complex analysis. function theory

8/18/2019 Complex Analysis. Function Theory

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Complex AnalysisMathematics 113. Analysis I: Complex Function Theory


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Complex Analysis

Mathematics 113. Analysis I: Complex Function Theory

For the students of Math 113

Harvard University, Spring 2013


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FOREWORD

Welcome to Mathematics 113: Complex Function Theory! These are the compiled lecture

notes 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

http://localhost/var/www/apps/conversion/tmp/scratch_6/[email protected]://localhost/var/www/apps/conversion/tmp/scratch_6/[email protected]://localhost/var/www/apps/conversion/tmp/scratch_6/[email protected]://localhost/var/www/apps/conversion/tmp/scratch_6/[email protected]://math.harvard.edu/~acotton/math113.htmlhttp://math.harvard.edu/~acotton/math113.htmlhttp://math.harvard.edu/~acotton/math113.htmlhttp://math.harvard.edu/~acotton/math113.htmlhttp://courses.fas.harvard.edu/0405http://www.flickr.com/photos/syntopia/6811008466/sizes/o/in/photostream/http://www.flickr.com/photos/syntopia/6811008466/sizes/o/in/photostream/http://www.flickr.com/photos/syntopia/6811008466/sizes/o/in/photostream/http://courses.fas.harvard.edu/0405http://math.harvard.edu/~acotton/math113.htmlhttp://math.harvard.edu/~acotton/math113.htmlhttp://localhost/var/www/apps/conversion/tmp/scratch_6/[email protected]://localhost/var/www/apps/conversion/tmp/scratch_6/[email protected]://localhost/var/www/apps/conversion/tmp/scratch_6/[email protected]


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TABLE OF CONTENTS

The contents of this compilation are arranged by lecture, and generally have the topics

covered in a linear manner. Additional notes are provided in the appendices.

Foreword . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2 Riemann Sphere, Complex-Differentiability, 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

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15 Möbius Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

16 Automorphisms of D and H . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

17 Schwarz-Christoffel 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

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CHAPTER 1

INTRODUCTION

Consider R but throw in i = √ −1 to get C = {a + bi : a, b ∈ R}. We will be looking atfunctions f : C → C which are complex-differentiable . Complex-differentiable, or holomor-phic , functions are quite a bit different from real-differentiable 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 ∈ 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 ∈ C has a multiplicative inverse.Proof. In fact, z−1 := aa2+b2 − ( ba2+b2 )i works.

For z ∈ C, z = a + bi, we define the complex conjugate z := a − bi = a + b(−i).

Claim. z + w = z + w, z w = zw for z , w ∈ C.Proof. Write it out.

We also define the norm |z| := √ a2 + b2 = zz. We can deduce a few properties of thenorm:

Claim. (multiplicative property) |z||w| = |zw| for z , w ∈ C.Proof. It suffices 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 in√

1 + i,etc.)? The Fundamental Theorem of Algebra tells us we’re done as soon as we throw in i;

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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 θ + cos 2θ + ... + cos(n − 1)θ. For z = cos θ + i sin θ,1 + z + ... + zn−1 = (n−1k=0 cos kθ + i sin kθ). Now multiply by z −

1 = (cos θ

−1) + i(sin θ)

to get(cos nθ − 1) + i sin nθ

(cos θ − 1) + i sin θ = (cos nθ − 1)(cos θ − 1) − sin nθ sin θ

(cos θ − 1)2 + sin2 θ .

Son−1k=0

cos kθ = (cos nθ − 1)(cos θ − 1) − i sin nθ sin θ

2 − 2cos θ .

This isn’t the best formula, but we’re happy with it.

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CHAPTER 2

RIEMANN SPHERE, COMPLEX-DIFFERENTIABILITY,

AND CONVERGENCE

Stereographic projection. Consider a sphere S 2 := {(u,v,w) : u2 + v2 + w2 = 1} in R3.We have a bijection ϕ : S 2 − N → C, where N is the north pole (point at ∞), given bystereographic projection. Indeed, we define the Riemann sphere Ĉ := C ∪{∞} ∼= S 2.

Diagram 1: Stereographic projection. Source: Wikipedia

Given (u,v,w) ∈ S 2 − N and (x, y) ∈ R2, the bijection is given by

ϕ : (u,v,w) →

u

1 − w , v

1 − w

, ϕ−1 : (x, y) →

2x

x2 + y2 + 1,

2y

x2 + y2 + 1, x2 + y2 − 1x2 + y2 + 1

.

Claim. Any circle on S 2

maps to a circle or line in C under ϕ and vice-versa.Proof. I hope you paid attention in lecture.

Now consider the map (u,v,w) → (u, −v, −w), which swaps 0 ∈ C with S , the southpole, and ∞ ∈ Ĉ with N , the north pole. What does it do to z = x + iy? We see thatz → u−iv1+w .

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Claim. This is 1z .Proof. Trivial.

In fact, 1

z is just rotating the sphere around.

Claim. The inversion 1z sends (circles and lines) to (circle and lines).

If we have a function f : C → C and we think it’s “nice” at ∞, how do we quantifythat? We simply use the inversion 1z . In particular, we look at f (

1z ). Suppose f (∞) = ∞;

then 1f ( 1z )

sends 0 to 0.

For example, consider a polynomial of degree n, f (z) = c0 + ... + cnzn and cn = 0. Take

1

f ( 1z ) =

1

c0 + ... + cnz−n =

zn

c0zn + ... + cn,

which is well-defined in a neighborhood of 0 ∈ C and clearly, sends 0 → 0.

Differentiation. Lets talk about something we all like, differentiation. :-)

Definition. Given f (z) : C ⊃open U → C, z ∈ U , we say f is complex-differentiable (or holomorphic ) at z if

f (z) := limh→0

f (z + h) − f (z)h

exists for h ∈ C.

Proposition. If f , g are holomorphic, then f g and f + g is as well, with (f g) = f g + fg

and (f + g) = f + g.Proof. Also trivial.

Example. Consider complex polynomials in z, which are holomorphic: f (z) =

ckzk.Then f (z) exists and f (z) =

n−1(k + 1)ck+1z

k.

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-differentiable functions. Thereare “different severities” of being real differentiable. Let f : R ⊃open U → R:

1. real differentiable: f (x) exists ∀x in domain2. C 1: f (x) exists and is continuous3. C k: f, f ,...,f (k) exists and are continuous

4. C ∞: (smooth) all derivatives exists and are continuous5. real-analytic: f (x) =

∞ ckxk is a convergent power seriesThe punchline is that homolorphic functions are always C ∞ and analytic! (This is some-thing that should really be appreciated.) Indeed, we oftentimes interchange holomorphic,C ∞, and analytic in the complex world.

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To show this, we will go back to convergence. Consider a sequence of functions {f k(z)}defined on E ⊂ C that is containable in a compact set. We say that it is pointwise conver-gence if limk→∞ f k(z) exists ∀z; the pointwise limit is defined as f (z).

