IB Complex Analysis (Full)

Part IB — Complex Analysis

Based on lectures by I. Smith

Notes taken by Dexter Chua

Lent 2016

These notes are not endorsed by the lecturers, and I have modified them (often

significantly) after lectures. They are nowhere near accurate representations of what

was actually lectured, and in particular, all errors are almost surely mine.

Analytic functions

Complex differentiation and the Cauchy–Riemann equations. Examples. Conformal

mappings. Informal discussion of branch points, examples of log z and z

. [3]

Contour integration and Cauchy’s theorem

Contour integration (for piecewise continuously differentiable curves). Statement and

proof of Cauchy’s theorem for star domains. Cauchy’s integral formula, maximum

modulus theorem, Liouville’s theorem, fundamental theorem of algebra. Morera’s

theorem. [5]

Expansions and singularities

Uniform convergence of analytic functions; local uniform convergence. Differentiability

of a power series. Taylor and Laurent expansions. Principle of isolated zeros. Residue

at an isolated singularity. Classification of isolated singularities. [4]

The residue theorem

Winding numbers. Residue theorem. Jordan’s lemma. Evaluation of definite integrals

by contour integration. Rouch´e’s theorem, principle of the argument. Open mapping

theorem. [4]

Contents

0 Introduction

1 Complex differentiation

1.1 Differentiation

1.2 Conformal mappings

1.3 Power series

1.4 Logarithm and branch cuts

2 Contour integration

2.1 Basic properties of complex integration

2.2 Cauchy’s theorem

2.3 The Cauchy integral formula

2.4 Taylor’s theorem

2.5 Zeroes

2.6 Singularities

2.7 Laurent series

3 Residue calculus

3.1 Winding numbers

3.2 Homotopy of closed curves

3.3 Cauchy’s residue theorem

3.4 Overview

3.5 Applications of the residue theorem

3.6 Rouch´es theorem

0 Introduction

Complex analysis is the study of complex differentiable functions. While this

sounds like it should be a rather straightforward generalization of real analysis,

it turns out complex differentiable functions behave rather differently. Requir-

ing that a function is complex differentiable is a very strong condition, and

consequently these functions have very nice properties.

One of the most distinguishing results from complex analysis is Liouville’s

theorem, which says that every bounded complex differentiable function

C → C

must be constant. This is very false for real functions (e.g.

sin x

). This

gives a strikingly simple proof of the fundamental theorem of algebra — if the

polynomial

has no roots, then

is well-defined on all of

, and it is easy to

show this must be bounded. So p is constant.

Many things we hoped were true in real analysis are indeed true in complex

analysis. For example, if a complex function is once differentiable, then it is

infinitely differentiable. In particular, every complex differentiable function has

a Taylor series and is indeed equal to its Taylor series (in reality, to prove these,

we show that every complex differentiable function is equal to its Taylor series,

and then notice that power series are always infinitely differentiable).

Another result we will prove is that the uniform limit of complex differentiable

functions is again complex differentiable. Contrast this with the huge list of

weird conditions we needed for real analysis!

Not only is differentiation nice. It turns out integration is also easier in

complex analysis. In fact, we will exploit this fact to perform real integrals by

pretending they are complex integrals. However, this will not be our main focus

here — those belong to the IB Complex Methods course instead.

1 Complex differentiation

1.1 Differentiation

We start with some definitions. As mentioned in the introduction, Liouville’s

theorem says functions defined on the whole of

are often not that interesting.

Hence, we would like to work with some subsets of

instead. As in real analysis,

for differentiability to be well-defined, we would want a function to be defined

on an open set, so that we can see how

U → C

varies as we approach a point

∈ U from all different directions.

Definition

(Open subset)

A subset

U ⊆ C

is open if for any

x ∈ U

, there is

some ε > 0 such that the open ball B

(x) = B(x; ε) ⊆ U .

The notation used for the open ball varies form time to time, even within

the same sentence. For example, instead of putting

as the subscript, we could

put

as the subscript and

inside the brackets. Hopefully, it will be clear from

context.

