’t have this isolated zero of f . ⌅
48
CHAPTER 13
INTEGRALS AND SOME GEOMETRY
Recall the residue theorem: let ↵k
be the singularities of f . ThenZ
�
f(z)dz = 2⇡iX
Res(f ;↵k
)n(�,↵k
).
There are many applications of this. If P,Q are polynomials, Q(x) 6= 0 for real x, anddegQ � degP + 2, then
Z 1
�1
P (x)dx
Q(x)dxdx = 2⇡i
nX
k=1
Res
✓P (z)
Q(z),↵
k
◆.
You can show this with the usual residue calculation (yes, the ones that we beat to deathin our review session...).
Example. Z 1
�1
1
x6 + 1dx
The poles are where z6 = �1, e.g. at ei✓ for ✓ = ⇡
6
, 3⇡
6
, ....Recall that, for f(z) = A(z)/B(z) with B(↵) = 0, A(↵) 6= 0, if B0(↵) 6= 0 then this is a
simple pole, and the residue is given by Res(f ;↵) = A(↵)/B0(↵).So using this we get that the residues are
Res(f(z); ei⇡/6) =1
6e5⇡i/6
Res(f(z); ei⇡/2) =1
6e⇡i/2
Res(f(z); e5i⇡/6) =1
6e⇡i/6
Then Z 1
�1
1
x6 + 1dx =
2⇡i
6(e�5⇡i/6 + e�i⇡/2 + e�i⇡/6) =
2⇡
3.
49
Geometry of complex functions. Recall the open mapping theorem: if f is holomor-phic and nonconstant, and if U is open in C, then f(U) is open, i.e. given ✏, 9� : B
�
(z) ✓f�1(B
✏
(w)) = {x 2 C : f(x) 2 B✏
(w)}, i.e. |z � ↵| < � =) |f(z)� f(w)| < �. Open meansthat 8�, 9✏ : f(B
�
(z)) ◆ B✏
(w), so |f(z)� �| < ✏ =) 9↵ : |z � ↵| < �, f(↵) = �.Proof. Let ↵ 2 C. Wlog f(↵) = 0; take C inside B
�
(↵). Take C inside B�
(↵); min f(C)exists and it’s not zero, so call it z
✏
. Let w 2 B✏
(0). Then for z 2 C, |f(z) � w| �
|f(z)| � |w| � 2✏ � ✏ = ✏. For z = ↵, |f(↵) � w| = | � w| < ✏. Hence min |f(z) � w| forz 2 B
�/2
(↵) is not achieved on the boundary. Thus, by the minimum modulus theorem, 9a zero, e.g. f(z)� w = 0 in B.
Theorem. (Schwarz Lemma) Let D be the unit disk. If f : D ! D is holomorphic(extending continuously to the boundary) and f(0) = 0, then
(|f(z)| |z|
|f 0(0)| 1
with equality in either i↵ f(z) = ei✓z.Proof. Define
g(z) =
(f(z)
z
z 6= 0
f 0(0) z = 0,
which is holomorphic on the circle of radius �; |g(z)| = |f(z)||z|
1
�
. By the maximum modulus
principle, we know in fact |g(z)| 1 for all z 2 D. Let � ! 1.So if a max is achieved on the interior, then f(z) is constant, i.e. g(z) is constant
if |f(z)| = |z| for any z 2 D1
(0), z 6= 0, or |f 0(0)| = 1. This means g(z) = ei✓; theng(z) = f(z)/z (f 0(0)), so f(z) = ei✓z. ⌅
Corollary. If a map from D ! D, f(0) = ↵, has a maximum value of |f 0(0)| amongmaps D ! D, then f is surjective.
Proof. Consider h(z) = B↵
(f(z)). h(0) = f 0(0)B0↵
(0), h(z)D ! D, h(z) : 0 7! 0, |h0(0)| 1.
The conclusion is that
|f 0(1)| 1
B0↵
(↵) = 1� |↵|2.
Notice that this is achieved for inverse of B↵
, which is B↵
.
B↵
(B↵
(z)) = z.
Also note that
B↵
(z) =z � ↵
1� ↵z,
and
B0↵
(z) =1
1� |↵|2,
so B0↵
(0) = 1� |↵|2, B↵
(0) = 0, |B↵
(z)| 1 for |z| 1. ⌅
We used h0(0) 1. We also know |h(z)| |z| and h(z) = B↵
(f(z)); |B↵
(f(z))| |z|.
50
We can therefore conclude that
|f(z)| |B↵
(z)|
if f(0) = ↵, f : D ! D.
Riemann mapping theorem. (casually stated) Given a simply connected region U ✓ C,U 6= C, 9f : D ! U where D denotes the open unit disk, which is a holomorphic bijectionwith holomorphic inverse (and essentially unique).
51
CHAPTER 14
FOURIER TRANSFORM AND SCHWARZ REFLECTIONPRINCIPLE
We will go over contour integration via Fourier transforms and infinite sums, §11, theSchwarz reflection principle §7.2, and the Mobius transformations §13.2.
The Fourier Transform. Let f : R ! R or R ! C. The Fourier transform of f isgiven by
f(y) =
Z 1
�1f(x)e�2⇡ixydx.
The inversion formula is
f(x) =
Z 1
�1f(y)e2⇡iyxdy.
This is writing f in terms of e2⇡iyx, and has applications in physics, probability theory (prov-ing CLT via generating functions), etc. Look at Wikipedia for a summary of how useful it is.
Example. f(x) = 1
1+x
2 .
f(y) =
Z 1
�1
1
1 + x2
e�2⇡iyxdx.
Assume y 0. Let C(R) = C1
(R) + C2
(R) denote the semicircle contour with C1
(R)denoting the part on the real axis and C
2
(R) the semicircle part. Then consider
Z
C1(R)+C2(R)
1
1 + z2e�2⇡iyzdz = 2⇡i · Res(g
y
(z); i)
where gy
(z) is the integrand. With a simple bounding argument, we note that
Z
C1(R)
gy
(z) =
ZR
�R
1
1 + x2
e�2⇡iyxdx.
52
Also, Res(gy
(z); i) = e
2⇡y
2i
. Thus f(y) = ⇡e2⇡y for y 0. Now for y � 0 we use the same
contour reflected over the real axis. Then Res(gy
(z);�i) = e
�2⇡y
�2i
=)
ZR
�R
1
1 + x2
e�2⇡iyxdx = 2⇡i · n(C,�i) · Res(gy
(z); i) = ⇡e�2⇡y.
So
f(y) = ⇡e�2⇡|y|.
⇤
The applications of Fourier transformations abound. Frankly Prof. Cotton-Clay hasn’tused it much, but natural scientists do. A very nice source for getting to know the complex-analytic details of Fourier transforms is Stein and Shakarchi.
Example. Verify thatP1
n=1
1
n
2 = ⇡
2
6
.
We can do this by showingP1
n=�1,n 6=0
1
n
2 = ⇡
2
3
. This is left as an exercise to the reader(or see the textbook). Note that this is ⇣(2), and the method that you use for computingthis generalizes to ⇣(2n), n 2 Z+; this was first computed by Euler. ⇤
Conformal mappings. These are functions that preserve angles. The Schwarz reflec-tion principle is useful as a tool here.
Theorem. (Schwarz reflection) Let H denote the upper half-plane, {z 2 C : Im(z) > 0}.Suppose that f is holomorphic in a region D ✓ H, that f is continuous on D [ L whereL = @D [ R, and that f is real on L. Then
g(z) :=
(f(z) z 2 D [ L
f(z) z 2 D
is holomorphic on D [ L [D.Proof. First, g is continuous because both parts agree on L. Second, g is holomorphic
on D, and on D:
limh!0
g(z + h)� g(z)
h= lim
h!0
f(z + h)� f(z)
h= f 0(z)
exists, so g is holomorphic on D. Then use Morera’s theorem to show that g(z) is holomor-phic. ⌅
Mobius transformations. This answers the question: what are the injective, holomorphicmaps C ! C?
Recall that if f : C ! C is holomorphic (meromorphic on C), then f is a rational
function. Solving ↵ = P (z)
Q(z)
, we have P (z) � ↵Q(z) = 0, and the LHS is a polynomial of
degree max(degP, degQ). For f(z) to be injective, we need max deg = 1. So
f(z) =az + b
cz + d,
53
where (a, b) and (c, d) are linearly independent; equivalently, ad� bc 6= 0. Maps of the formabove are called Mobius transformations. It is obvious that the inverse f�1 is given by
f�1(w) =�dw + b
cw � a
with ad� bc 6= 0.
Proposition. If
f(z) =az + b
cz + d, g(z) =
↵z + �
�z + �,
then letting ✓A BC D
◆=
✓a bc d
◆✓↵ �� �
◆,
we have
f(g(z)) =Az +B
Cz +D.
Proof. This is just a simple check. ⌅
54
CHAPTER 15
MOBIUS TRANSFORMATIONS
Today we’ll talk about Mobius transformations §13.2, cross ratios, and automorphisms ofD and H. Recall that we defined a Mobius transformation to be a function f(z) of thefollowing form:
f(z) =az + b
cz + d.
These are the automorphisms of C, i.e. the holomorphic maps C ! C with holomorphicinverse. We showed that composition translates into matrix multiplication:
fM1(fM2(z)) = F
M1M2(z),
where M ’s are the matrix forms of the transformation.These 2⇥ 2 invertible matrices with complex coe�cients are known as the general linear
group of order 2 over C:
GL2
C =
⇢✓a bc d
◆: a, b, c, d 2 C, ad� bc 6= 0
�.
GL2
C has Lie group structure (it is an algebraic group endowed with manifold structure,e.g. the group operations are C1). Let M = {mobius transformations}. Then, moddingout by complex scalars 6= 0, we have in fact that
M =GL
2
CC⇥ = PGL
2
C,
where PGL2
C is the projective linear group of order 2 over C, which is also a Lie group.We mod out by scalar multiples of C because
f
✓a bc d
◆= f
✓�a �b�c �d
◆,
has no other coincidences; e.g. multiplication by a complex scalar gives the same transfor-mation. We now claim that the generators for M are as follows:
Generators for M:
55
(1) Dilations/rotations: z 7! az.(2) Translations: z 7! z + b.(3) Inversion: z 7!
1
z
.
Claim. These generate M.Proof. We get the map z 7! az + b from dilation and translation. Suppose c 6= 0; we
want a map of the form z 7!
az+b
cz+d
. Well,
z 7!
(1,2)
cz + d 7!
(3)
1
cz + d7!
(1,2)
az + b
cz + d.
Let’s show that we can achieve the latter map. We can write
az + b
cz + d= A
✓1
cz + d
◆+B =
A+Bcz +Bd
cz + d= �
✓ad� bc
c
◆✓1
cz + d
◆+
a
c,
for which we want B = a
c
and A = b � a
c
d = �
ad�bc
c
. Note that setting c = 0 we havef(z) = a
d
z + b
d
by (1,2). ⌅
Corollary. {Mobius transformations} : {circles and lines} ! {circles and lines}.Proof. All of the generators do. ⌅
Let’s examine the dimensions of the groups we have above. Note that GL2
C has 4 complexdimensions, or 8 real dimensions (by the obvious identification C ⇠= R2). This is becausethere are 4 degrees of freedom in choosing elements in the matrix; another way would be toshow this is to note that GL
2
C is open in Mat2
C under the continuous determinant map,the latter of which has complex dimension 4.
Since we are eliminating one degree of freedom by imposing the condition det(A) 6= 0for A 2 M, we see that dimC M = 3, or dimR M = 6. Given this, we can say somethingabout the action on the 3 points {1, 0, 1}:
Lemma. Given f 2 M and f 6= id, f has at most 2 fixed points (points w 2 C : f(w) = w)in {1, 0, 1}.
Proof. Let c 6= 0, so that f(z) = az+b
cz+d
and setting f(z) = z we have az + b = cz2 + dz,which maps 1 7!
a
c
6= 1 because c 6= 0. For c = 0, we have 1 7! 1, and az + b = z has 1 solution. ⌅
Proposition. There exists a unique Mobius transformation f(z) sending ↵1
,↵2
,↵3
to1, 0, 1 respectively, and f(z) is given by
f(z) =(z � ↵
2
)(↵3
� ↵1
)
(z � ↵1
)(↵3
� ↵2
).
Note that if any ↵j
= 1, we cross out both terms in which it appears.Proof. The existence of f(z) is trivial. Uniqueness follows thus: if we have two transfor-
mations f, g satisfying the above properties, we consder g � f�1, which sends 1 7! 1, 0 7!
0, 1 7! 1. So three points are fixed, which implies g � f�1(z) = z =) f(z) = g(z). ⌅
This suggests that dimC M = 3 is no mistake. Indeed, we can define the cross-ratio
56
[↵1
: ↵2
: ↵3
: ↵4
] as the number w 2 C where ↵4
is sent if ↵1
7! 1,↵2
7! 0, and↵3
7! 1 by a Mobius transformation. Note that this number is equal to
(↵4
� ↵2
)(↵3
� ↵1
)
(↵4
� ↵1
)(↵3
� ↵2
).
Proposiiton. The cross-ratio is preserved by Mobius transformations, i.e. if f(z) = az+b
cz+d
with ad� bc 6= 0, then [↵1
: ↵2
: ↵3
: ↵4
] = [f(↵1
); f(↵2
); f(↵3
); f(↵4
)].Proof. Let g be a Mobius transformation sending ↵
1
7! 1,↵2
7! 0,↵3
7! 1. We have[↵
1
;↵2
;↵3
;↵4
] = g(↵4
). Then g � f�1 sends f(↵1
) 7! 1, f(↵2
) 7! 0, f(↵3
) 7! 1, andg � f�1(f(↵
4
)) = g(↵4
). ⌅
Proposition. There exists a non-unique Mobius transformation sending z1
7! w1
, z2
7! w2
and z3
7! w3
, and it satisfies, for w = f(z),
(w � w2
)(w3
� w1
)
(w � w1
)(w3
� w2
)=
(z � z2
)(z3
� z1
)
(z � z1
)(z3
� z2
).