Suppose f k(z) is continuous. It’s uniformly convergent if ∀, ∃N : ∀n ≥ N : |f n(z) −f (z)| ≤ , ∀z ∈ E .

Lemma. If {f k(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) =∞

0 ckzk. We need a condition on |ck| for this to converge.

Definition. Given a sequence {ak}∞1 , we define the limit supremum

lim supk→∞

ak = limn→∞

supk≥n

ak

= lim

k→∞ak

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 < ∞, or L = ∞, then:1. If L = 0, then f (z) =

ckzk converges for all z .

2. If 0 < L R, and R iscalled the radius of convergence . (We do not know for |z| = R.)

3. If L = ∞, ckzk diverges ∀z = 0.Proof. L = 0: ∀ > 0, ∃N : ∀k > N, |ck|1/k < . Let = 1z |z|. Then |ck|1/k < 12 |z|,

Re|ck| < 12k 1|z|k. We want |ck|1/k|z| < 12 , so |ck||z|k ≤ 12k

so

ckzk converges. We can do

this with the M -test, by considering partial sums, etc.

0 < L < ∞: lim|ck|1/k

= L. Let |z| = R(1 − 2δ ), then lim|z||ck|1/k

= 1 − 2δ for R = 1

L .So ∃N : |z||ck|1/k N , and

ckzk converges because |ck|zk 1+δ and |ckzk| > 1. The last condition is proved in the same manner.

Example. ∞

0 zk = 11−z . Deduce this from the theorem above.

Example. ∞

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) =∞

0 ckzk converges for |z| < R. Then:

1. ∞1 kckz

k−1 converges for |z| < R.2. f (z) exists and equals

∞1 kckzk−1.Proof. (1) It suffices to check lim|ck|1/k ≤ 1R =⇒ lim(k + 1)1/l|ck|1/k ≤ 1R . Note that

lim(k + 1)1/k|ck|1/k = lim(k + 1)1/klim|ck|1/k.(2) R = ∞: subtract and show that limits → 0.

f (z + h) − f (z)h

−

kckzk−1 =

1

h

ck[(z + h)

k − zk] −

kckzk−1

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=

ck[1

h(z + h)k − 1

hzk − kzk−1].

Recall the binomial theorem,

(z + h)k =

kl

hlzk−l,

which we can use to rewrite the expression above as

=

∞0

ck[k

l=1

k

l

zk−lhl−1 − kzk−1] =

∞0

k2

k

l

zk−lhl−1

= h∞0

k2

k

l

zk−lhl−2.

We also have

klzk−lhl−2 ≤

k

l|

z

|k−1 = (

|z

|+ 1)k,

so

ck(|z| + 1)k converges because R = ∞ for

ckzk.

R < ∞: Let |z| = R − 2δ, |h| < δ ; then |z + h| < R − δ . Then

f (z + h) − f (z)h

−

kckzk−1 = h

∞0

ck

k2

k

l

zk−lhl−2.

Write k

l

=

k(k − 1)...(k − l + 1)l!

.

For l ≥ 2 we can bound

kl

≤ k2

k

l−2

. So

2

kl

zk−lhl−2

≤ k2|z|2 (|z| + |h|)kand the equation above translates into

= h

|z|2

k2ck(|z| + |h|)k,

which converges and → 0 because of the h in the front.

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CHAPTER 3

POWER SERIES AND CAUCHY-RIEMANN EQUATIONS

Power series. Last time we were dealing with power series, f (z) =∞

k=0 ckzk. We defined

the radius of convergence R = 1L , where L = lim|ck|1/k and either L = 0, 0 < L 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) =

ckzk is a convergent power series and f (zn) = 0 for a sequence

{zn}∞n=1 with zn → 0, zn = 0, then ck = 0 ∀k and Re(f (z)) ≡ 0.Proof. c0 = f (0) = limn→∞ f (zn) = 0. Now form g1(z) =

f (z)z = c1 + c2z + c3z

2 + ...(exercise: show that this holds for some radius of convergence). Check c1 = g1(0) =

limn→∞ g1(zn) = limn→∞f (z)

z = 0, and induct on gi.

Proposition. If f (z) =

akzk, g(z) =

bkzk and these agree on some set accumu-lating at 0, then ak = bk ∀k, i.e. f (z) = g(z).

Proof. Consider

(ak − bk)zk and apply the lemma. Show that lim|ak|1/k ≤ L andlim|bk|1/k ≤ L =⇒ lim|ak − bk|1/k ≤ L.

Note: To center the power series at w ∈ C, consider ck(z − w)k, which shifts the centerof the power series from 0 ∈ C to w and maintains the radius of convergence at R.

Complex-differentiability. We refer to complex-differentiability as either holomorphic or 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.

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Definition. f : C → C is holomorphic at z ∈ C if

limh→0

f (z + h) − f (z)h

exists for h ∈ C. Equivalently, ∃ a number f (z) :

0 = limh→0

f (z + h) − f (z)h

− f (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 proof follows 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)

x2

+y2 , z

= 0, and 0 for z = 0.

Show that this is not differentiable 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.

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An equivalent (geometric) formulation of the CR equations. Consider the Jacobianof f : R2 → R2:

J f := ∂u

∂x∂u∂y

∂v∂x

∂v∂y

and the rotation matrices

cos θ − sin θsin θ cos θ

.

So multiplication by i is multiplication by the matrix (aside: see the relation to complex structure )

0 −11 0

.

The equivalent form is that J f I = I J f ; e.g. the Jacobian matrix commutes with rotationby π/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) = n

k=0 ckzk. Look

at complex-valued polynomials of real variables x, y:

p(x, y) =n

m=0

k+l=m,

k,l≥0

ck,lxkyl (3.3)

with ck,l = ∂p∂ kx∂ ly (0, 0), and take a “new basis” as z = x + iy and z = x − iy. Write the

polynomial in z , z :

p(z, z) = m=0

k+l=m,

k,l≥0

c̃k,lzkzl. (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) = lz kzl−1,

∂

∂z(zkzl) = kz k−1zl.

Proof. We simply check:

∂

∂z(zkzl) =

1

2[kzk−1zl + lzkzl−1] +

i

2[kizk−1zl + (−i)lzkzl−1] = lz kzl−1,

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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) =

c̃k,lzkzl is holomorphic iff c̃k,l = 0 for l > 0 (e.g.

no zs). Alternatively, p(x, y) =

ck,lxkyl is holomorphic iff it can be written as

p(z) =

ckzk.

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 2uxu + 2vxv = 0 and2uyu + 2vyv = 0. Applying the CR equations, we get 2uxv − 2vxu = 0. Adding equationsgives 2ux(u

2 + v2) = 0, and similarly 2vx(u2 + v2) = 0. So either u2 + v2 = 0 (f = 0), or all

partials = 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.