This is good, but we also want to rule out some silly cases, such as functions

defined on subsets that look like this:

This would violate results such as functions with zero derivative must be constant.

Hence, we would require our subset to be connected. This means for any two

points in the set, we can find a path joining them. A path can be formally

defined as a function γ : [0, 1] → C, with start point γ(0) and end point γ(1).

Definition

(Path-connected subset)

A subset

U ⊆ C

is path-connected if for

any

x, y ∈ U

, there is some

: [0

→ U

continuous such that

(0) =

and

γ(1) = y.

Together, these define what it means to be a domain. These are (usually)

things that will be the domains of our functions.

Definition

(Domain)

A domain is a non-empty open path-connected subset

of C.

With this, we can define what it means to be differentiable at a point. This

is, in fact, exactly the same definition as that for real functions.

Definition

(Differentiable function)

Let

U ⊆ C

be a domain and

U → C

be a function. We say f is differentiable at w ∈ U if

(w) = lim

z→w

f(z) −f (w)

z − w

exists.

Here we implicitly require that the limit does not depend on which direction

we approach

from. This requirement is also present for real differentiability, but

there are just two directions we can approach

from — the positive direction

and the negative direction. For complex analysis, there are infinitely many

directions to choose from, and it turns out this is a very strong condition to

impose.

Complex differentiability at a point

is not too interesting. Instead, what we

want is a slightly stronger condition that the function is complex differentiable

in a neighbourhood of w.

Definition

(Analytic/holomorphic function)

A function

is analytic or holo-

morphic at

w ∈ U

is differentiable on an open neighbourhood

(

w, ε

) of

(for some ε).

Definition

(Entire function)

C → C

is defined on all of

and is

holomorphic on C, then f is said to be entire.

It is not universally agreed what the words analytic and holomorphic should

mean. Some people take one of these word to mean instead that the function

has a (and is given by the) Taylor series, and then take many pages to prove

that these two notions are indeed the same. But since they are the same, we

shall just opt for the simpler definition.

The goal of the course is to develop the rich theory of these complex dif-

ferentiable functions and see how we can integrate them along continuously

differentiable (C

) paths in the complex plane.

Before we try to achieve our lofty goals, we first want to figure out when

a function is differentiable. Sure we can do this by checking the definition

directly, but this quickly becomes cumbersome for more complicated functions.

Instead, we would want to see if we can relate complex differentiability to real

differentiability, since we know how to differentiate real functions.

Given

U → C

, we can write it as

, where

u, v

U → R

are

real-valued functions. We can further view

and

as real-valued functions of

two real variables, instead of one complex variable.

Then from IB Analysis II, we know this function

U → R

is differentiable

(as a real function) at a point (

c, d

)

∈ U

, with derivative

Du|

(c,d)

= (

λ, µ

), if and

only if

u(x, y) − u(c, d) −(λ(x − c) + µ(y − d))

k(x, y) − (c, d)k

→ 0 as (x, y) → (c, d).

This allows us to come up with a nice criterion for when a complex function is

differentiable.

Proposition.

Let

be defined on an open set

U ⊆ C

. Let

id ∈ U

and

write

. Then

is complex differentiable at

if and only if

and

viewed as a real function of two real variables, are differentiable at (c, d), and

= v

= −v

These equations are the Cauchy–Riemann equations. In this case, we have

(w) = u

(c, d) + iv

(c, d) = v

(c, d) − iu

(c, d).

Proof. By definition, f is differentiable at w with f

(w) = p + iq if and only if

lim

z→w

f(z) − f(w) − (p + iq)(z − w)

z − w

= 0. (†)

If z = x + iy, then

(p + iq)(z −w) = p(x − c) − q(y −d) + i(q(x − c) + p(y − d)).