Proof. Existence comes from the formula and the cross-ratios. Uniqueness follows asbefore. ⌅
Let’s use the above propositions above to find a Mobius transformation f sending D ! Hbijectively. If we visualize how f sends D ⇢ C ! H ⇢ C, we can also visualize how it mapsC. In particular, it should send �1 7! 1, 0 7! i, 1 7! 0, i 7! 1,�1 7! �1. Let w = f(z).Then using the equation above we have
(w � i)(0�1)
(w �1)(0� i)=
(z � 0)(1 + 1)
(z + 1)(1� 0)=) iw + 1 =
2z
z + 1=) w = i
�z + 1
z + 1.
Claim. This sends D to H.Proof. Let’s go with a more geometric proof. Note that f(i) = 1. We have a subclaim:
given 3 points in C, there exists a unique circle or line through the 3 points.Proof of subclaim. Send the 3 points to 1, 0, 1 by a Mobius transformation. There’s a
unique circle/line through these, namely R. We showed that Mobius transformations sendscircles and lines to circles and lines. ⇤
Now f sends the unit circle to the real line by the formula f(eit) = tan(t/2), where0 t < 2⇡. We make another subclaim, namely that f sends points inside D to pointsinside H.
Proof of subclaim. Use the following proposition on curves that are circles and omits �1(a “Pac-Man” curve). ⇤
Proposition. Let f(z) be a Mobius transformation, ↵ 2 C, � a contour curve, and ↵ /2 theinterior of �. Suppose n(f(�); f(1)) = 0. Then n(�;↵) = n(f(�); f(↵)).
Proof. Under dilations,
n(�;↵) =1
2⇡i
Z
�
dz
z � ↵=
1
2⇡i
Z1
0
�0(t)dt
�(t)� ↵dt.
The equalities for the other generators follow similarly. In particular, note that for inversionwe want to show the formula n( 1
�
; 1
↵
) = n(�;↵) + n( 1�
; 0), from which we can conclude the
57
proposition. This is a simple check using the definition of winding number and the chain rule.
Maps from regions into other regions. We know that log maps to a horizontal strip.In particular, log(z) : H ! {z : 0 < Im(z) < ⇡}, so
1
⇡log(z) : H ! R⇥ [0, 1].
We then see thatD !
f
H !
g
R⇥ [0, 1]
for f(z) = i 1�z
1+z
, g(z) = 1
⇡
log(z). Thus we have an automorphism from the unit disc to astrip given by
1
⇡log(i
1� z
1 + z) : D ! R⇥ [0, 1].
58
CHAPTER 16
AUTOMORPHISMS OF D AND H
We were looking at H = {z 2 C : Im(z) > 0} and D = {z 2 C : |z| < 1}.
Claim. f(z) = �i z�1
z+1
maps D to H, and g(z) = �z+i
z+i
maps H to D. Each are one-to-one, holomorphic, and inverses of each other.
Proof. Check the last part; we will show that f maps D to H. Note that, under f ,1 7! 0,�1 7! 1, i 7! �i( i�1
i+1
) = 1. Hence, like last time, we see that the unit circle ismapped to the real line, since the 3 points determine a circle or a line. (Also note 0 7! i,�i 7! �1.) Apparently, the real axis is mapped to the imaginary axis. Now fill up D bycircular arcs through �1 and 1. These map to straight lines “through 0 and 1”, all ⇢ Hbecause they intersect the upper half of the unit circle, which is the image of the imaginaryaxis ✓ D.
Another proof is the use the argument principle. Notice that the winding of f(�(t))around ↵ 2 C is the number of zeroes of f(z) = ↵... but this argument is sketchier, so weleave it to you to fill in the gaps. ⌅
Here’s an unnecessary but amusing claim:
Claim. Let �(t) be a contour curve and 0,↵ /2 im(�). Then
n
✓1
�,1
↵
◆= n(�,↵) + n
✓1
�, 0
◆= n(�,↵)� n(�, 0).
Proof. We have
n(1/�, 0) =
Z
1/�
dz
z � 0=
Z1
0
d( 1
�t
)1
�(t)
� 0=
Z1
0
��0(t)
�(t)dt = �n(�, 0).
On the other hand, we have
n(1/�, 1/↵) =
Z1
0
d( 1
�(t)
)1
�(t)
�
1
↵
=
Z1
0
��0(t)/�(t)2
1/�(t)� 1/↵=
Z1
0
�↵�0(t)
↵�(t)� �(t)2dt =
Z1
0
↵�0(t)
�(t)[�(t)� ↵]dt
=
Z1
0
��0(t)
�(t)dt+
Z1
0
�0(t)
�(t)� ↵dt = n(1/�, 0) + n(�,↵),
59
which gives the result. ⌅
Automorphisms of D and H. What are the injective, surjective, holomorphic mapsD ! D with holomorphic inverses? Any such map is in the automorphism group of D,denoted Aut(D). Recall that the Schwarz lemma tells us that if f : D ! D is such thatf(0) = 0, we have |f(0)| 1 and |f(z)| |z| for all z 2 D, with equality in either wheref(z) = ei✓z.
Corollary. If f : D ! D, f(0) = 0, and f is an automorphism of the disk (there ex-ists a holomorphic inverse f�1 : D ! D), then f(z) = ei✓z.
Proof. |f(z)| |z| =) |f�1(z)| |z| so |z| |f(z)|, e.g. equality is obtained andf(z) = ei✓z. ⌅
Recall that B↵
(z) = ↵�z
1�↵z for |↵| < 1. This maps D ! D and 0 7! ↵,↵ 7! 0. HenceB↵
(B↵
(z)) maps D ! D, 0 7! 0,↵ 7! ↵, and their composition is the identity map.
Theorem. The automorphisms of D are precisely the maps
f(z) = ei✓✓↵� z
1� ↵z
◆.
Proof. Suppose f is an automorphism of D with f(↵) = 0. Then f(B↵
(w)) takes 0to 0 and f(B
↵
(w)) = ei✓w. Take w = B↵
(z) and B↵
(w) = z. Then f(z) = ei✓B↵
(z) =ei✓( ↵�z
1�↵z ). ⌅
Corollary. The automorphisms of D are the Mobius transformations taking D to itself.
Corollary. The automorphisms of H are the Mobius transformations taking H to itself.
Theorem. The automorphisms of H are precisely the maps f(z) = az+b
cz+d
with ad� bc > 0and a, b, c, d 2 R.
Proof. Take x1
7! 1, x2
7! 0, x3
7! 1 in R [ {1} via a Mobius transformation forx1
, x2
, x3
all distinct (note that there is a unique such Mobius transformation). Write itdown:
f(z) =(z � x
2
)(x3
� x1
)
(z � x1
)(x3
� x2
).
The collection of these is the collection of maps f(z) = az+b
cz+d
, ad� bc 6= 0. Now check when
f(i) 2 H, f(i) = (bd+ac)+i(ad�bc)
c
2+d
2 . We see that f maps into H i↵ ad� bc > 0. ⌅
Here is a classification of the automorphisms of D (and H), up to conjugation (e.g. forf, g : D ! D, g�1fg : D ! D is allowed):
• Identity: f(z) = z
• Elliptic automorphisms: f(z) = ei✓z (rotation on D)• Parabolic: f(z) = z + b on H• Hyperbolic: f(z) = az on H, for a 2 R
Claim. Every automorphism of D (or H) is conjugate to precisely one of these.
60
Proof. Assume f is not the identity. We claim that every automorphism of H (or D)fixes either one point inside H, one point on R, two points on R, and no others. To see this,note that every Mobius transformation fixes one or two points. If one is in H, then the fullMobius transformation is given by the Schwarz reflection
g(z) =
8><
>:
f(z) z 2 Hsomething in R [ {1} z 2 R [ {1}
f(z) z 2 H.
So if we have one in H, then there can be no more. Either this happens, or one/two pointsare on R.
If one fixed point is in H, we conjugate to D and see it’s at ↵ 2 D; conjugate by B↵
to get it at 0, then get a rotation. In particular, let f : H ! H, f(�) = f(�). Take ourF : D ! H, F�1 : H ! D. F�1
� f � F then fixes F�1(�) = ↵. Then taking B↵
(which isits own inverse), we have B
↵
� F�1
� f � F �B↵
sends 0 7! 0,D ! D, so it equals ei✓z.If we fix one point on R, we conjugate to the disk, then conjugate by rotation so that the
fixed point on the unit circle is at �1. Then we conjugate back to H so that the fixed pointis at 1 (of Mobius transformation); i.e. f(z) = az + b. We need b 6= 0, else 0 7! 0. We alsohave a > 0 by ad � bc > 0. If a 6= 0, we fix another point so that a = 1 and f(z) = z + b.If we fix two points in R, we again place one at 1 and conjugate the other by A(z + 1) forappropriate A; then f(z) = az, a > 0. ⌅
In fact, these are the symmetries of the hyperbolic plane (taking the line y = 0 to be asort of “line at infinity”). As an aside, the hyperbolic metric on H is given by
ds2 =dx2 + dy2
y2,
for y > 0. The arc length of �(t) in the hyperbolic metric isR�
ds =Rb
a
p
x
0(t)
2+y
0(t)
2
y(t)
dt for
�(t) = (x(t), y(t)) : [a, b] ! H.
61
CHAPTER 17
SCHWARZ-CHRISTOFFEL AND INFINITE PRODUCTS
Today we’ll be talking about Schwarz-Christo↵el functions §13.3 and infinite products andEuler’s formula for sin(⇡✓), §17.3.
Schwarz-Christo↵el maps. These are holomorphic maps from H to a single polygon.We’ll see next time how it follows from the Riemann mapping theorem. But first o↵ we cannote that there are only a limited number of ways to do this.
Let’s draw a polygon with vertices vi
, interior angles ⇡�i
, and exterior angles ⇡↵i
(⇡↵i
+ ⇡�i
= ⇡). Note that �1 < ↵i
< 1, andP⇡↵
i
= 2⇡ =)P↵i
= 2.
Proposition. Given angles ⇡↵1
, ...,⇡↵n
and that a1
, ..., an�1
are points on R, then
f(z) =
Zz
0
dz
(z � a1
)↵1 ...(z � an�1
)↵n
�1
gives a holomorphic map from H to a polygon with angles ↵i
, and has a continuous exten-sion to R with f(a
i
) = vi
, f(1) = an
.
Remarks on the function. This is a contour integral taken on H [ R. We can take abranch of 1
(z�a
j
)
↵
j
= (z � aj
)�↵j which is positive for z real, > aj
and extends downward,
e.g. is defined on C � {aj
� ir}, r > 0. We can require 0 6= aj
, but in fact this still makes
sense if 0 is one of those:R1
0
1
x
�
dx converges for � < 1. Let’s examine the argument of
f 0(z) =Q
n�1
j=1
(z � aj
)�↵j on R:
Claim. Arg(f 0(z)) is constant on the connected segments of R � [{↵j
}. Also, Arg(f 0(z))increases by ⇡↵
j
when preserving aj
.Proof. This follows from:
Claim. Let z�↵ be defined on C � {negative imaginary axis}, and be and positive realfor z real and positive. Then Arg(x�↵) = 0 for x > 0. If x is real, Arg(x�↵) = �⇡↵, x < 0.
Proof. Write z = rei✓. Then z�↵ = r�↵e�i↵✓ gives the desired branch, which is in partic-ular our negative real axis. So if x is negative, i.e. � = rei⇡, then we get x�↵ = r�↵e�i↵⇡,i.e. Arg(x�↵) = �↵⇡. We can conclude that on the boundary (R) of H, f(z) maps to
62
straight lines, turning by angles ⇡↵k
to L. ⌅
Claim. This closes up, e.g.R1�1
Q(x� a
j
)↵jdx = 0.Proof. Use a semicircle contour, but skip over the poles on the real line. Then use a
combination of the closed curve theorem and bounding on the arc. ⌅
Example. sin z maps (�⇡
2
, ⇡2
) ⇥ (0,1) to H. We can view the domain as a triangle withexterior angles ⇡/2,⇡/2,⇡. So, in some sense, sin�1 z : H ! this strip. We can then write
sin�1(z) =
Zz
0
dzp
1� z2=
Zz
0
dz
(1 + z)1/2(1� z)1/2,
for some branch of the root function. (This is because d
dx
(sin�1 x) = 1p1�x
2 .) ⇤
Infinite products. We want to segue into the question, when does the infinite prod-uct
Q1k=1
ak
converge?
Lemma.Q
ak
converges i↵P
log(ak
) converges, for Re(ak
) > 0.Proof. Let p
n
:=Q
n
k=1
ak
, sn
=P
n
k=1
log(ak
). Then log pn
= sn
and pn
= esn . e andlog are continuous, so convergence for one gives convergence for another. ⌅
Lemma.Q(1 + a
k
) converges ifP
|ak
| converges.Proof. If
P|a
k
| converges, then past some pointN , |ak
| < 1/z. So we check:P1
k=N
log(ak
)converges.
| log(1 + ak
)| =
����ak �
a2k
2+
a3k
3�
a4k
4+ ...
���� |ak
|
✓1 +
1
2+
1
4+ ...
◆ 2|a
k
|.
SoP1
N
| log(1 + ak
)| P
2|ak
| converges, henceP1
N
log(1 + ak
) converges; finally, thisimplies that
Q1N
(1 + ak
) converges. ⌅
Proposition. Suppose fk
(z) is holomorphic on a regionD, withP
|fk
(z)| uniformly conver-gent on compact subsets of D. Then
Qk=1
(1+ fk
(z)) is uniformly convergence on compactsubsets and hence convergent.
Proof. Use the proof of the lemma in uniform convergence along with Morera’s theorem.⌅
Example. (Euler, 1734) We have the infinite product formula for sin(⇡z):
sin(⇡z) = ⇡z
1Y
k=1
✓1�
z2
k2
◆.