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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, ddx ex = ex,... or (2) ex =

xn

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

differential equation has an ez solution). So define ez = ex(cos y + i sin y), for z =

x + iy.

(2) Let

f (z) =

∞n=0

zn

n!.

The radius of convergence is ∞, since limn→∞( 1n! )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 (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− z2

2! +z4

4! +...)+i(z− z3

3! +z5

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, cos0 = 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−iz2i

.

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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+iθ+i(2πn), so the complex log is multi-valued. Write z = reiθ = r(cos θ+i sin θ).

We define the principal branch of log z as follows. Consider C − R≤0. log z has imag-inary part in (−π, π). Write log z = log |z| + iArg(z) for Arg(z) ∈ (−π, π); this is ourdefinition of log z for the principal branch, which has range −π


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Lemma. (complex chain rule) Let g(t) = f (γ (t)), f (z) holomorphic. Then g(t) =γ (t)f (γ (t)).

Proof. As usual, with limits or whatnot.

Example. Compute S 1

zkdz

for k ∈ Z, and S 1 the unit circle in C oriented counterclockwise.Explanation. Let γ (t) = eit = cos t + i sin t. By chain rule, γ (t) = ieit. Then

S 1

zkdz =

2π0

eiktieitdt = i

2π0

ei(k+1)tdt.

If k = −1, this is

S

1

1

zdz = i

2π

0

1dt = 2πi.

If k = 1, this is 2π0

ei(k+1)tdt = ei(k+1)t

i(k + 1)

2π

0

= 0.

So z k for k ∈ Z, k = −1 has an antiderivative in C − {0}: zk+1k+1 . 1/z does not; we’d wantlog z, but it’s multivalued.

Proposition. If f (z) holomorphic with F (z) holomorphic and F (z) = f (z), then 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 ∃F (z) : F (z) = f (z). Along the way, we’ll prove the closed curve theorem .

For motivation, we recall Green’s theorem :

ba

(P, Q) · r(t)dt =

D

∂Q

∂x − ∂ P

∂y

dxdy

for r(t) : [a, b] → R2 a closed curve around a region D. This requires that Qx, P y becontinuous (or something like that). The theorem we’ll prove does not need this requirement.

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CHAPTER 5

THE CLOSED CURVE THEOREM AND CAUCHY’S

INTEGRAL FORMULA

Today we’ll be talking about the closed curve theorem and Cauchy’s integral theorem.Recall from last lecture that, for a parameterization γ of the contour C , we had

C

f (z)dz =

f (γ (t))γ (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 (z) = f (z), then

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 γ , consider splitting into real and imaginary parts:

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

γ (t) = a(t) + ib(t) f (γ (t))γ (t)dt =

(u + iv)(ȧ + iḃ)dt =

[(uȧ − vḃ) + i(uḃ + vȧ)]dt

= (u, −v) · (ȧ, ḃ)dt + i (v, u) · (ȧ,

ḃ)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

D

− ∂v

∂x − ∂ u

∂y

dxdy + i

D

∂u

∂x − ∂ v

∂y

dxdy = 0,

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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

D curl(f )dA +

D div(f )dA,

where f is thought of as a vector field on R2 induced by f . The equivalent formulation of the 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 + 12 bz

2, so F (z) = f (z). So we have

C f (z)dz = F (final) −

F (initial) = 0.

Theorem. (closed curve theorem) Suppose f (z) is holomorphic in an open disk D . Let C be any closed contour curve in D. Then

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

| T f (z)dz

|= C . Then

∃T i with boundary Γi

such that | Γi f (z)dz| ≥ c/4. Call this T (1) with boundary Γ(1). Keep subdividing to get(dropping the parentheses) T 1, T 2, T 3,... and Γ1, Γ2, Γ3,.... such that | Γk f (z)dz| ≥ c4k .Note that ∩∞k=1T k = {point}; call the point z0. f (z) is holomorphic at z0, so we have

limz→z0

f (z) − f (z0)z − z0 − f (z0) = 0.

Thusf (z) = f (z0) + f

(z0)(z − z0) + (z)(z − z0)with (z) → 0 as z → z0. Now consider

Γk

f (z)dz = Γk

[f (z0) + f (z0)(z

−z0) + (z)(z

−z0)]dz =

Γk

(z)(z

−z0)dz.

Given > 0, ∃N s.t. for k ≥ N , (z) < for z ∈ T k. Let the perimeter of T be L and theperimeter of T k be L2k . The max distance between two points in T

k would then be ≤ L2k .Thus

c

4k ≤

Γk(z)(z − z0)dz

≤ L2k L2k = L2

4k ,

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where the first term after the equality is length of Γk, the second is an upper bound on (z),and the third is an upper bound on |z−z0|. We had c = |

Γ

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 Γ, then

P 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 Ω ⊂ C), then ∃F (z) holomorphic on D (or Ω) s.t.F (z) = f (z).

Definition. An open region Ω ⊆ C is polygonally connected (in topology, connected) if ∀z, w ∈ Ω, ∃ piecewise linear curve in Ω connecting z, w. A region is simply polygonally connected (simply connected) if any closed polygonal curve is the boundary of a union of polygons in Ω.

Proof of antiderivative theorem. Let p ∈ D (or Ω ⊆ C). Let F (z) = z 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 Ω). We check that

F (z) − F (z0)z − z0 =

1

z − z0

zz0

f (z)dz.

Note F (z) − F (z0)z − z0 − f (z0) =

1z − z0 z

z0

(f (z) − f (z0))dz

because zz0

f (z0)dz = f (z0)

zz0

1dz = f (z0)(z − z0).

Then we have 1z − z0 z

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 (z0)| (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 Ω), then

C f (z)dz = 0 for C being any closed contour in D (or Ω).

Proof. Take F (z), the antiderivative of f (z), so that

C 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 = wf (w) z = w

for w ∈ C.

Claim. g(z) satisfies the closed curve theorem.

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Proof. g(z) is continuous because f (w) exists. The goal is to prove the above resultfor triangles for g (z); then it will follow ∃G(z) : G(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 ∈ Γ = ∂T , then we cut off small pieces {T i} so

Γi

f (z)dz = 0 (because they do notcontain w). The length of Γi 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 ∈ C, and f (z) holomorphic on the closed boundary C of theopen disk D. Then we have the formula

f (w) = 1

2πi

f (z)

z − w dz

Proof. Suppose C is a circle centered at w. The circle is parameterized by γ (t) = w +Reitfor 0 ≤ t ≤ 2π. We know from the closed curve theorem that

0 =

C

f (z) − f (w)z − w dz,

because the integrand is g(z) defined on an interior satisfying the closed curve theorem.Now split this up:

C

f (z)

z − w dz = f (w)

C

1

z − w dz,

C

1

z − w dz = 2π

0

iReitdt

Reit = 2πi

=

⇒f (w) =

1

2πi

f (z)

z − wdz.