So, breaking into real and imaginary parts, we know (†) holds if and only if

lim

(x,y)→(c,d)

u(x, y) − u(c, d) − (p(x − c) − q(y − d))

(x − c)

+ (y − d)

= 0

and

lim

(x,y)→(c,d)

v(x, y) − v(c, d) − (q(x − c) + p(y − d))

(x − c)

+ (y − d)

= 0.

Comparing this to the definition of the differentiability of a real-valued function,

we see this holds exactly if u and v are differentiable at (c, d) with

Du|

(c,d)

= (p, −q), Dv|

(c,d)

= (q, p).

A standard warning is given that

U → C

can be written as

where

and

−v

at (

c, d

)

∈ U

, we cannot conclude that

is complex

differentiable at (

c, d

). These conditions only say the partial derivatives exist,

but this does not imply imply that

and

are differentiable, as required by the

proposition. However, if the partial derivatives exist and are continuous, then

by IB Analysis II we know they are differentiable.

Example.

(i)

The usual rules of differentiation (sum rule, product, rule, chain rule,

derivative of inverse) all hold for complex differentiable functions, with the

same proof as the real case.

(ii)

A polynomial

C → C

is entire. This can be checked directly from

definition, or using the product rule.

(iii)

A rational function

p(z)

q(z)

U → C

, where

U ⊆ C \ {z

(

) = 0

}

, is

holomorphic on any such U. Here p, q are polynomials.

(iv) f

(

) =

|z|

is not complex differentiable at any point of

. Indeed, we can

write this as f = u + iv, where

u(x, y) =

+ y

, v(x, y) = 0.

If (x, y) 6= (0, 0), then

+ y

, u

+ y

If we are not at the origin, then clearly we cannot have both vanish, but

the partials of

both vanish. Hence the Cauchy–Riemann equations do

not hold and it is not differentiable outside of the origin.

At the origin, we can compute directly that

f(h) − f(0)

|h|

This is, say, +1 for

h ∈ R

and

−

1 for

h ∈ R

−

. So the limit as

h →

0 does

not exist.

1.2 Conformal mappings

The course schedules has a weird part where we are supposed to talk about

conformal mappings for a lecture, but not use it anywhere else. We have to put

them somewhere, and we might as well do it now. However, this section will be

slightly disconnected from the rest of the lectures.

Definition

(Conformal function)

Let

U → C

be a function holomorphic at

w ∈ U . If f

(w) 6= 0, we say f is conformal at w.

What exactly does

(

)

= 0 tell us? In the real case, if we know a function

: (

a, b

)

→ R

is continuous differentiable, and

(

)

= 0, then

is locally

increasing or decreasing at

, and hence has a local inverse. This is also true in

the case of complex functions.

We write

, then viewed as a map

→ R

, the Jacobian matrix is

given by

Df =





Then

det(Df) = u

− u

= u

+ u

Using the formula for the complex derivative in terms of the partials, this shows

that if

(

)

= 0, then

det

(

Df|

)

= 0. Hence, by the inverse function theorem

(viewing

as a function

→ R

is locally invertible at

(technically, we

need

to be continuously differentiable, instead of just differentiable, but we

will later show that

in fact must be infinitely differentiable). Moreover, by

the same proof as in real analysis, the local inverse to a holomorphic function is

holomorphic (and conformal).

But being conformal is more than just being locally invertible. An important

property of conformal mappings is that they preserve angles. To give a precise

statement of this, we need to specify how “angles” work.

The idea is to look at tangent vectors of paths. Let

, γ

: [

−

→ U

continuously differentiable paths that intersect when

= 0 at

(0) =

(0).

Moreover, assume γ

(0) 6= 0.

Then we can compare the angles between the paths by looking at the difference

in arguments of the tangents at w. In particular, we define

angle(γ

, γ

) = arg(γ

(0)) − arg(γ

(0)).

Let

U → C

and

w ∈ U

. Suppose

is conformal at

. Then

maps our two

paths to

f ◦ γ

: [

−

→ C

. These two paths now intersect at

(

). Then the

angle between them is

angle(f ◦ γ

, f ◦γ

) = arg((f ◦γ

)

(0)) − arg((f ◦ γ

)

(0))

= arg



(f ◦ γ

)

(0)

(f ◦ γ

)

(0)



= arg



(0)



= angle(γ

, γ

using the chain rule and the fact that f

(w) 6= 0. So angles are preserved.