Theorem. (Weierstrass product) There exists an entire, holomorphic function with zeroesonly at {↵
k
}
k=1
if ak
! 1. Furthermore, the order of zero is the number of times repeatedin the list.
Proof. First attempt to go about it: f(z) =Q(z � a
k
) =) f(z) =Q(1� z
k
a
k
) converges
ifP
1
|ak
| converges.
63
Claim. IfP
1
|ak
|2 converges, then
f(z) =Y✓
1�z
ak
◆ez/ak
�.
Proof. This converges ifP
|(log(1� z
a
k
)+ z
a
k
)| does. Suppose |ak
|2
> |z| for |z| R; thisis possible for k > N since a
k
! 1. Then we can bound
|z|2
a2k
✓1 +
1
2+
1
4+ ...
◆ 2
|z|2
|ak
|
2
,
i.e. supposeP
1
|ak
|2 converges.
Back to the full proof of Weierstrass’s theorem: Let
f(z) =Y✓
1�z
ak
◆E
k
(z)
�,
for
Ek
(z) = ez
a
k
+
z
2
2ak
+...+
z
k
ka
k
k .
Check via a diagonal argument that this works; in particular, consult your textbook. ⌅
Proposition. (Euler’s formula)
sin⇡z = ⇡zY✓
1�z2
k2
◆
Proof. Take the quotient
Q(z) =⇡zQ⇣
1� z
2
k
2
⌘
sin⇡z.
This is an entire function with no zeroes. We’ll argue that this has growth at most Aez;let’s claim that it follows that Q(z) is constant.
We note that Q(z) is even. So, taking the log of Q(z), we haveRz
0
Q
0(z)
Q(z)
dz. This has
growth at most constant = |z|3/2 for large values. So logQ(z) is linear, i.e. Q(z) = AeBz
with B = 0 since Q is even, i.e. Q(z) is constant. Q(z) is constant because limz!0
sin(⇡z)
⇡z
= 1
and limz!0
Q1k=1
(1� z
2
k
2 ) = 1.Now let’s talk about the growth. 1
sin⇡z
is bounded on a square with sides from �N�1/2to N + 1/2 (exercise), and likewise on the edges. To bound the rest...actually, we’ll savethat for next time because we’re out of time.
64
CHAPTER 18
RIEMANN MAPPING THEOREM
Guest lecture: Joe Rabino↵Note: These notes are particularly messy since the scribe wasn’t paying too much attention,so use in discretion.
You’re lucky that we cover the Riemann mapping theorem, which is one of the most in-teresting results you’ll learn in this class. The context is the following:
Let regions R1
, R2
( C be open and simply connected, with R1
connected and C � Rconnected. Then there exists ' : R
1
! R2
that is onto, one-to-one, and holomorphic.Thus '�1 : R
2
! R1
is also holomorphic. As far as complex analysis is concerned, theseregions are “identically the same.” Now let’s introduce a bit of reductionism here. Assume : R
2
! U = {x 2 C : |x| < 1}, so that we have the following diagram:
R1
U
R2
'
'i
i
We need R1
6= C: if ' : C ! U , then ' is bounded, then constant by Liouville’s theorem.
Theorem. Let R ( C be open, R 6= ;, and R 6= C. Choose a z0
2 R. Then there exists aunique, one-to-one, holomorphic map ' : R ! U such that '(z
0
) = 0 with '0(z0
) 2 R�0
.Proof. Uniqueness: suppose '
1
,'2
: R ! U both satisfy the theorem. Then ' : '2
�'�1
1
:U ! U is one-to-one, onto, and holomorphic, and '(0) = '
2
('�1(0)) = '2
(z0
) = 0.By the Schwarz lemma, we know that |'(z)| |z| for all z 2 U , and if ' is onto,
'(z) = ei✓z. Then '�1(0) = ei✓ 2 R�0
=) ei✓ = 1 =) '(z) = z. This implies
'2
= '�1
1
= Id =) '2
= '1
, as '0(0) = '
02(0)
'
01(0)
> 0. So uniqueness is easy. Let’s work
out an example for the other parts:
Example. Let R = U , ' : R = U ! U . If '(0) = 0, then by Schwarz’s lemma, we
65
know that ' is onto () |'0(0)| = 1 =) |'0(0)| is the largest possible among all mapsU ! U if 0 7! 0.
If ' : U ! U,'(0) = ↵ 6= 0, then B↵
(z) = ↵�z
1�↵z is such that B↵
: U ! U , B↵
(0) = ↵,B↵
(↵) = 0; e.g. B↵
swaps 0 and ↵. Then |(' � B↵
)0(0)| = |'0(↵)||B0↵
(0)| () |'0(↵)| isconstant. It turns out that ' is one-to-one and onto =) |'0(↵)| is maximized.
In general, we want to find ' : R ! U that’s one-to-one and holomorphic, with '(z0
) =0, |'0(z
0
)| maximized. Even more precisely, we will take
F = {f : R ! U : f is one-to-one, holomorphic, and f 0(z0
) = 0}.
We’ll show(a) F 6= ;
(b) supf2F f 0(z
0
) = M < 1
(b’) There exists f 2 F : f 0(z0
) = M .(c) If ' = f from above, then ' : U ! U is one-to-one and onto, '(z
0
) = 0, and'0(z
0
) > 0.In this case, the function we will cook up looks like B
↵
.
First we want f : R ! U that is one-to-one and holomorphic. Suppose there existsD(p
0
, �) := {z : |z � p0
| < �} such that D(p0
, �) \ R � ;. Let f(z) = �
z�p0. Then this
is one-to-one since it’s a Mobius transform, if z 2 R =) |f(z)| = �
|z�p0| <�
�
= 1.
What is no such disc exists? Choose any p0
/2 R, and define g(z) =q
z�p0
z0�p0with
g(Z0
) = 1, one-to-one since g(z)2 is. Do this by choosing the appropriate square root. Howdo we choose a branch of p ? Well,
rz � p
0
z0
� p0
= exp
✓1
2log
✓z � p
0
z0
� p0
◆◆
and
log
✓z � p
0
z0
� p0
◆= log(z � p
0
)� log(z0
� p0
) =
Zz
z0
d⇣
⇣ � p0
.
This makes sense by the closed curve theorem, as 1
⇣�p0is holomorphic on R.
Claim. There exists � > 0 such that |g(z) + 1| > � for all z 2 R.
Proof. If not, then 9z0
, z1
, ... 2 R such that g(zi
) ! �1. Thenq
z
i
�p0
z0�p0! �1, and
squaring z
i
�p0
z0�p0= 1 =) lim(z
i
� p0
) = lim(z0
� p0
) =) lim zi
= z0
. But by continuity of g,
lim g(zi
) = g(z0
) = 1, a contradiction. This proves (a).
Proof of (b). Since R is open, D(z0
, 2�) ✓ R for some � > 0. If f 2 F , then
|f 0(z0
) =
�����1
2⇡i
Z
C(z0,�)
f 0(z)
(z � z0
)2dz
����� 1
2⇡
Z|f 0(z)|
|z � z0
|
2
dz
1
2⇡
Z1
�2dz =
1
�
=) supf2F
|f 0(z)| 1
�< 1
66
For (b’), choose f1
, f2
, ... 2 F such that limn!1 |f
i
(z0
)| = M . Suppose that we knewthat f
i
! ' uniformly, where ' is holomorphic on R.
Claim. ' 2 F .Proof. f 0
i
(z) ! '0(z0
) = M . Also, for all z 2 R, |'(z)| = lim |fn
(z)| 1 so ' : R ! U =closed ball.
By the open mapping theorem, '(R) is open, so '(R) ✓ U .
Why is ' one-to-one? We need the fact that a limit of one-to-one functions is again one-to-one. Suppose not. Then 9z
1
, z2
2 R such that '(z1
) = '(z2
) = a 2 U , and let D1
, D2
be discs around z1
, z2
such that D1
\D2
= ;. Now ' � a is not constant, since ' � a hasa zero on D
i
. Thus fi
� a has a zero on D1
for infinitely many i, which implies that fi
� ahas no zeroes on D
2
for all such i. Thus fi
(z0
) ! a. Apply Hurwitz to the subsequence ofsuch f
i
on D2
, so '� a has no zeroes on D2
, which implies '(z1
) = a, a contradiction. Sowe’ve checked that ' is one-to-one, holomorphic, and onto.
We need to show that the fi
converge uniformly on compact subsets K ✓ R. We’ll only usethat |f
i
| 1 for all i. This is Montel’s theorem.
Theorem. If f1
, f2
, ... : R ! U is any sequence of functions, then there is some subse-quence f
i1 , fi2 , ... that converges uniformly on compact subsets K ✓ R.Proof. Let {⇠
1
, ⇠2
, ...} ✓ R be a countable dense subset, e.g. {x + iy 2 R : x, y 2 Q}.Since {f
n
(⇠i
)}1n=i
✓ U is bounded, there is a subsequence {f1,n
} ✓ {fn
} such that f1,n
(⇠i
) !some number '(⇠
i
). Likewise, there is {f2,n
} ✓ {fn
} such that f2,n
(⇠i
) ! '(⇠i
)...For every m we get {f
m,n
} such that fm,n
(⇠i
) ! '(⇠i
) for all i m. Let 'n
= fn,n
, the“diagonal” elements, so that {'
n
}
n 6=m
✓ {fm,n
}
n�m
. Thus 'n
(⇠m
) ! '(⇠n
) for all m � 1.
Claim. {'n
}
1n=1
converges on all of R.Proof. Let D = D(x0, 3d) ✓ R be some disk, and let K = D(z0, d) ⇢ D(z0, 3d) ✓ R be a
compact subset of R. Moreover, for all z 2 K, C(z, d) ✓ D. Since |'n
| < 1, 8z 2 K we have
|'0n
(z)| =
�����1
2⇡i
Z
C(z,d)
'n
(⇠)
(⇠ � z)2d⇠
����� 1
d.
=) 8z1
, z2
2 K,
|'n
(z1
)� 'n
(z2
)| =
����Z
z2
z1
'n
(z)dz
���� ����Z
z2
z1
|'n
(z)|dz
���� ����Z
z2
z1
dz
d
���� =1
d|z
1
� z2
|.
Hence {'n
} is equicontinuous on K in that 8n, z1
, z2
2 K, if |z1
� z2
| < d✏, then |'n
(z1
)�'n
(z2
)| < ✏.For z 2 K,n,m � 0, k � 0,
|'n
(z)� 'm
(z)| |'n
(z)� 'n
(⇠k
)|+ |'n
(⇠k
)� 'm
(⇠k
)|+ |'m
(⇠k
)� 'm
(z)|.
If we choose ⇠k
2 K such that |z � ⇠k
| < 1
3
✏d, and choose N s.t. 8n,m � N , |'n
(⇠k
) �'m
(⇠k
)| < ✏/3, then|'
n
(z)� 'm
(z)| < ✏/3 + ✏/3 + ✏/3 = ✏.
67
Thus {'n
(z)} is Cauchy, and 'n
(z) ! '(z). We’ve proven 'n
! ' pointwise.
Claim. ' is uniformly continuous on K if |z1
� z2
| < d✏.
|'(z1
)� '(z2
)| = lim |'n
(z1
)� 'n
(z2
)| ✏.
Uniform convergence on K? Suppose 'n
! '. Let ✏ > 0, and Sj
:= {z 2 K : |'n
(z) �'(z)| < ✏ for all n > j}. S
j
is open: indeed, if z1
2 Sj
and |z1
� z2
| < d� for � > 0, then
|'n
(z1
)� '(z2
)| |'n
(z1
)� 'n
(z1
)|+ ... |'n
(z1
)� '(z)|+ 2� < ✏/2
for all n > j. Taking � small enough, z2
2 Sj
. This implies D(z, �) ✓ Sj
.
1[
j=1
Sj
= K
Compactness implies K + Sj
for some j implies uniform convergence. All you do now isfiddle around with B
↵
’s.
68
CHAPTER 19
RIEMANN MAPPING THEOREM AND INFINITEPRODUCTS
We will continue with the Riemann mapping theorem, which we got through most of lasttime:
Theorem. (Riemann mapping) Let R ⇢ C be open and simply connected (for us, use C�Ris connected, i.e. can apply closed curve theorem). Also suppose R 6= C. Then 9! map f thatis holomorphic, with holomorphic inverse, s.t. f : R ! D, with f(↵) = 0,R�0
3 f 0(↵) > 0.
Improperly Stated Generalization. (Uniformization theorem) Every simply connected Rie-mann surface is conformal to, and has a holomorphic map with holomorphic inverse to,precisely one of
(1) C = S2
(2) C(3) D
The nice thing is that these correspond to di↵erent geometries. D is the hyperbolic geome-try, S2 you’ve done on homework, and C is Euclidean. We don’t have the tools to do thisproperly yet.
Proof of the Riemann mapping theorem so far. Let
F = {f : R ! D : f is holomorphic, 1-1, and f 0(↵) > 0}.
We show that(A) F 6= ;.(B) sup
f2F |f 0(↵)| < 1 and 9g 2 F achieving this sup, e.g. |g0(↵)| = supf2F |f 0(↵)|.
(C) Given this g achieving the sup, g is surjective and g(w) 6= 0 for w 2 R (so that weget a holomorphic inverse).We can also draw pictures for A and B, which we showed last time with the help of thefollowing:
Lemma. (Montel’s theorem) A uniformly bounded sequence of holomorphic functionshas a subsequence which converges on compact subsets of the domain, and the limit is
69
holomorphic (using Morera’s theorem).For more info here, see also the Arzela-Ascoli theorem.
Let’s finish up with C.
Claim. This limit g has the property that g(↵) = 0 and is surjective (any 1-1 functionf can have f 0(w) = 0 for any w. ).