Theorem. Suppose f (z) is holomorphic on the open disk Dk(0) of radius R around 0 ∈ C.Then there exists a convergent power series

∞k=0 ckz

k with lim|ck|1/k ≤ 1R such thatf (z) =

∞k=0 ckz

k.Proof. From Cauchy’s integral formula, we have

f (z) = 1

2πi

f (w)

w − z dw.

Let |z| < R̃ < R, and write

1w − z = 1w 11 − zw = 1w

1 + zw + z

2

w2 + ...

| zw | = |z|R̃


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= 1

2πi

C

f (w)

w dw +

1

2πi

C

f (w)

w2 dw

z +

1

2πi

C

f (w)

w3 dw

z2 + ...,

i.e.

C k = 12πi

̃C

f (w)wk + 1

dw.

We check this as follows:

|C k| ≤ 12π

2π R̃M 1

R̃k+1 = M/Rk,

where the terms of the right of the inequality come as follows: 2π R̃ is the length of R and

M is an upper bound on f in DR̃(0). So lim|C k|1/k ≤ lim M 1/k

R̃ = 1

R̃, and

f (k)(0)

k! =

1

2πi

C

f (w)

wk+1dw

for C containing 0.

Thus we have the result that f holomorphic =⇒ f analytic =⇒ f infinitely differentiable.

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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

2πi

C

f (w)

w − z dw,

for C a curve containing z . For f (w) =

ckwk, note that we can write

f (0) = 1

2πi

C

f (w)

w dw =

1

2πi

C

∞k=1

ckwk−1dw =

1

2πi

C

c0w

dw = c0.

Let’s use this formula to compute some stuff. The general method is to decompose a closed

contour (over which the integral is zero) into a sum of directed edges, then evaluating theintegral over the other edges to give us results.

Claim. ∞0

sin x

x dx =

π

2.

Proof. Let C R 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 eix

x :

0 =

Rr

eix

x dx +

C R

eiz

z dz +

−r−R

eix

x −

C r

eiz

z dz.

Now parameterize C r with z = reiθ for θ ∈ [0, π] and likewise with z = Reiθ to see

0 =

Rr

eix

x dx +

π0

eiReiθ

Reiθ iReiθdθ +

−r−R

eix

x dx −

π0

eireiθ

reiθ ireiθdθ.

=

Rr

eix

x dx +

π0

ieiReiθ

dθ +

−r−R

eix

x dx −

π0

ieireiθ

dθ.

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But note that the second integral → 0 as R → ∞ and the fourth integral → πi as r → 0(show this yourself), so we’re done as sin xx (the imaginary part of the remaining two inte-grals) is symmetric over the y-axis and its integral from 0 to ∞ would then be equal to theimaginary part of our answer divided by 2. Namely, we have

∞0

sin x

x dx =

π

2 as desired.

Claim (added by Felix, for more practice)

∞0

sin(x2)dx =

∞0

cos(x2)dx =

√ 2π

4 .

Proof. We first show that ∞−∞ e

−x2dx =√

π as follows:

∞−∞

e−x2

dx

2=

∞−∞

∞−∞

e−(y2+z2)dydz =

2π0

∞0

re−r2

drdθ = π,

with y = r cos θ and z = r sin θ.

Using what we showed above, we integrate f (z) = e−z2

on the contour defined byC = {Reit : 0 < t < π/4}. Observe that we can bound the integral by

0 ≤

C

f (z)dz

≤ π/4

0

e−R2(cos2(t)−sin2(t))Rdt =

π/40

e−R2 cos(2t)Rdt,

but also (from brute force bounding) we know that R π/4

0 e−R

2 cos(2t)dt → 0 as R → ∞(∗). This is good, because we can invoke Cauchy’s theorem to see that

C f (z) for C =

{reit : 0 < r < R, 0 < t < π/4} = γ 1 + γ 2 + γ 3 (where γ 1 is the path from 0 to R, γ 2 is theeighths-circle, and γ 3 is the line from Re

iπ/4 back to 0), a closed boundary, is zero. LettingR → ∞, we saw from above that γ 2 f (z) = 0, so what we have left is (I’ll drop the f (z)’sto make writing easier)

γ 1

+

γ 3

= ∞

0

e−z2 +

γ 3

= √ π/2 + γ 3

= 0 ⇒ √ π/2 = − γ 3

.

So to finish, we evaluate − γ 3 :−

γ 3

=

R0

e−(eiπ/4t)2eiπ/4dt = eiπ/4

R0

e−it2

= eiπ/4 R

0

cos(t2) − i sin(t2)dt.

Thus we see that

√ π/2 = eiπ/4

∞0

cos(t2)−i sin(t2)dt ⇒ ∞

0

cos(t2)−i sin(t2)dt =√

π/2√ 2/2 +

√ 2/2i

=

√ 2π

4 −

√ 2π

4 i.

Equating real and imaginary parts, we see that ∞0 sin(x2)dx = ∞0 cos(x2)dx = √ 2π4 , as

desired.(∗) Brute force bounding: For more exercise, we’ll explicitly show that

R

π/40

e−R2 cos(2t)dt → 0

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as R → ∞ here. We have

R π/4

0

e−R2 cos(2t)dt = R

π/4

0

e−R2 sin(2t)dt = R

π/6

0

e−R2 sin(2t)dt + R

π/4

π/6

e−R2 sin(2t)dt.

Then, since we can bound sin x ≥ (1/2)x for 0 ≤ x ≤ 1/3 ( ddx sin x ≥ 1/2 for x ≤ 1/3) andsin x ≤ 1/2 for π/6 ≤ π/4, we see that

R

π/60

e−R2 sin(2t)dt + R

π/4π/6

e−R2 sin(2t)dt ≤

π/60

Re−R2tdt +

π/4π/6

Re−R2/2dt

= π

12Re−R

2/2 + 1

R(1 − e−πR

2

6 ) → 0,as R → ∞. Integrals are fun stuff.

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 C 0(R) be the circle of radois R about 0. Then

|f (a) − f (b)| = 12π

C 0(R)

f (z)

z − a dz −

C 0(R)

f (z)

z − b dz

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

2π

C 0(R)

f (z)

z − a dz −

C 0(R)

f (z)

z − b dz = 12π

C 0(R)

f (z)(b − a)(z − a)(z − b) dz

≤ 12π 2πRM |b − a|14 R

2

= 4M |b − a|

R → 0

as R → ∞.

Let’s look at polynomials P (z) =n

k=0 ckzk, cn = 0. What can we say about them?

Claim. limz→∞ P (z) = ∞ in the sense that ∀M, ∃r : ∀|z| > R, |P (z)| > M .Proof. Should be obvious.