What else can we do with conformal maps? It turns out we can use it to

solve Laplace’s equation.

We will later prove that if

U → C

is holomorphic on an open set

, then

: U → C is also holomorphic. Hence f is infinitely differentiable.

In particular, if we write

, then using the formula for

in terms

of the partials, we know

and

are also infinitely differentiable. Differentiating

the Cauchy–Riemann equations, we get

= v

= −u

In other words,

+ u

= 0,

We get similar results for

instead. Hence

(

) and

(

) satisfy the Laplace

equation and are hence harmonic (by definition).

Definition

(Conformal equivalence)

and

are open subsets of

and

f : U → V is a conformal bijection, then it is a conformal equivalence.

Note that in general, a bijective continuous map need not have a continuous

inverse. However, if we are given further that it is conformal, then the inverse

mapping theorem tells us there is a local conformal inverse, and if the function

is bijective, these patch together to give a global conformal inverse.

The idea is that if we are required to solve the 2D Laplace’s equation on

a funny domain

subject to some boundary conditions, we can try to find a

conformal equivalence

between

and some other nice domain

. We can

then solve Laplace’s equation on

subject to the boundary conditions carried

forward by

, which is hopefully easier. Afterwards, we pack this solution into

the real part of a conformal function

, and then

g ◦f

U → C

is a holomorphic

function on

whose real part satisfies the boundary conditions we want. So we

have found a solution to Laplace’s equation.

You will have a chance to try that on the first example sheet. Instead, we

will focus on finding conformal equivalences between different regions, since

examiners like these questions.

Example.

Any M¨obius map

(

) =

az+b

cz+d

(with

ad−bc 6

= 0) defines a conformal

equivalence

C ∪ {∞} → C ∪ {∞}

in the obvious sense.

(

)

= 0 follows from

the chain rule and the invertibility of A(z).

In particular, the M¨obius group of the disk

M¨ob(D) = {f ∈ M¨obius group : f(D) = D}



z − a

¯az −1

∈ M¨ob : |a| < 1, |λ| = 1



is a group of conformal equivalences of the disk. You will prove that the M¨obius

group of the disk is indeed of this form in the first example sheet, and that these

are all conformal equivalences of the disk on example sheet 2.

Example.

The map

z 7→ z

for

n ≥

2 is holomorphic everywhere and conformal

except at z = 0. This gives a conformal equivalence

z ∈ C

∗

: 0 < arg(z) <

↔ H,

where we adopt the following notation:

Notation. We write C

∗

= C \ {0} and

H = {z ∈ C : Im(z) > 0}

is the upper half plane.

Example.

Note that

z ∈ H

if and only if

is closer to

than to

−i

. In other

words,

|z − i| < |z + i|,



z − i

z + i



< 1.

z 7→

z−i

z+i

defines a conformal equivalence

H → D

, the unit disk. We know

this is conformal since it is a special case of the M¨obius map.

Example. Consider the map

z 7→ w =



z +



assuming z ∈ C

∗

. This can also be written as

w + 1

w − 1

= 1 +

w − 1

= 1 +

z +

− 2

= 1 +

− 2z + 1



z + 1

z − 1



So this is just squaring in some funny coordinates given by

z+1

z−1

. This map is

holomorphic (except at z = 0). Also, we have

(z) = 1 −

+ 1

So f is conformal except at ±1.

Recall that the first thing we learnt about M¨obius maps is that they take

lines and circles to lines and circles. This does something different. We write

z = re

iθ

. Then if we write z 7→ w = u + iv, we have

u =



r +



cos θ

v =



r −



sin θ

Fixing the radius and argument fixed respectively, we see that a circle of radius

ρ is mapped to the ellipse



ρ +





ρ −



= 1,

while the half-line arg(z) = µ is mapped to the hyperbola

cos

−

sin

= 1.