Proof. g sends 0 7! 0. If not, g(↵) = w 6= 0, and Bh
� g : ↵ 7! 0. Then |(Bw
� g)0(↵)| =|B0
w
(g(↵))g0(↵)| = 1
1�|w|2 |g0(↵)| > 1, and there is a larger such g.
g is surjective. Suppose not, that that it misses w. Wlog w = 0, and consider (Bw
� g),which sends ↵ 7! w and misses 0. Then |(B
w
�g)0(↵)| = |B0w
(g(↵))||g0(↵)| = (1�|w|2)|g0(↵)|.Now consider the following claim:
Claim. 9 a holomorphic branch ofp
Bw
� g (which maps R ! D)Proof. We can write
log(Bw
� g)(z)� log(Bw
� g)(↵) =
Zz
↵
(Bw
� g)0(z)
(Bw
� g)(z)dz = ⇤.
Then take e12 (⇤).
We have (p � Bw
� g) : w 7!
p
w by construction. Then (p � Bw
� g)0(↵)| = |
p0(w)|(1 �
|w|2)|g0(↵)| = 1
2|w|1/2 (1 � |w|2)|g0(↵)|. Also, (Bpw
�
p
� Bw
� g) : ↵ 7! 0, and (Bpw
�
p
�
Bw
� g)0(↵) = |B0pw
(p
w)|... = 1
1�|w|1
2|w|1/2 (1� |w|2)|g0(↵)|.
Set |w| = r2, for r 2 R�0
(with w not 0 b/c g0(↵) > 0). We claim that
✓1
1� r2
◆✓1
2r
◆(1� r4) > 1
for 0 < r < 1. To see this, note that
1 + r2 � 2r = (r � 1)2 > 0
for 0 < r < 1, so 1+r
2
2r
> 1.
So g is surjective. Hence g is injective. We claim that g0(w) 6= 0 for any w. To seethis, suppose g0(w) = 0. Wlog suppose w = 0. Then g0(0) = 0 and g(z) = zkh(z) for k � 2,with h holomorphic and h0(0) 6= 0. Hence
Z
C
✏
(0)
g0(z)
g(z)dz = Z(g)
inside D✏
(0) = k. Take � with |�| small; then
k =
Z
C
✏
(0)
g0(z)
g(z)� �dz = Z(g � �)
in D✏
(0). If so, g(z)� � is 0 at inverse images of �, but all of the same small ball around 0is an inverse image of something. So g0(0) = 0 and g is constant. ⌅
70
Here are some consequences of the Riemann mapping theorem:(1) Every connected and simply connected region in R2 is topologically equivalent (in
the sense of homeomorphic) to a disk. For example, consider taking out a Cantor set in R2.(2) See moduli spaces of curves and their relations to string theory.
Back to infinite products and the gamma function. Let’s finish the proof of thefollowing theorem:
Theorem.
sin(⇡z) = ⇡z
1Y
k=1
✓1�
z2
k2
◆.
Corollary. Set z = 1/2, and
1 =⇡
2
1Y
k=1
✓10
1
(2k)2
◆=⇡
2
1Y
k=1
✓(2k � 1)(2k + 1)
(2k)(2k)
◆.
Then⇡
2=
✓2 · 2
1 · 3
◆✓4 · 4
3 · 5
◆✓6 · 6
5 · 7
◆....
This has a taste of number theory.
Proof of the theorem. The gist of the proof is to write
Q(z) =⇡zQ1
k=1
(1� z
2
k
2 )
sin(⇡z),
which has growth order of at most something like Ae|z|3/2
. We conclude that Q(z) is aconstant, which taking limits we see to be = 1. See the textbook for a more thorough proof.⌅
This is presumably the flavor of analytic number theory, where we can’t show thingsdefinitely but can use complex analysis to give a (poor) bound that translates into somethingnice. We’ll continue with the gamma function �(z) next time.
71
CHAPTER 20
ANALYTIC CONTINUATION OF GAMMA AND ZETA
Analytic continuation of gamma and zeta. So we have these two functions:
�(z) =
Z 1
0
e�ttz�1dt, ⇣(z) =1X
n=1
1
nz
The first is holomorphic for Re(z) > 0 and the second for Re(z) > 1; this is just anapplication of Morera’s theorem. We’ll talk about the gamma function for a while; whatwe’d like is for these functions to be holomorphic in all of the complex plane. It turnsout that the properties of the gamma function will be important for understanding primenumbers. Let’s start with gamma—recall what happened last time. We have
�(z) =
Z 1
0
e�ttz�1dt =
Z1
0
e�ttz�1dt+
Z 1
1
e�ttz�1dt.
The right integral uniformly converges on BR
(0), i.e. it converges on compact subsets of C.Now look at the left integral. We saw that it converges uniformly for compact subsets ofRe(z � 1) > �1, i.e. Re(z) > 0.
How do we extend this to all of C? The extension will have various poles, and is mero-morphic in C.
Claim. �(z + 1) = z�(z).Proof. We integrate by parts. Consider
�(z) =
Z 1
0
e�ttz�1dt.
Take u = tz�1, dv = e�tdt =) du = (z � 1)tz�2dt, v = �e�t. ThenZ 1
0
e�ttz�1dt = �e�ttz�1
��10
+
Z 1
0
(z � 1)e�ttz�2dt = (z � 1)�(z � 1),
as desired. ⌅
Corollary. �(n) = (n� 1)! for n 2 Z�1
.
72
Proof. Check �(1) =R10
e�tdt = 1 from the claim. ⌅
From the above, we can see that
�(z) =�(z + 1)
z
is meromorphic (holomorphic) for Re(z+1) > 0, i.e. Re(z) > �1 with a pole only at z = 0.Repeating this k times, we see that
�(z) =�(z + k + 1)
z(z + 1)...(z + k)
is meromorphic for Re(z + k + 1) > 0, i.e. Re(z) > �k + 1, with poles only at nonnegativeintegers. This gives a continuation of � into the left side of C.
The residues at the poles is given by
Res(�(z); 0) =�(1)
1= 1,
Res(�(z);�k) =�(1)
(�k)(�k + 1)...(�1)=
(�1)k
k!.
These poles are indeed simple by our calculations above. We will show another nice claim:
Claim. �(z)�(1� z) = ⇡
sin(⇡z)
.
Corollary. � has no zeroes.Proof. This property follows directly from the claim. sin(⇡z) is nonzero for z not an
integer; for z an integer, we already know that � has a pole if z 2 Z0
or �(z) = (z�1)! 6= 0if z 2 Z�1
. ⌅
To prove the initial claim, we’ll need to show another fact, that for
e�t = limn!1
✓1�
t
n
◆n
we have
�(z) = limn!1
Zn
0
✓1�
t
n
◆n
tz�1dt,
e.g. we can switch the lim andR. Then
limn!1
Zn
0
✓1�
t
n
◆n
tz�1dt = limn!1
1
nn
Zn
0
(n� t)ntz�1dt,
and from integration by parts we haveZ
n
0
(n� t)ntz�1dt =tz
z(n� t)n
����n
0
+
Zn
0
tz
zn(n� t)n�1dt.
So
limn!1
1
nn
Zn
0
(n� t)ntz�1dt = limn!1
1
nn
n
z
Zn
0
tz(n� t)n�1
73
= limn!1
1
n2
n(n� 1)...1
z(z + 1)...(z + n� 1)
Zn
0
tz+n�1dt
= limn!1
✓nz
z
✓1
z + 1
◆...
✓n
z + n
◆◆since
tz+n
z + n
����n
0
=nnnz
z + n.
This gives
1
�(z)= lim
n!1
z
nz
✓z + 1
1
◆...
✓z + n
n
◆= lim
n!1
z
nz
nY
k=1
⇣1 +
z
k
⌘,
which looks something like the expression for sin(⇡z) we saw before. Let’s rewrite the lastexpression as
limn!1
z
nz
nY
k=1
⇣1 +
z
k
⌘= lim
n!1zn�z exp
⇣z +
z
2+
z
3+ ...+
z
n
⌘ 1Y
k=1
h⇣1 +
z
k
⌘e�z/k
i,
and the product by itself converges as shown before. Rewriting once again with n�z =e�(logn)z,
limn!1
zn�z exp⇣z +
z
2+
z
3+ ...+
z
n
⌘ 1Y
k=1
h⇣1 +
z
k
⌘e�z/k
i
= limn!1
exp
"z
nX
k=1
1
k� log n
!#z
1Y
k=1
h⇣1 +
z
k
⌘e�z/k
i.
Also,
limn!1
"nX
k=1
✓1
k
◆� log n
#= � ⇡ 0.577,
the Euler-Mascheroni constant. Then
1
�(z)= e��zz
1Y
k=1
h⇣1 +
z
k
⌘e�z/k
i.
From this we see that
1
�(z)�(1� z)=
1
�z�(z)�(�z)=
1
�ze��ze�zz(�z)
1Y
k=1
✓1�
z2
k2
◆= z
1Y
k=1
✓1�
z2
k2
◆=
sin(⇡z)
⇡,
which is a cute equation. It also gives the values
�
✓1
2
◆=
p
⇡, “(1/2)!” = �
✓3
2
◆=
p
⇡
2.
The zeta function. We turn to ⇣(z), defined as
⇣(z) =1X
n=1
1
nz
, Re(z) > 1.
74
We claim the following:
Claim. ⇣(z) has a meromorphic extension to all of C with a simple pole only at 1.
We’ll show next time that ⇣(z) has no zeroes on Re(z) = 1, and from this deduce the primenumber theorem, which relates to the growth rate of the number of prime numbers. Youmay have also heard of the Riemann hypothesis, which says that the nontrivial zeros of ⇣ areall on the line Re(z) = 1/2; this gives us a stronger estimate of the growth rate of the primes.
Claim. For Re(z) > 1,
⇣(z) =Y
p prime
✓1
1� p�z
◆.
Proof. We first write
1
1� 1/pz= 1 +
1
pz+
1
p2z+
1
p3z+ ...
This gives us something of the form 1 + 1/2z + 1/3z + .... If we write n = p↵11
...p↵k
k
, then
1
nz
=1
p↵11
z...
1
p↵k
k
z
in the product. Then
�����
NX
n=1
1
nz
�����
������
Y
p<N
✓1 +
1
pz+ ...+
1
pMz
◆������
for some M . Moreover,������
Y
p<N
✓1 +
1
pz+ ...+
1
pMz
◆������
������
Y
p<N
✓1
1� p�z
◆������
������
Y
p prime
✓1
1� p�z
◆������.
In the limit, we thus have
�����
1X
n=1
1
nz
�����
������
Y
p prime
✓1
1� p�z
◆������
For the reverse direction, we have
Y
p<N
✓1 +
1
pz+ ...+
1
pMz
◆
NX
n=1
1
nz
,
etc. Then ✓1�
1
2z
◆⇣(z) = 1 +
1
3z+
1
5z+ ...
and Y
p<N
✓1�
1
pz
◆⇣(z) =
X
n has no prime factors<N
1
nz
! 1 as N ! 1,
75
giving the desired equality. ⌅
Now let’s extend ⇣(z) to all of C (with a pole). With the change of variables s = nt,we have Z 1
0
e�nttz�1dt =
Z 1
0
e�s
sz�1
nz�1
ds
n= n�z�(z).
Then
⇣(z) =1X
n=1
1
�(z)
Z 1
0
e�nttz�1dt =1
�(z)
Z 1
0
1X
n=1
e�nt
!tz�1dt =
1
�(z)
Z 1
0
tz�1
et � 1dt
for Re(z) > 1. This is defined on the entire complex plane:
⇣(z) =1
�(z)
Z1
0
tz�1
et � 1dt+
Z 1
1
tz�1
et � 1dt
�.
The right integral is holomorphic by Morera’s theorem and the fact that it converges uni-formly on compact subsets of C. For the first integral, we break the integrand up. Notethat the function f(z) = 1
e
z�1
has a simple pole at z = 0 near the real line, and is otherwiseholomorphic. Its Laurent expansion around z = 0 converges on A(0, 2⇡), and uniformly andabsolutely on A(", 2⇡ � "):
1
et � 1=
1
t+ c
0
+ c1
t+ c2
t2 + ...
Note that c�1
= 1. Rewriting the integral with the Laurent expansion, we have
Z1
0
tz�1
et � 1dt =
1X
n=�1
Z1
0
tz�1+ncn
dt,
and we can commute theP
andRby absolute convergence of the integrand. Then we can
simplify:1X
n=�1
cn
z + n
�=
1
z � 1+
c0
z+
c1
z + 1+ ....
[Alternatively, theR1
0
is holomorphic and converges since
Z1
0
xs�1
ex � 1dx =
1X
m=0
Bm
m!(s+m� 1),
where Bm
denotes the m-th Bernoulli number defined by
x
ex � 1=
1X
m=0
Bm
m!xm.
ThenB0
= 1, and since z
e
z�1
is holomorphic for |z| < 2⇡, we must have lim supm!1 |B
m
/m!|1/m =1/2⇡.] Anyway, we can claim
Claim. This gives a holomorphic function on all of C.
76
Claim.P1
n=1
c
n
z+n
converges uniformly for z in BR
(0) � [
bRcn=�1
B✏
(�n). The same ar-
gument also shows it’s meromorphic by excluding the term 1
z+k
and B✏
(�k).
The punchline here is that there exists a meromorphic function on C which equalsP1
n=1
1
n
z
for Re(z) > 1 and has a pole only at z = 1.
77
CHAPTER 21
ZETA FUNCTION AND PRIME NUMBER THEOREM
Functional equations towards the prime number theorem. Last time we had
⇣(z) =1
�(z)
Z 1
0
tz�1
et � 1dt, Re(z) > 1,
where1
et � 1=
1
t+ c
0
+ c1
t+ ...
We used this to see that ⇣(z) extends to a meromorphic function in all of C with a simplepole at z = 1 (and this is the only pole). We can see this by breaking the integral down into
1
�(z)
Z 1
0
tz�1
et � 1dt =
1
�(z)
✓Z1
0
tz�1
et � 1dt+
Z 1
1
tz�1
et � 1dt
◆,
and noting that the right hand integral is holomorphic by Morera’s theorem and uniformconvergence on compact subsets of C, while the left hand integral is also holomorphic andconverges.