Theorem. (Fundamental Theorem of Algebra) Given a nonconstant polynomial P (z) =nk=0 ckz

k, cn = 0, n ≥ 1, ∃w ∈ C : p(w) = 0.Proof. Suppose not. Then 1P (z) =: f (z) is a bounded entire function. Choose R such

that1

R >

n−1k=0

|ck||cn|

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and R > 1 so that for |z| > R we have1

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| → ∞. So f (z) is bounded by C for |z| > R, and it’s similarlybounded on any compact region D0(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 (α)

z−α z = αf (α) z = α

for any α. Notice that g(z) is holomorphic, since we saw

∃G(z) that is holomorphic with

G(z) = g(z), and we showed that holomorphic implies C ∞ 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| → ∞ we see that the growth rate(growth order ≥ N ) would not match with f (z)−f (α)z−a or f (α) (growth order N − 1).

(Also, as said in lecture, we can see the last statement as follows: look inside a largeball, then outside the large ball. Inside, it’s bounded by some constant; outside, because of the 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 − α) is apolynomial of degree ≤ N . This completes the inductive step.

Uniqueness theorem. It turns out that holomorphic functions are determined up to

their values in a certain region; this is the content of the uniqueness theorem.

Definition. A region Ω 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 ∈ D : ∃ 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 ∈ B, then ∃ a disk containing z with only finitelymany such w’s. Let δ


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f : Ĉ → Ĉ entire, then f : ∞ → a for a finite implies that f (z) is bounded, i.e. a constant.If f : ∞ → ∞, then f (z) is a polynomial.

Proposition. If f (z) is entire and f (z) → ∞ as z → ∞, 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 finitelymany 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) = ∞ =⇒ |g(z)| ≥ 1|z|m , where m =

mi.

So 1g(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 Dα(R), then ∀r : 0 < r < R, wehave

f (α) = 1

2π

2π0

f (α + reiθ)dθ.

Corollary. This is true as well for u = Re(f ), v = Im(f ) in place of f .Proof. Use the integral formula.

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CHAPTER 7

MEAN VALUE THEOREM AND MAXIMUM MODULUS

PRINCIPLE

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

2π

2π0

f (z + reiθ)dθ.

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 DR(z), then its absolute value |f | hasno maximum in DR(z).

To show this, suppose that w = f (z) is any value in DR(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) = 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

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in the textbook.

Corollary. If f is holomorphic in DR(z), then on Dr(z), r < R, the maximum value

of |f (z)| on Dr(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 z0; if z0 ∈ ∂E , we’re done. Else if z0 is an interior point, then |f (z0)| is alsothe maximum of |f (z)| in a disk |z − z0| < δ contained in E . But this is not possible unlessf (z) is constant in the component of the interior of E which contains z0. 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 is

always attained at a boundary point.

Proposition (“anti-calculus”) If |f (z)| achieves its max value on Dr(z) at w, then f (w) =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 ∀ > 0, ∃δ > 0 :f (Bδ(z)) ⊆ B(f (z)).

Sketch of another proof. If f (z) = 0, then

J f =

ux uyvx vy

is nonsingular. Then det J f = uxvy − uyvx = u2x + v2x = |f (z)|2 = 0. The inverse functiontheorem gives the result.

If f (z) = 0, then consider f (Bδ(0)) = Bf (δ)(0). Write f (z) = zk(ck + ck+1z + ...). Whenf has a zero of order k at 0, i.e. f (0) = f (0) = ... = f (k−1)(0) = 0, it turns out thatf (z) = [g(z)]k, g(z) = Az + ... i.e. g(0) = 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 ∀ p, q ∈ 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:

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Proposition. If U ⊆ C is open, then U is connected iff 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 Ω 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 γ we know fromCauchy’s integral theorem that

γ

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.

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CHAPTER 8

GENERALIZED CLOSED CURVE THEOREM AND

MORERA’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 connected

subset) in C and Ĉ − D is connected, then the closed curve theorem holds, i.e. C

f (z)dz = 0

for f holomorphic on D and C a contour curve in D .

Definition. A set S in Ĉ ∼= S 2 is connected if A, B open in Ĉ with A ∪ B ⊃ S andA ∩ B = ∅.The method for proving the general closed curve theorem is:

1. Ĉ − 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 ) := {ni=1 aiT i}, for T i a triangle in U , ai ∈ Z, and any n.• E (U ) := {ni=1 aiE i}, for E i an edge in U (recall that an edge means a pair of points

{P, Q} with P = Q).• P (U ) := {

ni=1 aiP i}, for P i a point in U .

• ∂T := E 1 + E 2 + E 3 for T a triangle.• ∂E = Q − P ; ∂ : E (U ) → P (U ) and ∂ : T (U ) → E (U ).• e ∈ E (U ) is closed if ∂ e = 0, e.g. e = E 1 + ... + E n is closed iff ±E i form loops.

Definition. e1 is homologous to e2, denoted e1 ∼ e2, if ∃t : ∂t = e1 − e2. This is anequivalence relation.

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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 . Then e

f (z)dz =

ai

E i

f (z)dz.

Define F (z) = z

z0f (z)dz. This makes sense because

e1

f (z)dz =

e2

f (z)dz.

if e1 ∼ e2, since then e1−e2

f (z)dz =

bj

Γ

f (z)dz

where Γ is the boundary of the triangle and the integral is zero. So the antiderivativetheorem implies the closed curve theorem as before:

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 iff U is connected.

Claim. Let D be a region in C ⊂ Ĉ. If Ĉ − D is connected, then D is p.s.c.

Theorem. (Morera) If f (z) is continuous on a region D and

Γf (z)dz = 0 for Γ 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→∞ f n = f is holomorphic.2. Showing that some functions defined by sums or integrals are holomorphic, e.g. the

Riemann zeta function and the gamma function :

ζ (z) :=

∞n=1

1

nz

Γ(z) :=

∞0

e−ttz−1dt

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3. Showing functions that are continuous and holomorphic on all but some set (e.g. somepoint) are holomorphic.

Proposition. Suppose {f n} is holomorphic in an open set D, and f n → f uniformly oncompact subsets of D . Then f = limn→∞ f n is holomorphic.

Proof. It suffices to work in B(α) ⊂ D. B/2(α) is compact, so by assumption f n → f is uniform here. Note that f n → f uniformly implies f is continuous. Also

Γ

f (z)dz =

Γ

limn→∞

f n(z)dz = limn→∞

Γ

f n(z)dz = 0

(Uniform convergence implies that we can swap

and lim, as we all know from elementaryanalysis.) By Morera’s theorem, f is holomorphic.

The Riemann zeta function. The zeta function ζ (z) is defined as

ζ (z) =

∞n=1

1nz ,

for Re(z) > 1. We claim that ζ (z) is convergent for Re(z) > 1, and indeed, is uniformlyconvergent for Re(z) > R > 1. (Hence by Morera’s theorem ζ (z) is holomorphic.)