We can do something more interesting. Consider a off-centered circle, chosen to

pass through the point −1 and −i. Then the image looks like this:

−1

−i

f(−1)

f(−i)

Note that we have a singularity at

(

−

1) =

−

1. This is exactly the point where

f is not conformal, and is no longer required to preserve angles.

This is a crude model of an aerofoil, and the transformation is known as the

Joukowsky transform.

In applied mathematics, this is used to model fluid flow over a wing in terms

of the analytically simpler flow across a circular section.

We interlude with a little trick. Often, there is no simple way to describe

regions in space. However, if the region is bounded by circular arcs, there is a

trick that can be useful.

Suppose we have a circular arc between α and β.

Along this arc,

θ − φ

arg

(

z − α

)

− arg

(

z − β

) is constant, by elementary

geometry. Thus, for each fixed µ, the equation

arg(z − α) − arg(z − β) = µ

determines an arc through the points α, β.

To obtain a region bounded by two arcs, we find the two

−

and

that

describe the boundary arcs. Then a point lie between the two arcs if and only if

its µ is in between µ

−

and µ

, i.e. the region is



z : arg



z − α

z − β



∈ [µ

−

, µ

]



This says the point has to lie in some arc between those given by µ

−

and µ

For example, the following region:

−1 1

can be given by

U =



z : arg



z − 1

z + 1



∈

, π



Thus for instance the map

z 7→ −



z − 1

z + 1



is a conformal equivalence from

. This is since if

z ∈ U

, then

z−1

z+1

has

argument in



, π



, and can have arbitrary magnitude since

can be made

as close to

−

1 as you wish. Squaring doubles the angle and gives the lower

half-plane, and multiplying by −1 gives the upper half plane.

z 7→

z−1

z+1

z 7→ z

z 7→ −z

In fact, there is a really powerful theorem telling us most things are conformally

equivalent to the unit disk.

Theorem

(Riemann mapping theorem)

Let

U ⊆ C

be the bounded domain

enclosed by a simple closed curve, or more generally any simply connected domain

not equal to all of

. Then

is conformally equivalent to

|z| <

} ⊆ C

This in particular tells us any two simply connected domains are conformally

equivalent.

The terms simple closed curve and simply connected are defined as follows:

Definition

(Simple closed curve)

A simple closed curve is the image of an

injective map S

→ C.

It should be clear (though not trivial to prove) that a simple closed curve

separates C into a bounded part and an unbounded part.

The more general statement requires the following definition:

Definition

(Simply connected)

A domain

U ⊆ C

is simply connected if every

continuous map from the circle

→ U

can be extended to a continuous

map from the disk

→ U

such that

F |

∂D

. Alternatively, any loop

can be continuously shrunk to a point.

Example.

The unit disk is simply-connected, but the region defined by 1

|z| <

2 is not, since the circle

|z|

= 1

5 cannot be extended to a map from a disk.

We will not prove this statement, but it is nice to know that this is true.

If we believe that the unit disk is relatively simple, then since all simply

connected regions are conformally equivalent to the disk, all simply connected

domains are boring. This suggests we will later encounter domains with holes to

make the course interesting. This is in fact true, and we will study these holes

in depth later.

Example. The exponential function

= 1 + z +

+ ···

defines a function

C → C

∗

. In fact it is a conformal mapping. This sends

the region

(

)

∈

[

a, b

]

}

to the annulus

≤ |w| ≤ e

}

. One is simply

connected, but the other is not — this is not a problem since

is not bijective

on the strip.

1.3 Power series

Some of our favorite functions are all power series, including polynomials (which

are degenerate power series in some sense), exponential functions and trigono-

metric functions. We will later show that all holomorphic functions are (locally)

given by power series, but without knowing this fact, we shall now study some

of the basic properties of power series.

It turns out power series are nice. The key property is that power series

are infinitely differentiable (as long as it converges), and the derivative is given

pointwise.