Functional equation. Define a theta function #(t) as
#(t) :=1X
n=�1e�⇡n
2t.
This has the property that#(t) = t�1/2#(1/t)
for t 2 R>0
, which is on the problem set and uses the Fourier transform.
Theorem.
⇡�z/2�(z/2)⇣(z) =1
2
Z 1
0
uz/2�1(#(u)� 1)du =
Z 1
0
uz/2�1 (u)du,
78
where
(u) :=#(u)� 1
2=
1X
n=1
e�⇡n2u.
Proof. First we rewrite
1
2
Z 1
0
uz/2�1(#(u)� 1)du =
Z 1
0
1X
n=1
uz/2�1e�⇡n2u
!du.
Note that Z 1
1
uz/2�1(#(u)� 1)du
converges nicely because #(u)� 1 is bounded by the decreasing exponential:�����
1X
n=1
e�⇡n2u
�����
�����
1X
m=1
e�⇡mu
����� =e�⇡u
1� e�⇡u.
So for u = 1, this is bounded and the integral converges. Near 0, we consider
1
2
Z1
0
uz/2�1(#(u)� 1)du.
and as u ! 0, #(u) is well-behaved and the integral converges. Take the original integral andswap the sum and integral (justifying the interchange of summation and integral yourself)to get Z 1
0
1X
n=1
uz/2�1e�⇡n2u
!du =
1X
n=1
Z 1
0
uz/2�1e�⇡n2udu
Let t = ⇡n2u, dt = ⇡n2du. Then
1X
n=1
Z 1
0
uz/2�1e�⇡n2udu =
1X
n=1
Z 1
0
tz/2�1e�tdt · (⇡n2)�z/2 =1X
n=1
�(z/2)⇡�z/2n�z
= �(z/2)⇡�z/2⇣(z),
giving us the desired result. ⌅
What do we do with this theorem? Let’s define the xi function
⇠(z) = �(z/2)⇡�z/2⇣(z).
Theorem. ⇠(z) is holomorphic for Re(z) > 1, extends to a meromorphic function on all ofC with poles at 0 and 1, and
⇠(z) = ⇠(1� z).
Proof. First we have #(u) = u�1/2#(1/u). Then
(u) =#(u)� 1
2=
u�1/2#(1/u)� 1
2= u�1/2
#(1/u)� 1
2+
1
2u1/2
� 1/2
79
= u�1/2 (1/u) +1
2u1/2
� 1/2.
We break the integral up into two pieces:
⇠(z) =
Z1
0
uz/2�1 (u)du+
Z 1
1
uz/2�1 (u)du.
Let t = �1/u, du = �1/u2du, du = �1/t2dt. Then
Z1
0
uz/2�1 (u)du =
Z 1
1
t�z/2+1
t2
✓t1/2 (t) +
t1/2
2�
1
2
◆dt
=
Z 1
1
t�z/2�1/2 (t)dt+
Z 1
1
✓t�z/2�1/2
�
1
2t�z/2�1
◆dt
=
Z 1
1
t�z/2�1/2 (t)dt+1
2
t�(z�1)/2
�
z�1
2
����1
1
+1
2
t�z/2
�z/2
����1
1
=1
z � 1�
1
z+
Z 1
1
t�z/2�1/2 (t)dt
=) ⇠(z) =1
z � 1�
1
z+
Z 1
1
(u)⇣uz/2�1 + u�z/2�1/2
⌘du,
and the whole thing is symmetric through z 7! 1� z. ⌅
The conclusion is that ⇠ is symmetric about x = 1/2. There are no zeroes with x > 1,and by symmetry there are no zeroes with x < 1. Note that �(z) has simple poles atnonnegative integers, while �(z/2) has simple poles at 0, �2,�4, .... Then
⇣(z) =⇡z/2⇠(z)
�(z/2)
has trivial zeroes at �2,�4, ..., and we’ll show that there are no zeroes right of x = 1 (also,there is neither a pole nor zero at 0).
Claim. ⇣(z) has no zeroes in {Re(z) > 1}.Proof. ⇣(z) =
Qp prime
(1� 1/pz) � 1 for Re(z) > 1. ⌅
Theorem. ⇣(z) 6= 0 for Re(z) = 1.Proof. Deferred to below.
Theorem. (Prime Number) Let ⇡(N) = # of primes N . Then
⇡(N) ⇠N
logN,
i.e.⇡(N)
N/ logN! 1 as N ! 1.
We won’t show this theorem today, but we will show that ⇣(z) has no zeroes for Re(z) = 1.
80
To do this we need a few other results:
Lemma.
log ⇣(z) =X
m2Z,p prime
1
mpmz
.
Proof.
⇣(z) =Y
p prime
✓1
1� 1/pz
◆
log ⇣(z) = �
X
p prime
log
✓1�
1
pz
◆=
X
p prime
1X
m=1
(p�z)m
m=
X
m2Z,p prime
1
mpmz
.
Note that this is much like the zeta function:
X
p prime
1X
m=1
(p�z)m
m=
X
m2Z,p prime
1
mpmz
=1X
n=1
cn
nz
,
where
cn
=
(1/m n = pm
0 otherwise
and cn
� 0. This is an instance of a Dirichlet series, which we won’t talk about in detail.(Take Math 229x for this.) ⌅
Lemma. 3 + 4 cos ✓ + cos 2✓ � 0.Proof. Trivial. ⌅
Lemma. For z = x+ iy, x > 1,
log |⇣(x)3⇣(x+ iy)4⇣(x+ 2iy)| � 0.
Proof. We have
log |⇣(x)3⇣(x+ iy)4⇣(x+ 2iy)| = 3 log |⇣(x)|+ 4 log |⇣(x+ iy)|+ log |⇣(x+ 2iy)|
= 3Re(log ⇣(x)) + 4Re(log ⇣(x+ iy)) + Re(⇣(x+ 2iy))
=X
n
n�x(3 + 4 cos(y log n) + cos(2y log n)) � 0,
as desired. ⌅
Proof of the theorem. We wish to show that ⇣(z) 6= 0 for Re(z) = 1. Suppose ⇣(x+2iy) = 0.Then consider ⇣(x)3⇣(x + iy)4⇣(x + 2iy) = A. If ⇣(1 + iy) = 0, then as x ! 1+, A ! 0, acontradiction. This is because ⇣(z) has a simple pole at z = 1, ⇣(z) has zero at z = 1 + iy,and ⇣(z) is holomorphic (no pole) at z = 1 + 2iy. ⌅Chebyshev/Tchebychev’s (x) function. (unrelated to (u) earlier) Define
(x) :=X
p
m
<x
log p =X
n<x
⇤(n) =X
p<x
�log x
log p
⌫log p,
81
for the von Mangoldt ⇤ function defined as
⇤(n) =
(log p n = pm
0 n 6= pm,
and p a prime.
Proposition. (x) ⇠ x () ⇡(x) ⇠ x
log x
.Proof. We will only show the forward direction. Note that
(x) =X
p<x
�log x
log p
⌫log p
X
p<x
log x = log x · ⇡(x).
So (x)
x
⇡(x) log x
x,
and if (x)/x ! 1, then
1 = lim inf⇡(x) log x
x.
For the opposite inequality, note that
(x) �X
p<x
log p �
X
x
↵
<p<x
log p � (⇡(x)� ⇡(x↵)) log(x↵)
for ↵ 2 (0, 1). Hence
(x)/x+↵⇡(x↵) log x
x� ↵
⇡(x) log x
x,
i.e. (x)
x+↵ log x
x�↵ �
↵⇡(x) log x
x.
So if (x)x
! 1,
1 � ↵ lim sup⇡(x) log(x)
x
for all ↵ 2 (0, 1), so
1 � lim sup⇡(x) log x
x
and �(x)/x ! 1 =) ⇡(x) log x
x
! 1. ⌅
82
CHAPTER 22
PRIME NUMBER THEOREM
Prime number theorem. Recall from last time that if ⇡(x) is the number of primes x,then the prime number theorem states that ⇡(x) ⇠ x
log x
. We’ll go about proving this today.A bit of history: while the prime number theorem was proved in the 1800s, an elementary
proof of it was still in the air in the 1940s. Selberg and Erdos actually had a bitter fight overit: Selberg came up with a formula that implied the Prime Number Theorem but didn’twork out the details until Erdos had already did, and they disagreed on who got the creditfor it. Eventually Selberg won out, much to Erdos’s chagrin. There is an even simpler proofusing basic complex analysis, given by Newman, one of the authors of our textbook.
Anyways, recall that we had
log ⇣(z) =X
p prime,n�1
1
npnz
for Re(z) > 1. Since
X
p prime,n�2
1
npnz=
X
p prime
✓1
2p2z+
1
3p3z+ ...
◆
is holomorphic where Re(z) > 1/2,
X
p prime
1
pz= log ⇣(z)�
X
p prime,n�2
1
npnz
is holomorphic where Re(z) > 1/2. Note that log(z � 1)⇣(z) is entire, holomorphic on asubset of Re(z) � 1, so that
Pp prime
1
p
z
+ log(z� 1) is as well, along withP
p prime
� log p
p
z
+1
z�1
.Let � be defined as
�(z) =X
p prime
log p
pz.
Lemma. �(z)� 1
z�1
is holomorphic where Re(z) � 1.
83
Proof. See above.
We also consider the function ✓(x) =P
px
log p for real x.
Theorem. ✓(x) ⇠ x =) ⇡(x) ⇠ x.Proof. We have ✓(x) =
Ppx
log p
Ppx
log x = ⇡(x) log x. ⌅
Chebyshev showed that ✓(x) C · x in limit, i.e. lim sup ✓(x)
x
< c, which we will showlater. For the moment, we turn to proving another useful theorem first.
Theorem. Suppose
g(z) =
Z 1
0
e�ztf(t)dt
for f bounded and integrable, and is holomorphic for Re(z) � 0. Then
g(0) =
Z 1
0
f(t)dt
is well-defined.Proof. Suppose T > 0 and let g
T
(z) :=RT
0
e�ztf(t)dt. Let C be the boundary of theregion {z : |z| R,Re(z) � � for small � = �(R) > 0 and large R so that g(z) is holomorphicon the region and its boundary. Applying Cauchy’s integral formula, we get
g(0)� gT
(0) =1
2⇡i
Z
C
(g(z)� gT
(z))eTz
✓1 +
z2
R2
◆(1/z)dz.
Now bound this: on the part of C on the upper half-plane (denote this as C1
), we can write
|g(z)� gT
(z)| =
����Z 1
T
e�tzf(t)dt
���� B
Z 1
T
|e�tz
|dt =Be�TRe(z)
Re(z),
where B = maxt�0
|f(t)| and since |eTz
| = eTRe(z) and |1 + z
2
R
2 | = 2Re(z)
R
, we see that|
RC1
|
B
R
. For the other piece of the contour C2
, we can replace the integral for gT
(sincegT
is entire) by the contour of a semicircle in the lower half plane C 02
(which, similar tothe above, has |
RC
02|
B
R
), while the other integral ! 0 as T ! 1 by a simple bounding
argument. Taking R ! 1 concludes the proof. ⌅
Now we can get a bound for ✓(x) in the flavor of Chebyshev:
Lemma. ✓(x) (4 log 2)x.Proof.
�2n
n
�by definition contains every prime number from n+1, ..., 2n in its factoriza-
tion; thusQ
n<p2n
p
�2n
n
�. But we can also expand
(1 + 1)2n =
✓2n
0
◆+
✓2n
1
◆+ ...+
✓2n
2n
◆=)
✓2n
n
◆ 4n.
Thus Y
n<p2n
p 4n =)X
n<p2n
log p n log 4.
84
If n takes on values from 1, 2, 4, ..., 2M , where M is the highest such that 2M+1
� x, we get
✓(x) X
1<p2
M+1
< 2M+1 log 4 < 2x log 4 = (4 log 2)x,
as desired. ⌅
We’re ready to finish o↵ the proof of the prime number theorem. If Re(z) > 1, then
�(z) =X
p
log p
pz= z
Z 1
0
e�zt
· ✓(et)dt,
and from the above, we have limx!1
✓(x)
x
2 = 0. This implies
�(z + 1)
z + 1=
Z 1
0
e�(z+1)t✓(et)dt =)�(z + 1)
z + 1�
1
z=
Z 1
0
e�(z+1)t(✓(et)e�t
� 1)dt.
We can then take g(z) =R10
f(t)e�ztdt and f(t) = ✓(et)e�t
� 1, so that the lemma weproved above tells us Z 1
0
f(t)dt =
Z 1
1
✓(x)� x
x2
dx < 1.
This furthermore implies that ✓(x) ⇠ x:
Lemma. If g(x) is nondecreasing and
Z 1
1
g(x)� x
x2
dx < 1,
then g(x) ⇠ x.Proof. Almost trivial. ⌅
So we’ve proved the prime number theorem.
85
CHAPTER 23
ELLIPTIC FUNCTIONS
Elliptic Functions. These are doubly periodic, meromorphic functions on C. We won’ttalk too much about the history of the word “elliptic.” If w
1
, w2
are two nonparallel vectorsin C (w1
w2/2 R), then an elliptic function f are periodic in both directions:
f(z) = f(z + w1
) and f(z) = f(z + w2
).
These had origins in determining elliptic arclengths. By precomposing with multiplicationby a nonzero complex number ↵, we may assume w
1
= 1, w2
= ⌧ = w2w1
: that is, takeg(z) = f �m
↵
(z), g(z) = f(↵z). Taking ↵ = w1
gives g(z + 1) = f(w1
z + w1
) = g(z) andg(z + ⌧) = f(w
1
z + w2
) = f(w1
z) = g(z).Wlog, we have ⌧ 2 H, i.e. Im(⌧) > 0. Then Im(⌧), Im(1/⌧) have opposite signs because
Im(⌧) 6= 0 or else it’s collinear with 1.The objective now is to study these meromorphic functions f on C such that f(z+1) =
f(z) and f(z + ⌧) = f(z). The fundamental domain for translation by the lattice
⇤ = {a+ b⌧ : a, b 2 Z}
is then given by the parallelogram
P = {a+ b⌧ : (a, b) 2 [0, 1)2}.