Proof of claim. First note that 1nz = e−(log n)z. We first show uniform convergence. Let

f N (z) :=N

n=1

1

nz =

N n=1

e−(log n)z.

We bound ∞

n=N +1 |e−(log n)z| for the whole region Re(z) ≥ R > 1:

|e−(log n)z

|=

|e− log neRe(z)

| ≤R

|e− log n

|So we’re left with

∞n=N +1

1nR , R > 1. The integral test enures convergence, telling us that

the tail end is small.

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C 2 − C 1, we see that C 2−C 1

f (z)dz = 0

for f (z) holomorphic on C∗. An image of C 2 (the outside counterclockwise contour) and C 1(the inside counterclockwise contour) is as follows:

Diagram 1: C 1 and C 2 in C∗.

Now let’s pick off with Morera’s theorem, which as you recall states

Theorem. (Morera) If f is continuous on an open set D and

Γ f (z)dz = 0 for Γ be-ing the boundary of a triangle in D, then f (z) is holomorphic.

Corollary. If f n(z) is holomorphic on D open and f n → 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

ζ (z) =

∞n=1

1

nz.

We claim that ζ (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

Γ(z) =

∞0

e−ttz−1dt.

We claim that Γ(z) is holomorphic for Re(z) > 0.Remark. Γ(z) is a natural extension of the usual factorial function in that Γ(n + 1) = n!.

Show this yourself using integration by parts and induction.Proof of the claim. We show that

∞0

e−ttzdt is holomorphic in Re(z) > −1. To thatextent we work in −1 < R1 ≤ Re(z) ≤ R2; note that any compact subset of {z : Re(z) > −1}is contained in one of these sets. Define

f n(z) := 1

1/n

e−ttzdt + n

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,

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the last term of which has norm 1. Then

|f n(z)/2 − f (z)/2| ≤ ∞

n

e−ttR2dt,

which converges for n = 1. So ∀ > 0, ∃N : ∀n > N , ∞n

e−ttR2dt < /2.

For t ≤ 1, |e−ttz| = e−ttRe(z) ≤ e−ttR1 and 1

1/n

e−ttzdt − 1

0

e−ttzdt

= 1/n

0

e−ttzdt

≤ 1/n

0

e−ttR1dt

for R1 > −1. Let s = 1/t and ds = − 1t2 dt, so that we get ∞

n e−1/ss−2−R1ds. We need

−2

−R1 <

−1, i.e.

−1 < R1. Thus the sequence converge uniformly, so

∀ > 0,

∃N :

∀n >

N, ∞n e−1/ss−2−R1ds < /2.

Corollary. If f is continuous on D and f is holomorphic on D except at α ∈ D, thenf is holomorphic on D .

Proof. Let Γ = ∂T . Then Γ

f (z)dz = 0

if 0 /∈ T . If α ∈ T , we split the contour so that α is located at the corners of the triangle.If α is at a corner, then

0 = liml→∞

Γl

f (z)dz =

Γ

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 of the different types of singularities and how they pertain to classes of functions is importantfor complex anaysis in general.

Definition. Given α ∈ C, a deleted neighborhood D of α is a neighborhood −{α}, e.g.{z : 0


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The following proposition allows us to recognize if something has a removable singularity.

Proposition. If lim

z→αf (z)(z

−α)

exists and is equal to 0, then f (z) has a removable singularity at α.Proof. Define h(z) as

h(z) =

f (z)(z − α) z = α0 z = α.

This is continuous, and holomorphic except at α. So h(z) is holomorphic in a neighborhood

of α by the corollary to Morera’s theorem. Moreover, we see that h(α) = 0, so g(z) = h(z)z−αis holomorphic and equal to f (z) away from α. (If h(α) = 0 and h is holomorphic, thenh(z)z−α 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 α and ∃n :lim

z→α f (z)(z − α)n = 0.

Then letting k = 1 be the least such n, f (z) has a pole of order k at α.Proof. Again, let

h(z) =

f (z)(z − α)k+1 z = α0 z = α.

h(z) is continuous on a neighborhood and holomorphic on a deleted neighborhood, so it’sholomorphic on a neighborhood of α again by the corollary to Morera’s theorem. h(α) = 0

so g(z) = h(z)z−α is holomorphic in D∪{α} and g(z) = f (z)(z−α)k on D. Then f (z) = g(z)(z−α)k .(Note limz

→α g(z)

= 0 by assumption; also notice it exists).

For essential singularities, we’ve shown that it must be the case that limz→α f (z)(z − α)ndoes 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 < C 1 1|z|N + C 2 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 α and D is anydeleted neighborhood of α, then f (D) =

{f (z) : z

∈ D

} is dense in C (i.e for any > 0,

w ∈ C, B(w) ∩ f (D) = ∅).Proof. Suppose not. Then ∃B(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 α, and bounded there.Thus g(z) = 1f (z)−w is holomorphic on a neighborhood of α.

Now consider 1g(z) = f (z) − w, so that f (z) = wg(z)+1g(z) is a ratio of two holomorphicfunctions; we see that the singularity must be removable, or f (z) must have a pole.

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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 ∈ C.We claim that f (B(0) − {0}) is C∗.To see this, note that f (z) is the composition of 1z and then ez. Under 1z , B(0) − {0}gets inverted to fill C outside the ball. ez is 2πi-periodic, so we can draw horizontal linesin the complex plane at π + 2πn, n ∈ 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 outsideB0(1/) so that f (B(0) − {0}) = C∗.

Diagram 2: The mappings 1z : B(0) − {0} → C − (B(0) − {0}) and ez : {z : π ≤ Im (z) ≤ 3π} → C − R≥0.

Laurent expansions. Simply put, Laurent expansions are Taylor series of the formf (z) =

∞k=−∞ ckz

k.

Theorem. If f (z) is holomorphic on the annulus A(R1, R2) := {z : R1


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Example. The following Laurent series expansion converges on A(0, ∞):

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 ( 11+z ) = − 1(1+z)2 and 11+z = 1 − z + z2 − z3 + .... We can do the same

near z = −1.

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CHAPTER 10

MEROMORPHIC FUNCTIONS AND RESIDUES

Last time, we introduced the Laurent series for f (z) as f (z) = ∞

−∞ ck(z − α)k. This isdefined on Aα(R1, R2), the annular region centered at α.

Theorem. If f (z) is holomorphic on Aα(R1, R2), then f (z) = ∞

k=−∞ ck(z − α)k andthe series converges on Aα(R1, R2).

Proof. Do this yourself by considering the function

g(w) =

f (w)−f (z)

w−z w = zf (w) w = z

and using the closed curve theorem.

Remark. Laurent series are unique to every function. We define the principal part of f (z) =

near α as the sum

−1k=−∞ ck(z − α)k and observe the following:

(1) α is a removable singularity iff ck = 0 for k


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Theorem. Let f be meromorphic in C and at infinity, and suppose that limz→∞ f (z) = ∞for some w ∈ C. (The limit means ∀M, ∃R : |f (z)| > M when |z| > R.) Then f (z) is arational function.