We begin by recalling some facts about convergence from IB Analysis II.

Definition

(Uniform convergence)

A sequence (

) of functions converge

uniformly to

if for all

ε >

0, there is some

such that

n > N

implies

(z) − f(z)| < ε for all z.

Proposition. The uniform limit of continuous functions is continuous.

Proposition

(Weierstrass M-test)

For a sequence of functions

, if we can

find (

)

⊆ R

such that

(

)

| < M

for all

in the domain, then

converges implies

(x) converges uniformly on the domain.

Proposition.

Given any constants

}

n≥0

⊆ C

, there is a unique

R ∈

, ∞

]

such that the series

z 7→

∞

n=0

(

z − a

)

converges absolutely if

|z − a| < R

and diverges if

|z − a| > R

. Moreover, if 0

< r < R

, then the series converges

uniformly on {z : |z − a| < r}. This R is known as the radius of convergence.

So while we don’t necessarily get uniform convergence on the whole domain,

we get uniform convergence on all compact subsets of the domain.

We are now going to look at power series. They will serve as examples, and

as we will see later, universal examples, of holomorphic functions. The most

important result we need is the following result about their differentiability.

Theorem. Let

f(z) =

∞

n=0

(z − a)

be a power series with radius of convergence R > 0. Then

(i) f is holomorphic on B(a; R) = {z : |z − a| < R}.

(ii) f

(z) =

(z − 1)

n−1

, which also has radius of convergence R.

(iii) Therefore f is infinitely complex differentiable on B(a; R). Furthermore,

(n)

(a)

Proof.

Without loss of generality, take

= 0. The third part obviously follows

from the previous two, and we will prove the first two parts simultaneously. We

would like to first prove that the derivative series has radius of convergence

so that we can freely happily manipulate it.

Certainly, we have

|nc

| ≥ |c

. So by comparison to the series for

, we can

see that the radius of convergence of

n−1

is at most

. But if

|z| < ρ < R

then we can see

|nc

n−1

= n



n−1

→ 0

n → ∞

. So by comparison to

n−1

, which converges, we see that the

radius of convergence of

n−1

is at least

. So the radius of convergence

must be exactly R.

Now we want to show

really is differentiable with that derivative. Pick

z, w such that |z|, |w| ≤ ρ for some ρ < R as before.

Define a new function

ϕ(z, w) =

∞

n=1

n−1

j=0

n−1−j

Noting



n−1

j=0

n−1−j



≤ n|c

|ρ

we know the series defining

converges uniformly on

{|z| ≤ ρ, |w| < ρ}

, and

hence to a continuous limit.

If z 6= w, then using the formula for the (finite) geometric series, we know

ϕ(z, w) =

∞

n=1



− w

z − w



f(z) − f(w)

z − w

On the other hand, if z = w, then

ϕ(z, z) =

∞

n=1

n−1

Since ϕ is continuous, we know

lim

w→z

f(z) − f(w)

z − w

→

∞

n=1

n−1

So f

(z) = ϕ(z, z) as claimed. Then (iii) follows from (i) and (ii) directly.

Corollary. Given a power series

f(z) =

n≥0

(z − a)

with radius of convergence

R >

0, and given 0

< ε < R

, if

vanishes on

(

a, ε

then f vanishes identically.

Proof.

vanishes on

(

a, ε

), then all its derivatives vanish, and hence the

coefficients all vanish. So it is identically zero.

This is obviously true, but will come up useful some time later.

It is might be useful to have an explicit expression for

. For example, by

IA Analysis I, we know

R = sup{r ≥ 0 : |c

→ 0 as n → ∞}

lim sup

But we probably won’t need these.

1.4 Logarithm and branch cuts

Recall that the exponential function

= exp(z) = 1 + z +

+ ···

has a radius of convergence of

∞

. So it is an entire function. We have the usual

standard properties, such as e

z+w

= e

, and also

x+iy

= e

(cos y + i sin y).