Now we have some facts about f :
1. Fact. If f is holomorphic, then f is constant.
Proof. f is bounded by its maximum on P .
2. Fact.P↵2P
Res(f ;↵) = 0.
Proof. First, if there are no poles on @P , thenZ
@P
f(z)dz � 2⇡iX
↵2P
Res(f ;↵).
By taking the integral along P , we putZ
@P
f(z)dz =
Z1
0
f(t)dt+
Z1
0
f(1 + ⌧ t)⌧dt�
Z1
0
f(1 + ⌧ � t)dt�
Z1
0
f(⌧ � t)dt.
86
The integrals cancel in pairs by f being elliptic, and we are left with 0. If there is apole along @P , then the claim is that 9↵, ✏ : P +↵✏ has no pole where ↵ = (1+ ⌧) 2 Cis not collinear with 1 and ⌧ . To see this, suppose not; then we get an accumulationof poles at distinct points. Now note that P + ↵✏ is also a fundamental domain, andwe can reason similarly to the above.
Remark on terminology: The order of an elliptic function is the total order of its poles inthe fundamental domain P .
Corollary. There are no elliptic functions with orders 0 or 1.Proof. If f has order 0, then it is holomorphic, giving a contradiction. If f has order 1,
then there is one simple pole, and the residue is 6= 0, giving another contradiction. ⌅
Proposition. If f is an elliptic function of order k, then TFAE:(a) f has k total order of zeroes in P .(b) f(z)� c has total order of zeroes equal to k.Proof. (a) =) (b). Use the argument principle. (b) =) (a) follows similarly. ⌅
Weierstrass } functions. These are the order 2 elliptic functions with double polesat the lattice points (e.g. 0). Here’s a first attempt to define such a function:
X
w2⇤
1
(z + w)2=X
m,n2Z
1
(z +m+ n⌧)2.
The problem here is that this is not absolutely convergent.
Lemma.P
(m,n) 6=(0,0)
1
(|m|+|n|)r < 1 for r > 2.Proof.
X
m2Z
1
(|m|+ |n|)r=
1
|n|r+ 2
1X
m=1
1
(|m|+ |n|)r=
1
|n|r+ 2
1X
l=|n|+1
1
lr
1
|n|r+ 2
Z 1
|n|
1
lrdl
=1
|n|r+ 2
|n|�r+1
�r + 1.
Sum this over n: X
n
✓1
|n|r+ C
1
|n|r�1
◆
summable for r � 1 > 1, i.e. r > 2.
Corollary.P
(m,n) 6=(0,0)
1
(m+n⌧)
r
converges absolutely for any r > 2, ⌧ 2 H.Proof. Left as an exercise. ⌅
The right way to define such a function is
}(z) =1
z2+
X
w2⇤�{0}
✓1
(z + w)2�
1
w2
◆.
The claim is that this converges uniformly on compact subsets of C� ⇤ to a meromorphicfunction with double poles at ⇤.
87
Claim. } is the function we’re looking for.Proof. Let |z| < R. Then
}(z) =1
z2+
X
|w|<2R
✓1
(z + w)2�
1
w2
◆,
which is a finite sum. We add on another term:
1
z2+
X
|w|<2R
✓1
(z + w)2�
1
w2
◆+
X
|w|>2R
✓1
(z + w)2�
1
w2
◆.
The claim is thatP
|w|>2R
⇣1
(z+w)
2 �
1
w
2
⌘with w 2 ⇤ converges uniformly to a holomorphic
function on |z| < R. to see this, note that 1
(z+w)
2 �
1
w
2 = (z+w)
2�w
2
(z+w)
2w
2 = z
2+2zw
w
2(z+w)
2 , where
the denominator is like w4 and the numerator is like w where |z| is bounded. So the sumconverges uniformly on |z| < R. Thus the whole sum converges uniformly. ⌅
Proposition. } is elliptic: }(z + 1) = }(z) and }(z + ⌧) = }(z).Proof. Consider }0(z) on the swiss cheese disk. We have the limit fo holomorphic
functions, so the derivative converges uniformly. Note that
}0(z) = �2X
w2⇤
1
(z + w)2
is doubly periodic. Then we have that }(z + 1) = }(z) + a where a is a fixed constantbecause the derivative is 1-periodic, while }(z + ⌧) = }(z) + b where b is fixed because thederivative is ⌧ -periodic. The fact that a = b = 0 follows from the fact that } is even: inparticular, }(1/2) = }(�1/2) and }(�⌧/2) = }(⌧/2). ⌅
Remark. We note that } is an order 2 elliptic function, as it has a double pole at onlythe origin in P . Likewise, }0 is an order 3 elliptic function, since it has a triple pole at onlythe origin in P .
Theorem. Every elliptic function is a rational function in } and }0.Proof. First let’s understand the relationship between } and }0. We note that }0 is odd,
and }0(1/2) = }0(�1/2) = �}0(1/2), so that }0(!/2) = 0 and similarly for ⌧/2, (1 + ⌧)/2.These are all the zeroes of }0 since }0 is of order 3, while }0 has a pole of order 3 at theorigin.
For }, this means that }(z) � }(↵) = 0 for ↵ = 1/2, ⌧/2, (1 + ⌧)/2, and has a doubleroot (and nothing else since } is of order 2). }(z) � }(↵) = 0 has a simple root at ↵
j
� ↵distinct, for ↵ 6= 0, 1/2, ⌧/2, (1 + ⌧)/2 2 P .
Theorem. We have
}0(z)2 = 4(}(z)� e1
)(}(z)� e2
)(}(z)� e3
)
for e1
= }(1/2), e2
= }(⌧/2), e3
= }((1 + ⌧)/2).Proof. Look at the zeroes and the poles of each side. }0(z)2 has a pole of order 6 at 0,z
eroes of order 2 at 1/2, ⌧/2, (1+ ⌧)/2, and no others. The right hand side has poles of order
88
6 at 0, zeroes of order 2 at the same points, and no others. The function
}0(z)2
4(}(z)� e1
)(}(z)� e2
)(}(z)� e3
)
is holomorphic, doubly periodic, and hence constant. What’s the constant? If we Laurentexpand at 0, we get
}(z) =1
z2+ ... and }0(z) = �
�2
z2+ ...,
so
4(}(z)� e1
)(}(z)� e2
)(}(z)� e3
) = 4
✓1
z6+ ...
◆
and
}0(z)2 =4
z6+ ...
so the constant is 1. ⌅
Geometric interpretation. } is surjective from P to C with “order 2.” Note thatP/⇤ ⇠= T2.
A (sketch of a) geometric picture is as follows: consider skewering your donut on a stickthrough four points on the lattice; then taking the quotient by flipping over, we get half-torus that is one-to-one except for two points that are two-to-one.
Theorem. Every elliptic function that is doubly periodic under 1 and ⌧ is a rationalfunction in }
⌧
,}0⌧
.Proof. First, we do this for even elliptic functions. Let F denote one such, which has
even order (possibly 0) of zeroes and poles at 0. Hence F (z)}(z)m,m 2 Z has no zero orpole at 0. Let ↵
1
, ...,↵k
denote the zeroes of F with multiplicity; these come in pairs ±↵i
if ↵i
6= 1/2, ⌧/2, (1 + ⌧)/2. Likewise, list the poles �1
, ...,�k
of F ; the same comment holds.The claim is that
G(z) =
Qi
(}(z)� }(↵i
))Qj
(}(z)� }(�j
))
has the same zeroes and poles with multiplicity as F , so F = cG.Now for general elliptic F , let F = F
even
+ Fodd
. Then Feven
= F (z)+F (�z)
z
, Fodd
=F (z)�F (�z)
2
, and they are both elliptic. Feven
is a rational function in }, while Fodd
/}0 iseven and elliptic, and thus a rational function in }. The result follows. ⌅
Note that, for di↵erent ⌧ , P/⇤ are not generally the same torus.
Anyway, this story of elliptic curves has applications to number theory and cryptography,and algebraic geometry and other areas of math. Next time, we’ll start with the formula(}0)2 = 4}0
�g2
}�g3
for some g2
, g3
, and get into other uses and generalizations for ellipticfunctions.
89
CHAPTER 24
WEIERSTRASS’S ELLIPTIC FUNCTION AND ANOVERVIEW OF ELLIPTIC INVARIANTS AND MODULI
SPACES
Elliptic functions. There is a relationship between elliptic functions and elliptic integrals.Consider integrals of the form
Zdx
p
cubicor
Zdx
p
quartic.
These integrals show up when we want to compute the arc length of ellipses. The second is aSchwarz-Christo↵el transformation from H to a quadrilateral. As before, starting from theirdouble-periodicity and some transformation properties, we want to study the fundamentalfunction }(z).
Today, we’ll start with }(z), and consider the equation
(}0(z))2 = 4}(z)3 � g2
}(z)� g3
,
which arises when we consider the integral
Z 1
z
dzp4z3 � g
2
z � g3
=
Z}
0(z)
}(z)
}0(w)dwp}0(w)2
and substitute }(w) = z; this gives } as the inverse of the integral. Note that, under theelliptic integral given above, the three roots of the cubic in the denominator and 1 goto the vertices of the quadrilateral in the image, since the elliptic integral is basically aSchwarz-Christo↵el map. One can show here that an elliptic function is the inverse to theSchwarz-Christo↵el map.
Recap on what we know about }. First we recall that Weierstrass’s elliptic function }is defined for any fixed ⌧ as
}⌧
(z) =1
z2+X
w2⇤
⇤
✓1
(z + w)2�
1
w2
◆,
90
where ⇤⇤ = ⇤ � {0} and the lattice ⇤ for a fixed ⌧ is ⇤ = {n +m⌧ : m,n 2 Z} ✓ C. } ismeromorphic with double poles at lattice points. We also had
}0(z) = �2X
w2⇤
1
(z + w)3= �2
X
n,m2Z
1
(z + n+m⌧)3
We showed that if }(1/2) = e1
,}(⌧/2) = e2
, and }( 1+⌧2
) = e3
, then
}0(z)2 = 4(}(z)� e1
)(}(z)� e2
)(}(z)� e3
).
Today, we’ll compute }0(z) = 4}(z)3 � g2
}(z) � g3
. We do this by introducing Eisensteinseries:
Eisenstein series. These are functions given by
Ek
(⌧) =X
w2⇤
⇤
1
w2
=X
(m,n) 6=(0,0)
1
(m+ n⌧)k.
Here are a few facts, which we’ll leave to the reader to prove:
• Fact 1: If k � 3, then Ek
(⌧) is a holomorphic function of ⌧ in H: the sum convergesuniformly in ⌧ for Im(⌧) > � > 0.
• Fact 2: If k is odd, then Ek
(⌧) = 0.
• Fact 3: We have the following transformation relations:
Ek
(⌧ + 1) = Ek
(⌧) and Ek
(�1/⌧) = ⌧kEk
(⌧)
E.g. for the last relation we have (m+ n(�1/⌧))k = ⌧�k(m⌧ � n)k.
These properties tell us that E2k
is a weakly modular function of weight 2k. Note that, sincethe Eisenstein series converge depending on ⌧ , we can just write them as constants.
Proposition. If z is near 0, then
}(z) =1
z2+
1X
k=1
(2k + 1)E2k+2
z2k =1
z2+ 3E
4
z2 + 5E6
z4 + ...
Proof. We can write
}(z) =1
z2+X
w2⇤
⇤
✓1
(z � w)2�
1
w2
◆,
where we reindex the lattice under the transformation w 7! �w. Then
1
(z � w)2=
1
w2
1
(1� z/w)2
Noting that 1
1�x
= 1 + x + x2 + ... for |x| < 1 and 1
(1�x)
2 = @
@x
( 1
1�x
) = 1 + 2x + 3x2 + ...,
we have for |z/w| < 1 that
1
(z � w)2=
1
w2
✓1 + 2
z
w+ 3
⇣ z
w
⌘2
+ ...
◆.
91
So
}(z) =1
z2+X
w2⇤
⇤
1
w2
1X
n=1
(n+ 1)zn
wn
=1
z2+
1X
n=1
X
w2⇤
⇤
1
wn+2
!(n+ 1)zn
Since En+2
=P
w2⇤
⇤1
w
n+2 , we have
}(z) =1
z2+
1X
k=1
(2k + 1)E2k+2
z2k,
as En+2
= 0 when n is odd. ⌅
Theorem. If }0(z)2 = 4}(z)3 � g2
}(z)� g3
, then g2
= 60E4
and g3
= 140Eg
.Proof. We can write out some terms of the identity we derived above:
}(z) =1
z2+ 3E
4
z2 + 5E6
z4 + ...
}0(z) = �
2
z3+ 6E
4
z + 20E6
z3 + ...
}0(z)2 = 4}(z)3 + ... =4
z6� 24E
4
1
z2� 80E
6
+ ...
}(z)3 =1
z6+ 9E
4
1
z2+ 15E
6
+ ....
Now take the di↵erence
}0(z)2 � 4}(z)3 + 60E4
}(z) + 140E6
.
This is holomorphic near 0; each term is holomorphic away from ⇤ and is doubly-periodic,so the di↵erence is holomorphic, doubly-periodic, and takes 0 to 0. Then, from earlierproblems, we know that it’s 0, and the equality
}0(z)2 = 4}(z)3 � g2
}(z)� g3
gives g2
= 60E4
and g3
= 140Eg
as desired. ⌅
What is this explicit formula telling us? On the one hand, we have C/⇤, the “domainof } and }0” which we identify with a torus. Consider a map to C2 given by
z 7! (}(z),}0(z));
we can draw C2 (which we don’t really discuss in class) by drawing R ⇥ R on the axesin 2-space, and the curve it traces out is the set of solutions to the polynomial functiony2 = 4x3
� g2
x� g3
, a cubic curve. What you’re seeing is that this complex torus is a cubiccurve in C2. We can check injectivity and surjectivity and other nice things, but let’s getout of this story portion and move into....