Theorem. If f : Ĉ → Ĉ 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 effciently. Given a holomorphic function f (z)in a deleted neighborhood of α, by the Laurent expansion we have f (z) =

∞−∞ ck(z − α)k.

We define the residue of f (z) at the point α, Res(f ; α), as

Res(f ; α) = c−1.

From Cauchy’s integral formula, we notice that

1

2πi

C R(α)

f (w)dw = c−

1.

Also recall the expression for ck:

ck = 1

2πi

C R(0)

f (w)

wk+1dw.

We see from this that the “limsup stuff” shows that Laurent series are uniformly convergenton A(r1, r2) by comparison with geometric series.On a ball of radius , we can compute the residue of f at α by using the formula:

Res(f ; α) = c−1 = 1

2πi

C (0)

f (w − α)dw = 12πi

C (α)

f (w)dw.

There are a few ways to compute residues in general:(0) Compute the integral

C (α)

f (w)dw.

(1) Compute the Laurent series for f centered at α.(2) If f has a simple pole, then

c−1 = limz→α(

z − α)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 ; α) = A(α)B(α) .

We finish by working out an example. Consider ∞−∞

1

1 + x2dx.

We can compute this by first taking a semicircle contour of radius R, for which there is onepole contained inside (α = i). If we compute the residue at this pole, we can use the residue theorem to get the value of the integral (this is left to the reader as an exercise).

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CHAPTER 11

WINDING NUMBERS AND CAUCHY’S INTEGRAL

THEOREM

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 α, wehave

Res(f ; α) = 1

2πi

C (α)

f (z)dz = c−1,

where c−1 appears in the Laurent expansion for f around α, f = ∞

−∞ cnzn. Note that f

is holomorphic in C (α) except for the point at α.

Example. Res( 11+z2 ; 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 = −i2 .Method 2: If f (z) has a simple pole at α, i.e. if f (z) = A(z)B(z) for A(α) = 0, B(α) =

0, B(α) = 0, then Res(f ; α) = A(α)B(α) . Set Res( 11+z2 ; i) = 12i = −i/2.

An application of this is as follows. Consider the integral ∞−∞

1

1 + x2dx = lim

R→∞arctan(R) − lim

R→∞arctan(−R) = π/2 − (−π/2) = π.

We can also do this by contour integration over the contours C 1, C 2, C 3, where C 1 = [−R, R],C 2 is the CCW semicircle connecting R to −R, and C 3 is the CCW circle omitting the pointat i. Then

C 1+C 2−C 3f (z)dz = 0

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by the closed curve theorem, and by the estimation lemma we have

C 2

1

1 + z2dz ≤ πR ·

1

R2 =

π

R → 0

as R → ∞. Also, C 3

1

1 + z2dz = 2πiRes(

1

1 + z2; i) = 2πi(−i/2) = π.

So

limR→∞

R−R

1

1 + x2dx = π

after substituting for the values of the integrals over C 1, C 2, C 3.

Winding numbers. These intuitive give the number of times a path loops around apoint, and are also dealt with in topology.

Definition. Let γ : [0, 1] → C be a contour curve with γ (t) = α ∀t. Then the wind-ing number of γ about α, denoted n(γ, α), is the value

1

2πi

γ

1

z − α dz.

Theorem. n(γ, α) ∈ Z.Proof. We have

γ

1

z − α dz = 1

0

γ (t)γ (t) − α dt.

Note that the integrand on the RHS is ddt log(γ (t)−α), with a caveat that log is a multivaluedfunction. log z = log |z| + iArg(z) modulo 2πi. We’ll show that 1

0

γ (t)γ (t) − α dt = 2πik

for some k. This k = n(γ, α). Let F (s) = s

1γ (t)

γ (t)−α dt (secretly, log(γ (s)−α)− log(γ (1)−α).Then F (s) = γ

(s)γ (s)−α . We claim that

F (s) = γ (s) − αγ (0) − α .

To see this, let G(s) = (γ (s) − α)e−F (s). Then G(s) = −γ (s)e−F (s) + γ (s)e−F (s) = 0,so G(s) is a constant. Check that at s = 0, we get G(0) = (γ (0)

−α)e−F (0), so indeed

eF (0) = γ (0)−αγ (t)−α .

Now eF (1) = γ (1)−αγ (t)−α . F (1) = 1

0γ (t)

γ (t)−α dt and γ (1) = γ (0), so eF (1) = 1. So F (1) = 2πik

for k ∈ Z.

Corollary. (version of Jordan curve theorem) Let γ : [0, 1] → C be a closed curve. ThenC − image(γ ) is disconnected.

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Let’s call a simple closed curve one that doesn’t intersect itself. For us, if γ has n(γ, α) = 0or 1, then we call γ “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 have

n = 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 α1,...,αn ∈ D. Let γ be any contour curve in D. Then

γ

f (z)dz = 2πin

k=1

n(γ, αk)Res(f ; αk).

Proof. The gist is to repeatedly subtract the principal parts of f at each αk. This proof is

a bit long to type up, so it is left as an exercise to the reader.

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CHAPTER 12

THE ARGUMENT PRINCIPLE

Recall from last time that we had Cauchy’s residue theorem:

Theorem. Let D be a region with Ĉ −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 α1,...,αn, and let γ be a contour curve in D missing these points. Then

γ

f (z)dz = 2πin

k=1

Res(f ; αk)n(γ, αk).

The argument principle. This is essentially relating the number of zeroes and poles of f to (# of times f winds around 0). Let γ be a regular contour curve, e.g. if ∀α ∈ C − im(γ ),either n(γ, α) = 0 or n(γ, α) = 1. Intuitively, this means that the points are either “inside

of γ ” or outside; the former is given by the set {z ∈ C − im(γ ) : n(γ, z) = 1} and the latteris given by the set {z ∈ C − im(γ ) : n(γ, z) = 0}.

Theorem. If f is holomorphic in a simply connected region D, and γ a regular curvein D , then the following are equal:

(1) (# of zeroes of f with multiplicity) − (# of poles of f with multiplicity inside γ )(2) The winding number around zero of f (γ (t)), i.e. n(f ◦ γ, 0)(3) 12πi

γ

f (z)f (z) dz

Remark. This assumes that for f |γ = 0, there are no poles. The multiplicity is also calledthe order .

Proof. Let’s do (2) ⇐⇒ (3). For γ : [0, 1] → C, we have

n(f ◦ γ ; 0) = 12πi

f ◦γ

1z

dz = 12πi

10

γ (t)f (γ (t))f (γ (t))

dt = 12πi

γ

f (z)f (z)

dz.