So given

w ∈ C

∗

C \ {

}

, there are solutions to

. In fact, this

has infinitely many solutions, differing by adding integer multiples of 2

πi

. In

particular, e

= 1 if and only if z is an integer multiple of 2πi.

This means

does not have a well-defined inverse. However, we do want to

talk about the logarithm. The solution would be to just fix a particular range

allowed. For example, we can define the logarithm as the function sending

iθ

log r

iθ

, where we now force

−π < θ < π

. This is all well, except it

is now not defined at

−

1 (we can define it as, say,

iπ

, but we would then lose

continuity).

There is no reason why we should pick

−π < θ < π

. We could as well require

300π < θ < 302π. In general, we make the following definition:

Definition

(Branch of logarithm)

Let

U ⊆ C

∗

be an open subset. A branch of

the logarithm on

is a continuous function

U → C

for which

λ(z)

for

all z ∈ U.

This is a partially defined inverse to the exponential function, only defined

on some domain

. These need not exist for all

. For example, there is no

branch of the logarithm defined on the whole C

∗

, as we will later prove.

Example. Let U = C \ R

≤0

, a “slit plane”.

Then for each

z ∈ U

, we write

iθ

, with

−π < θ < π

. Then

(

) =

log

(

iθ

is a branch of the logarithm. This is the principal branch.

, there is a continuous function

arg

U →

(

−π, π

), which is why we can

construct a branch. This is not true on, say, the unit circle.

Fortunately, as long as we have picked a branch, most things we want to be

true about log is indeed true.

Proposition.

{z ∈ C

z 6∈ R

≤0

}

, the principal branch

log

U → C

holomorphic function. Moreover,

log z =

If |z| < 1, then

log(1 + z) =

n≥1

(−1)

n−1

= z −

− ··· .

Proof.

The logarithm is holomorphic since it is a local inverse to a holomorphic

function. Since e

log z

= z, the chain rule tells us

(log z) =

To show that

log

(1 +

) is indeed given by the said power series, note that

the power series does have a radius of convergence 1 by, say, the ratio test. So

by the previous result, it has derivative

1 − z + z

+ ··· =

1 + z

Therefore,

log

(1 +

) and the claimed power series have equal derivative, and

hence coincide up to a constant. Since they agree at

= 0, they must in fact be

equal.

Having defined the logarithm, we can now define general power functions.

Let

α ∈ C

and

log

U → C

be a branch of the logarithm. Then we can define

= e

α log z

on U . This is again only defined when log is.

In a more general setting, we can view

log

as an instance of a multi-valued

function on

∗

. At each point, the function

log

can take many possible values,

and every time we use

log

, we have to pick one of those values (in a continuous

way).

In general, we say that a point

p ∈ C

is a branch point of a multivalued

function if the function cannot be given a continuous single-valued definition in

a (punctured) neighbourhood

(

p, ε

)

\ {p}

for any

ε >

0. For example, 0 is

a branch point of log.

Example. Consider the function

f(z) =

z(z − 1).

This has two branch points,

= 0 and

= 1, since we cannot define a square

root consistently near 0, as it is defined via the logarithm.

Note we can define a continuous branch of f on either

or we can just kill a finite slit by

Why is the second case possible? Note that

f(z) = e

(log(z)+log(z−1))

If we move around a path encircling the finite slit, the argument of each of

log

(

)

and

log

(

z −

1) will jump by 2

πi

, and the total change in the exponent is 2

πi

. So

the expression for f(z) becomes uniquely defined.

While these two ways of cutting slits look rather different, if we consider this

to be on the Riemann sphere, then these two cuts now look similar. It’s just

that one passes through the point ∞, and the other doesn’t.

The introduction of these slits is practical and helpful for many of our

problems. However, theoretically, this is not the best way to think about multi-

valued functions. A better treatment will be provided in the IID Riemann

Surfaces course.