The space of elliptic curves. Now we want to consider all possible ⌧ ’s, and we notethat the lattice ⇤ doesn’t change under the action of PSL
2
Z ✓ Aut(H) = PSL2
R. This isjust the space of transformations of the form
⌧ 7!
a⌧ + b
c⌧ + d
92
for a, b, c, d 2 Z, generated by ⌧ 7! ⌧ + 1 and ⌧ 7! �1/⌧ ; it’s what you can do to ⌧ to notchange your lattice.
(The elliptic functions compose with dilations; e.g. they’re maps from coplex tori fromone direction to the other direction, just by dilation. You can’t just stretch it in onedirection, since then it won’t be holomorphic.)
So let me show you a fundamental domain D in H. This is the part of H above the circle|z| = 1 and bounded by the lines Re(z) = {�1/2, 1/2}; e.g. D = {z 2 H : |z| > 1,� 1
2
<Re(z) < 1
2
}. This is the moduli space of complex 1-tori (elliptic curves). In other words,D is the fundamental domain of the action of PSL
2
Z. H/PSL2
Z is the moduli space ofelliptic curves.
Let’s do some justice to why these things are nice. E2k
(⌧) is sort of a function onH/PSL
2
Z, defined on D, and transforms under the involution ⌧ 7! �1/⌧ :
E2k
(�1/⌧) = ⌧2kE2k
(⌧).
When we say these things are called modular forms, we mean that they “don’t change thecorrect way,” and the reason here is the power coe�cient. We can, however, write
limIm(⌧)!1
E2k
(⌧) = limIm(⌧)!1
X
(m,n) 6=(0,0)
1
(m+ n⌧)2k= 2
1X
m=1
1
m2k
= 2⇣(2k),
since for n 6= 0 the center experssion has the n⌧ term ! 0 for Im(⌧) ! 1. We also have
E4
(1) = 2⇣(2k).
Then
g2
(1) = 60E4
(1) = 60(2)(⇡2
90) =
4
3⇡4.
g3
(1) = 140E6
(1) = 140(2)(⇣(6)) =8
27⇡6.
Setting the modular discriminant � = g32
� 27g23
, we note that �(1) = 0; the right handside expression, g3
2
�27g23
, is the discriminant of 4x3
�g2
x�g3
. This is a weight 12 modularform, by which we mean that E
2k
is a transformation of weight 2k, and it’s modular iflim
Im(⌧)!1 E2k
(⌧) exists and holomorphic at 1, E2k
(⌧) is holomorphic on H, and E2k
(⌧ +1) = E
2k
(⌧). We have another modular form of weight 12, which is simply g32
. Define
j =1728g3
2
�.
This has weight 0, so it’s really defined as a function on H/PSL2
Z (e.g. a meromorphicfunction from H ! C invariant under the action of SL
2
Z), which is the moduli space of allelliptic curves. It’s called a j-invariant of elliptic curves. There’s a theorem that states:
Theorem. This is holomorphic on H, invertible under PSL2
Z, has j(1) = 1, withresidue 1 at 1, and gives a surjection H ! C that injects from the quotient H/PSL
2
Z.Proof. See Serre’s Course in Arithmetic. ⌅
I won’t say too much more about this subject; we’ve got j as an invariant in the mod-uli space, but we won’t talk about why number theorists and cryptographers are interestedin it. For this, consult any reference on elliptic curves. We’ll use the remainder of the timefor questions.
93
APPENDIX A
SOME THINGS TO REMEMBER FOR THE MIDTERM
Complex expressions:
1. Remember polar form: z = rei✓.
2. Stereographic projection is given by ' : S2
�{p} ! R2,' : (u, v, w) 7! ( u
1�w
, v
1�w
),'�1 :
(x, y) 7! ( 2x
x
2+y
2+1
, 2y
x
2+y
2+1
, x
2+y
2�1
x
2+y
2+1
).
3. cos z = e
iz
+e
�iz
2
, sin z = e
iz�e
�iz
2i
.
4. log z = log r + i✓ + i(2⇡n).
Holomorphic functions:
1. To show that a function f is holomorphic, use the Cauchy-Riemann equations andshow that the partials are continuous. You can also any rule from real analysis (thechain rule, the product rule, the quotient rule, etc.), show that f has a power seriesdevelopment, or use Morera’s theorem.
2. Recall that the Cauchy-Riemann equations are(
@u
@x
= @v
@y
�
@v
@x
= @u
@y
.
3. Morera’s theorem: If f(z) is continuous on a region D andR�
f(z)dz = 0 for � atriangle (or closed curve), then f(z) is holomorphic. This is useful for:
1. Showing limn!1 f
n
= f is holomorphic.
2. Showing some functions defined by sums or integrals are holomorphic.
3. Showing functions that are continuous and holomorphic on all but some set (e.g.some point) are holomorphic.
4. f is holomorphic() f complex analytic() f smooth; this means you can substituteany holomorphic function with its power series, and can di↵erentiate it endlessly.
5. The closed curve theorem tells us that the integral of a function that is holomorphicin the open disk D over a contour C ⇢ D is 0.
94
6. Per the antiderivative theorem, if f(z) is holomorphic on an open disk D, then ithas an antiderivative F (z) such that F 0(z) = f(z).
7. Cauchy’s integral formula allows us to determine the value of a holomorphic func-tion f at a point. In particular,
f(w) =1
2⇡i
Z
C
f(z)
z � wdz,
where f is holomorphic in an open disk D and C contains w.
8. The uniqueness theorem tells us that holomorphic functions are determined uniquelyup to their values in a certain region.
9. The mean value theorem allows us to evaluate a holomorphic function at point. Iff(z) is holomorphic on D
↵
(R), then 8r : 0 < r < R, we have
f(↵) =1
2⇡
Z2⇡
0
f(↵+ rei✓)d✓.
10. The maximum modulus principle: if f is holomorphic in any region, then itsmaximum is achieved at the boundary.
11. Harmonic functions are characterized by the fact that their Laplacian is zero, andpossess nice properties as you discovered in the problem sets.
Entire and meromorphic functions:
1. Entire functions are functions that are holomorphic in all of C.2. Liouville’s theorem tells us that any bounded entire function is constant. A gener-
alization is that if |f(z)| A+B|z|N and f(z) is entire, then f(z) is a polynomial ofdegree N .
3. Likewise, if f(z) is entire and f(z) ! 1 as z ! 1, then f(z) is a polynomial.
4. A nonconstant polynomial always has a root, per the fundamental theorem of algebra.
5. Remember the di↵erent types of singularities and why they’re useful, namely poles oforder k and essential singularities.
6. Meromorphic functions are holomorphic functions on C. In other words, they have afinite set of removable singularities, and can always be expressed as as a ratio of twoholomorphic functions.
7. A weak formulation of Picard’s little theorem tells us that if f is entire, then itsimage is dense.
Evaluating integrals:
1. There are generally two ways we’re learned to do nontrivial integrals:
1. Use the closed curve theorem to give a contour integral as zero, then decomposethe contour and bound the individual parts to find a specific integral.
2. Use the residue theorem by first calculating residues:
95
(0) Compute the integralRC
✏
(↵)
f(w)dw
(1) Compute the Laurent series for f centered at ↵, and take c�1
(2) If f has a simple pole, then c�1
= limz!↵
(z � ↵)f(z)
(3) If f(z) = A(z)/B(z) for B having a simple pole and A 6= 0, then c�1
= Res(f ;↵) =A(↵)/B0(↵).
The value of the integral over a curve containing these poles is 2⇡i⇥(sum of residues).
Series:
1. Know the definitions of radius of convergence, lim|an
|
1/n = L, and convergencetests/arguments for power series. These may be things that range from proving thata power series of a holomorphic function is uniformly convergent to finding the radiiof convergence of some functions.
2. Laurent series are power series that extend in both �1 and 1 directions, and aretypically defined on annuli (why?). To compute Laurent series, you can generally
1. Start with known power series and make the appropriate substitutions/adjustments.
2. Use known geometric series. Being able to decompose a function into its partialfraction expansion is a good trick to know here.
Others:
1. Know the topological aspects: polygonally connected (path connected), polygonallysimply connected (simply connected), what is meant by a region, etc. Note any openpath connected set is simply connected.
2. Know when to interchangePP
,PR
, @R, lim
R, etc. To deal with interchange of
sums, show absolute convergence. To deal with interchange of limits, show uniformconvergence. (This is covered in real analysis, and also in baby Rudin.)
3. A helpful tool is that you can bound the absolute value of any integral over CR
by2⇡R times the integrand; this is known as the ML theorem or the estimation lemma.
96
APPENDIX B
ON THE FOURIER TRANSFORM
This is an addendum to the class notes on Fourier transforms, and is based on Chapter 4 ofStein and Shakarchi’s Complex Analysis. It is important to note that a function f : R ! Cmust satisfy appropriate regularity and decay conditions to possess a Fourier transform. Ifit does (we’ll make this notion precise below), then the Fourier transform is defined by
f(⇠) =
Z 1
�1f(x)e�2⇡ix⇠dx, ⇠ 2 R. (B.1)
The Fourier inversion formula is then
f(x) =
Z 1
�1f(⇠)e2⇡ix⇠d⇠, x 2 R. (B.2)
There is a deep and natural connection between complex analysis and the Fourier transform.For example, given a function f initially defined on the real line, the possibility of extendingit to a holomorphic function is closely related to the rapid decay at infinity of its Fouriertransform f .
Proposition. The functions f(x) = e�⇡x2
and f(x) = 1
cosh⇡x
are their own Fourier trans-forms.
Proof. Show this yourself. ⌅
Definition. For ↵ > 0, denote by F↵
the class of all functions that satisfy the follow-ing two conditions:
(i) f is holomorphic in the horizontal strip S↵
= {z 2 C : |Im(z) < a}.(ii) There is a constant A > 0 :
|f(x+ iy)| A
1 + |x|1+✏
for any ✏ > 0 and all x 2 R and |y| < a. This condition is known as moderate decay. Aswe will soon see, F
↵
can be regarded as the class of functions with well-defined Fourier
97
transforms.
Intuitively, one can think of F↵
as those holomorphic functions of S↵
that are of “mod-erate decay” on each horizontal line Im(z) = y, uniformly in �a < y < a. Some functionsthat belong to F
↵
include
f(z) = e�⇡z2
for all ↵, and
f(z) =1
⇡
c
c2 + z2
for 0 < a < c.
Proposition. If f 2 F↵
, then f (n)
2 F�
for 0 < � < ↵.Proof. Use the Cauchy integral formula. ⌅
Proposition. If f 2 F↵
for some ↵ > 0, then
|f(⇠)| Be�2⇡�|⇠|
for some B and any 0 � < ↵.Proof. Use contour integration. ⌅
Let’s drop the ↵ in F↵
to make notation simpler. The previous proposition tells us that iff 2 F, then f has rapid decay at infinity. The Fourier inversion formula then works in thefollowing sense:
Proposition. If f 2 F, then the Fourier inversion formula holds, and we have
f(x) =
Z 1
�1f(⇠)e2⇡ix⇠d⇠
for all x 2 R.
Properties of the Fourier transform.
Now that we have some basis for knowing when a Fourier transform or inversion is well-defined, let’s talk about some of the things we do with them.
Theorem. (Basic properties) Let f, g 2 F. Then the following are true:(1) If h(x) = af(x) + bg(x) for a, b 2 C, then h = af + bg.(2) If h(x) = f(x� x0) for x0
2 R, then f(⇠) = e�2⇡ix
0⇠ f(⇠).
(3) If h(x) = e2⇡ix⇠0f(x) for ⇠0 2 R, then h(⇠) = f(⇠ � ⇠0).
(4) If h(x) = f(ax) for a 2 R� {0}, then h(⇠) = 1
|a| f⇣⇠
a
⌘.
(5) If h(x) = f(x), then h(⇠) = f(�⇠).(6) Let F denote the Fourier transform. F2(f)(x) = f(�x) and F
4(f) = f .Proof. These are all easy to verify.
98
Theorem. (Relation to di↵erentiation)
F
✓d
dx
◆m
((�ix)nf)
�= (i⇠)m
✓d
d⇠
◆n
f ,
where F denotes the Fourier transform. In particular, the Fourier transform of (�ix)f(x)is d
d⇠
f(⇠), and the Fourier transform of f 0 is (i⇠)f(⇠).Proof. Use di↵erentiation under the integral sign and integration by parts.
Theorem. (Relation to convolution) If f ⇤ g denotes the convolution of f and g defined by
(f ⇤ g)(⇠) :=
Z 1
�1f(⇠)g(⇠ � t)dt,
then [f ⇤ g(⇠) = f · g.
This latter theorem is important for functional analysis, but we won’t discuss it to make thisnote self-contained. We can also prove an important tool known as the Poisson summationformula:
Theorem. (Poisson summation formula) If f 2 F, thenX
n2Zf(n) =
X
n2Zf(n). (B.3)
Proof. Let f 2 F↵
and choose a � : 0 < � < ↵. The function 1
e
2⇡iz�1
has simple poles
with residue 1
2⇡i
at the integers. So f(z)
e
2⇡iz�1
has simple poles at the integers n, with residuesf(n)/2⇡i. Applying the residue formula to the contour �
N
below,
we getX
|n|N
f(n) =
Z
�
N
f(z)
e2⇡iz � 1dz.
Letting N ! 1, and recalling the moderate decay of f , we see that the sum converges toPn2Z f(n), and that the integral over the vertical segments goes to 0. So in the limit we
getX
n2Zf(n) =
Z
L1
f(z)
e2⇡iz � 1dz �
Z
L2
f(z)
e2⇡iz � 1dz, (B.4)
99
where L1
and L2
are the real line shifted down and up by b, respectively. Now we use thefact that if |w| > 1, then
1
w � 1= w�1
1X
n=0
w�n
to see that on L1
, where |e2⇡iz| > 1, we have
1
e2⇡iz � 1= e�2⇡iz
1X
n=0
e�2⇡inz.