The integrand is actually ∂ ∂z (log f ), which keeps track of how Arg(f ) changes and goesaround γ .

(3) ⇐⇒ (1). To do this we check the residues of f /f . The only singularities of f /f occur at the zeroes or poles of f . Near a zero or pole α of f , we have f (z) = (z − α)ng(z)

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for g(z) holomorphic near α, g(α) = 0. Also,f (z) = n(z − α)n−1g(z) + (z − α)ng(z),

sof (z)f (z)

= n

z − α + g (z)

g(z) ,

where we note that the residue at α of nz−α is n, and g(z) = 0 implies that the term on theright is holomorphic and is the residue at α. Hence

1

2πi

γ

f

f dz =

α pole of f

outside γ

n(γ, α) · order(f, α) −

α pole of f inside γ

n(γ, α) · order+poles(f, α)

And n(γ, α) → 1 in all these sums.

Moreover, we can say that the “zeroes of order n look like zn”; they wrap around 0 n

times CCW, and “the poles of order n look like 1zn ”; 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. (Rouché Theorem) (Assume f has no zeroes on γ .) Let f , g be holomorphic inthe unit disk (in γ ), and ||g(z)||


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where ρ is the curve traced out by the fact that 1 + g◦γ f ◦γ < 1, which has an antiderivativegiven by the principal branch of log z in Re(z) > 0.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Generalized Cauchy’s integral formula. Let γ be regular, and z be inside γ . Then

f (k)(z) = k!

2πi

γ

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 f n is holomorphic and f n → f uniformly on compact sets (henceholomorphic by Morera’s theorem). Then f n → f uniformly as well.

Proof. We have

f n(z) − f (z) = 1

2πi

γ

f n(w) − f (w)(w − z)2 dw

for γ = C r(α), z ∈ Dr(α). On Dr/2(α), we have

|f k(z) − f (z)| = 1

2π

γ

f n(w) − f (w)(w − z)2 dw

≤ 12π · 2πr · 4r2 · maxC r(α) |f n(w) − f (w)|,where the 4r2 comes as an upper bound on

1|w−z|2 because |w − z| ≥ r/2 =⇒ 1|w−z|2 ≤ 4r2 .

Now choose n large enough so that

|f n(w) − f (w)| < r

4

on C r(α). Then |f n(z) − f (z)| < for z ∈ Dr/2(α), n > N .Now cover the compact set in euqation by disks of half the radius. Because the set is

compact, finitely many suffiice. Take the largest N α among these.

Theorem. (Hurwitz) Suppose f n → 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 = 0, and let f (α) = 0 for some α ∈ D. α 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(α), > 0, with no zeroes of f in D(α) other than α. We have f

n → f uniformly,

and there are no zeroes on C (α) of f ; none of f n are zero, so 1f n

→ 1f uniformly on C (α)(show this yourself). Thus

f nf n

→ f

f

uniformly on C (α). It follows that

limn→∞

1

2πi

C (α)

f nf n

= 1

2πi

C (α)

f

f .

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Solim

n→∞ # of zeroes of f n in D(α) = # of zeroes of f in D(α) = 1,

a contradiction, so we can’t have this isolated zero of f .

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CHAPTER 13

INTEGRALS AND SOME GEOMETRY

Recall the residue theorem: let αk be the singularities of f . Then γ

f (z)dz = 2πi

Res(f ; αk)n(γ, αk).

There are many applications of this. If P, Q are polynomials, Q(x) = 0 for real x, anddeg Q ≥ deg P + 2, then

∞−∞

P (x)dx

Q(x)dxdx = 2πi

nk=1

Res

P (z)

Q(z), αk

.

You can show this with the usual residue calculation (yes, the ones that we beat to deathin our review session...).

Example. ∞−∞

1

x6 + 1dx

The poles are where z 6 = −1, e.g. at eiθ for θ = π6 , 3π6 , ....Recall that, for f (z) = A(z)/B(z) with B (α) = 0, A(α) = 0, if B (α) = 0 then this is a

simple pole, and the residue is given by Res(f ; α) = A(α)/B(α).So using this we get that the residues are

Res(f (z); eiπ/6) = 1

6e5πi/6

Res(f (z); eiπ/2) = 1

6eπi/2

Res(f (z); e5iπ/6) = 1

6eπi/6

Then ∞−∞

1

x6 + 1dx =

2πi

6 (e−5πi/6 + e−iπ/2 + e−iπ/6) =

2π

3 .

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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 ,

∃δ : Bδ(z)

⊆f −1(B(w)) = {x ∈ C : f (x) ∈ B(w)}, i.e. |z − α| < δ =⇒ |f (z) − f (w)| < δ . Open meansthat ∀δ, ∃ : f (Bδ(z)) ⊇ B(w), so |f (z) − β | < =⇒ ∃α : |z − α| < δ, f (α) = β .

Proof. Let α ∈ C. Wlog f (α) = 0; take C inside Bδ(α). Take C inside Bδ(α); min f (C )exists and it’s not zero, so call it z. Let w ∈ B(0). Then for z ∈ C , |f (z) − w| ≥|f (z)| − |w| ≥ 2 − = . For z = α, |f (α) − w| = | − w| < . Hence min |f (z) − w| forz ∈ Bδ/2(α) is not achieved on the boundary. Thus, by the minimum modulus theorem, ∃a 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)

| ≤1

with equality in either iff f (z) = eiθz.Proof. Define

g(z) =

f (z)

z z = 0

f (0) z = 0,

which is holomorphic on the circle of radius δ ; |g(z)| = |f (z)||z| ≤ 1δ . By the maximum modulusprinciple, we know in fact |g(z)| ≤ 1 for all z ∈ D. Let δ → 1.

So if a max is achieved on the interior, then f (z) is constant, i.e. g(z) is constantif |f (z)| = |z| for any z ∈ D1(0), z = 0, or |f (0)| = 1. This means g(z) = eiθ; theng(z) = f (z)/z (f (0)), so f (z) = eiθz.

Corollary. If a map from D → D, f (0) = α, has a maximum value of |f (0)| amongmaps D → D, then f is surjective.Proof. Consider h(z) = Bα(f (z)). h(0) = f

(0)Bα(0), h(z)D → D, h(z) : 0 → 0, |h(0)| ≤1.

The conclusion is that

|f (1)| ≤ 1Bα(α) = 1 − |α|2

.

Notice that this is achieved for inverse of Bα, which is Bα.

Bα(Bα(z)) = z.

Also note that

Bα(z) = z − α1

−αz

,

and

Bα(z) = 1

1 − |α|2 ,

so B α(0) = 1 − |α|2, Bα(0) = 0, |Bα(z)| ≤ 1 for |z| ≤ 1.

We used h(0) ≤ 1. We also know |h(z)| ≤ |z| and h(z) = Bα(f (z)); |Bα(f (z))| ≤ |z|.

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We can

complex analysis. function theory

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