If |w| < 1, then 1
w�1
= �
P1n=0
wn, so on L2
we have
1
e2⇡iz � 1= �
1X
n=0
e2⇡inz.
Substituting these equalities into equation (4) above, we see that
X
n2Zf(n) =
Z
L1
f(z)
e�2⇡iz
1X
n=0
e�2⇡inz
!dz +
Z
L2
f(z)
1X
n=0
e2⇡inz
!dz
=1X
n=0
Z
L1
f(z)e�2⇡i(n+1)zdz +1X
n=0
Z
L2
f(z)e2⇡inzdz =1X
n=0
f(n+ 1)+1X
n=0
f(�n) =X
n2Zf(n),
where we shift L1
and L2
back to the real line by replacing x 7! x� ib. ⌅
The Poisson summation formula gives many useful identities in complex analysis and be-yond (namely, number theory). For example, the Fourier transform of the function f(x) =
e�⇡t(x+a)
2
is f(⇠) = t�1/2e�⇡⇠2/te2⇡ia⇠. Applying the Poisson summation formula to this
pair gives the following identity:
1X
n=�1e�⇡t(n+a)
2
=1X
n=�1t�1/2e�⇡n
2/te2⇡ina.
This identity can be used to prove the following transformation law for the theta function#(t) :=
P1n=�1 e�⇡n
2t:
#(t) = t�1/2#(1/t),
for t > 0. It can also be used to prove the key functional equation of the Riemann zetafunction, which gives its analytic continuation:
⇣(s) = 2s⇡s�1 sin⇣⇡s
2
⌘�(1� s)⇣(1� s)
Finally, Fourier transforms can tell us much about a function. For example, the followingtheorem describes the nature of those functions whose Fourier transforms are supported;in particular, it relates the decay properties of a function with the holomorphicity of itsFourier transform:
Theorem. (Paley-Wiener) Suppose f is continuous and of moderate decrease on R. Thenf has an extension to the complex plane that is entire with |f(z)| Ae2⇡M |z| for someA > 0 if and only if f is supported in the interval [�M,M ].
100
Applications of the Fourier transform.
One can think of the Fourier transform as a mapping that converts a function f(t) in thetime domain (t has units of seconds) to a function f(!) in the frequency domain (! hasunits of hertz or something).
Fourier transforms can be used to solve di↵erential equations (c.f. below), and is clearlywidely used in physics, probability, engineering, and many other areas. For example, oneusually proves the central limit theorem in probability theory using Fourier transforms, andcan prove a mathematical formulation of the Heisenberg uncertainty principle. For all theFourier analysis you can get, Stein and Shakarchi’s Fourier Analysis is a good standardtextbook (and one you can find online...).
Examples involving the Fourier transform.
Here is a flavor of some of the problems one can ask about Fourier transforms.
Example 1. This example generalizes some of the properties of e�⇡x2
related to the factthat it is its own Fourier transform. Suppose f(z) is an entire function that satisfies
|f(x + iy)| ce�ax
2+by
2
for some a, b, c > 0. Let f(⇣) =R1�1 f(x)e�2⇡ix⇣dx. Then f
is an entire function of ⇣ that satisfies |f(⇣ + i⌘)| c0e�a
0⇣
2+b
0⌘
2
for some a0, b0, c0 > 0.
Proof. Note that f is clearly an entire function since we can just commute d
dz
with the
integral, and f(x)e�2⇡ix⇣ is entire. Thus f does not have any poles, and in particular wecan change the contour from the real axis to the line x � iy for some y > 0 fixed and�1 < x < 1 without a↵ecting the value of the integral. Assume ⇣ > 0, and observe that
f(⇣) =
Z 1
�1f(x)e�2⇡ix⇣dx =
Z 1
�1f(x� iy)e�2⇡i(x�iy)⇣dx =
Z 1
�1f(x� iy)e�2⇡(xi+y)⇣dx.
Then note����Z 1
�1f(x� iy)e�2⇡(xi+y)⇣dx
���� ����Z 1
�1ce�ax
2+by
2
e�2⇡(xi+y)⇣dx
���� =����Z 1
�1ce�ax
2�2⇡xi⇣+by
2�2⇡⇣ydx
���� .
Since |e�ax
2�2⇡xi⇣
| ! 0 as |x| ! 1, we see that f(⇣) = O(eby2�2⇡⇣y). Setting y = d⇣ where
d is a small constant, we have that f(⇣) = O(eb(d⇣)2�2⇡⇣(d⇣)) = O(e(bd
2�2⇡d)⇣
2
), as desired.Also observe that
|f(⇣ + i⌘)| =
����Z 1
�1f(x)e�2⇡ix(⇣+i⌘)dx
���� =����Z 1
�1f(x� iy)e�2⇡(xi+y)(⇣+i⌘)dx
����
=
����Z 1
�1f(x� iy)e�2⇡(x⇣i�x⌘+⇣y+y⌘i)dx
���� ����Z 1
�1ce�ax
2+by
2�2⇡(x⇣i�x⌘+⇣y+y⌘i)dx
����
Repeating the above argument, we see that f(⇣) = O(e(by2�2⇡⇣y)+(�ax
2+2⇡x⌘)). Now let y =
d⇣ and x = d0
⌘ for small d, d0
in the bound to see that |f(⇣+i⌘)| = O(e(bd2�2⇡d)⇣
2+(�ad
20+2⇡d0)⌘
2
).
So |f(⇣ + i⌘)| c0e�a
0⇣
2+b
0⌘
2
for a0 = 2⇡d � bd2, b0 = 2⇡d0
� ad20
(which are > 0 for smalld, d
0
), and some suitable scale factor c0 > 0. ⌅
101
Example 2. The problem is to solve the di↵erential equation
an
dn
dtnu(t) + a
n�1
dn�1
dtn�1
u(t) + ...+ a0
u(t) = f(t),
where a0
, a1
, ..., an
are complex constants, and f is a given function. Here we suppose thatf has bounded support and is smooth.
Proof. We do this in steps:(a) Let f(z) =
R1�1 f(t)e�2⇡iztdt. Observe that f is an entire function, and using
integration by parts show that |f(x+ iy)| A
1+x
2 if |y| a for any fixed a � 0.
(b) Write P (z) = an
(2⇡iz)n + an�1
(2⇡iz)n�1 + ... + a0
. Find a real number c so thatP (z) does not vanish on the line L = {z : z = x+ ic, x 2 R}.
(c) Set
u(t) =
Z
L
e2⇡izt
P (z)f(z)dz.
Check thatnX
j=0
aj
(d
dt)ju(t) =
Z
L
e2⇡iztf(z)dz
and Z
L
e2⇡iztf(z)dz =
Z 1
�1e2⇡ixtf(x)dx.
We then conclude by the Fourier inversion theorem thatP
n
j=0
aj
( d
dt
)ju(t) = f(t). Notethat the solution u depends on the choice of c.
(a) f is an entire function because we can commute d
dz
with the integral and the inte-grand is also an entire function. We know that f 2 C2, so now we use integration by parts:let u = f(t), du = f 0(t)dt, dv = e�2⇡iztdt, and v = �
1
2⇡iz
e�2⇡izt so that
|f(z)| =
����Z 1
�1f(t)e�2⇡iztdt
���� =�����
f(t)
2⇡ize�2⇡izt]1
t=�1 +
Z 1
�1
f 0(t)
2⇡ize�2⇡iztdt
����
Now let u = f 0(t), du = f 00(t)dt, dv = e
�2⇡izt
2⇡iz
dt, and v = �
1
(2⇡iz)
2 e�2⇡izt so that
|f(z)| =
�����f(t)
2⇡ize�2⇡izt]1
t=�1 +
✓�
f 0(t)
(2⇡iz)2e�2⇡izt]1
t=�1 +
Z 1
�1
f 00(t)
(2⇡iz)2e�2⇡iztdt
◆���� .
Since f is entire, we can (as in the previous exercise) change the contour of integrationsince f does not have any poles. So we change the contour from t to t � is, as in theargument on page 115 of the text. Rewriting, we have
|f(x+ iy)|
�����f(t� is)
2⇡i(x+ iy)e�2⇡i(x+iy)(t�is)]1
t=�1
����+����
f 0(t� is)
(2⇡i(x+ iy))2e�2⇡i(x+iy)(t�is)]1
t=�1
����
+
����Z 1
�1
f 00(t+ is)
(2⇡i(x+ iy))2e�2⇡i(x+iy)(t�is)dt
���� .
102
If f 0 and f 00 also decay, we can bound this again:
|f(x+ iy)|
�����M
0
2⇡i(x+ iy)e�2⇡s(x+iy)
����+����
M1
(2⇡i(x+ iy))2e�2⇡s(x+iy)
����
+
����Z 1
�1
M2
(2⇡i(x+ iy))2e�2⇡s(x+iy)dt
���� .
Note that the first two terms are bounded, and examining the integral we see that wecan extract the 1
z
2 term. Since y is also bounded, we have that |f(x + iy)| = O( 1
x
2 ), and
can therefore conclude that |f(x+ iy)| A
1+x
2 for a suitable scale factor A.(b) P is a polynomial of degree n, so it only has finitely many roots and we can choose
such a c so that the line L does not pass through a root. So we can choose c1
= 1, and letcn
= cn�1
+1 if P vanishes with the choice of cn�1
. Then c will be the cn
for which P doesnot vanish, and it’ll surely be found after n iterations.
We can find some constraints on c if we notice that we want |P (x + ic)| > 0 and canbound |P (x+ ic)| as follows:
|P (x+ ic)| � |an
||x+ ic|n �
n�1X
j=1
|aj
||x+ ic|j � |an
||x+ ic|n �
0
@n�1X
j=1
|aj
|
1
A|x+ ic|n�1
for |x+ ic| > 1 (we can repeat the same procedure with a di↵erent bound of |x+ ic| insteadof |x + ic|n�1 in the rightmost term if |x + ic| < 1, and again solve accordingly). Since we
want |an
||x+ ic|n � (P
n�1
j=1
|aj
|)|x+ ic|n�1 > 0, we see that |x+ ic| >P
n�1
j=1
|aj
||a
n
| . We also
know that |x+ ic| � |c|, so if |x+ ic| > 1 we can choose c > max(1,P
n�1
j=1
|aj
||a
n
| ) to satisfy theinequality and we’re done. In particular, one c for which the polynomial does not vanish isc = max(1,
Pn�1
j=1
|aj
||a
n
| ) + 1.
(c) We di↵erentiate under the integral sign to see that
nX
j=0
aj
(d
dt)ju(t) =
nX
j=0
aj
(d
dt)jZ
L
e2⇡izt
P (z)f(z)dz =
nX
j=0
aj
Z
L
(d
dt)je2⇡izt
P (z)f(z)dz
= an
Z
L
(2⇡iz)ne2⇡izt
P (z)f(z)dz+a
n�1
Z
L
(2⇡iz)n�1
e2⇡izt
P (z)f(z)dz+...+a
0
Z
L
(2⇡iz)0e2⇡izt
P (z)f(z)dz
=
Z
L
P (z)e2⇡izt
P (z)f(z)dz =
Z
L
e2⇡iztf(z)dz,
as desired.Furthermore, we claim that the equality
Z
L
e2⇡iztf(z)dz =
Z 1
�1e2⇡ixtf(x)dx
follows by observing that, as in the previous problem, e2⇡iztf(z) is clearly an entire functionand thus its integral is also entire by commuting d
dz
. So the integral does not have anypoles, and in particular we can change the contour of integration from the line L = {z :z = x+ ic, x 2 R} (where c is chosen so that P (z) does not vanish) to the real axis withouta↵ecting the value of the integral.
103
Finally, because we’re applying the Fourier inversion to the Fourier transform f of fand both functions satisfy the decay conditions in the hypothesis, we get from the Fourierinversion theorem that
Pn
j=0
aj
( d
dt
)ju(t) =R1�1 e2⇡ixtf(x)dx = f(t), and we are done. Thus
the solution to the di↵erential equation
an
dn
dtnu(t) + a
n�1
dn�1
dtn�1
u(t) + ...+ a0
u(t) = f(t),
is
u(t) =
Z
L
e2⇡izt
P (z)f(z)dz,
for P (z) = an
(2⇡iz)n + an�1
(2⇡iz)n�1 + ...+ a0
and c 2 R so that P (z) does not vanish onthe line L = {z : z = x+ ic, x 2 R}.
⌅
104
APPENDIX C
REFERENCES
We cite many sources in these notes. The main course textbook is:
J. Bak and D. J. Newman, Undergraduate Texts in Mathematics: Complex Analysis. Springer,2010.
One that is oftentimes referenced (and is a gem of a textbook), is
E. M. Stein and R. Shakarchi, Princeton Lectures in Analysis 2: Complex Analysis. Prince-ton UP, 2003.
And of course, the classic textbook is none other than Ahlfors:
L. V. Ahlfors, Complex Analysis: An Introduction to the Theory of Analytic Functionsof One Complex Variable. McGraw-Hill, 1979.
105
AFTERWORD
You survived! If you enjoyed what you saw in Math 113, you should consider Math 213a(graduate complex analysis) or Math 229x (analytic number theory). Math 113 providesadequate (dare I say ample) background for both in most years. 213a deals with complexanalysis, but on a graduate level: you’ll revisit some of the topological basics, and learnsome nice and beautiful mathematics like infinite product expansions and more on ellipticand meromorphic functions. Math 229x is number theory using complex analysis: you’llbegin with things like the prime number theorem and move on quickly. These courses canvary a bit, so do take our recommendation with caution.
If you aren’t pursuing further work in mathematics, we hope that you enjoyed the courseand remember how nice complex analysis is.
Thanks for being a great class, in terms of problem sets and everything else. Keep intouch with Andy or any of us CA’s if you’d like, and have a great summer!
Felix and Anirudha
106