2 Contour integration

In the remaining of the course, we will spend all our time studying integration of

complex functions. At first, you might think this is just an obvious generalization

of integration of real functions. This is not true. Starting from Cauchy’s theorem,

beautiful and amazing properties of complex integration comes one after another.

Using these, we can prove many interesting properties of holomorphic functions

as well as do lots of integrals we previously were not able to do.

2.1 Basic properties of complex integration

We start by considering functions

: [

a, b

]

→ C

. We say such a function is

Riemann integrable if

(

) and

(

) are individually, and the integral is

defined to be

f(t) dt =

Re(f(t)) dt + i

Im(f(t)) dt.

While Riemann integrability is a technical condition to check, we know that all

continuous functions are integrable, and this will apply in most cases we care

about in this course. After all, this is not a course on exotic functions.

We start from some less interesting facts, and slowly develop and prove some

really amazing results.

Lemma. Suppose f : [a, b] → C is continuous (and hence integrable). Then



f(t) dt



≤ (b − a) sup

|f(t)|

with equality if and only if f is constant.

Proof. We let

θ = arg

f(t) dt

and

M = sup

|f(t)|.

Then we have



f(t) dt



−iθ

f(t) dt

Re(e

−iθ

f(t)) dt

≤ (b − a)M,

with equality if and only if

(

)

and

arg f

(

) =

for all

, i.e.

constant.

Integrating functions of the form

: [

a, b

]

→ C

is easy. What we really care

about is integrating a genuine complex function

U ⊆ C → C

. However,

we cannot just “integrate” such a function. There is no given one-dimensional

domain we can integrate along. Instead, we have to make some up ourselves.

We have to define some paths in the complex plane, and integrate along them.

Definition

(Path)

A path in

is a continuous function

: [

a, b

]

→ C

, where

a, b ∈ R.

For general paths, we just require continuity, and do not impose any conditions

about, say, differentiability.

Unfortunately, the world is full of weird paths. There are even paths that fill

up the whole of the unit square. So we might want to look at some nicer paths.

Definition

(Simple path)

A path

: [

a, b

]

→ C

is simple if

(

) =

(

) only

if t

= t

or {t

, t

} = {a, b}.

In other words, it either does not intersect itself, or only intersects itself at

the end points.

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

Definition

(Contour)

A contour is a simple closed path which is piecewise

i.e. piecewise continuously differentiable.

For example, it can look something like this:

Most of the time, we are just interested in integration along contours. However,

it is also important to understand integration along just simple

smooth paths,

since we might want to break our contour up into different segments. Later, we

will move on to consider more general closed piecewise

paths, where we can

loop around a point many many times.

We can now define what it means to integrate along a smooth path.

Definition

(Complex integration)

: [

a, b

]

→ U ⊆ C

-smooth and

f : U → C is continuous, then we define the integral of f along γ as

f(z) dz =

f(γ(t))γ

(t) dt.

By summing over subdomains, the definition extends to piecewise

-smooth

paths, and in particular contours.

We have the following elementary properties:

(i)

The definition is insensitive to reparametrization. Let

: [

, b

]

→

[

a, b

]

such that

(

) =

a, φ

(

) =

. If

is a

path and

γ ◦ φ

, then

f(z) dz =

f(z) dz.

This is just the regular change of variables formula, since

f(γ(φ(t)))γ

(φ(t))φ

(t) dt =

f(γ(u))γ

(u) du

if we let u = φ(t).

(ii) If a < u < b, then

f(z) dz =

γ|

[a,u]

f(z) dz +

γ|

[u,b]

f(z) dz.

These together tells us the integral depends only on the path itself, not how we

look at the path or how we cut up the path into pieces.

We also have the following easy properties:

(iii) If −γ is γ with reversed orientation, then

−γ

f(z) dz = −

f(z) dz.

(iv) If we set for γ : [a, b] → C the length

length(γ) =

|γ

(t)| dt,

then



f(z) dz



≤ length(γ) sup

|f(γ(t))|.

Example.

Take

∗

, and let

(

) =

for

n ∈ Z

. We pick

: [0

]

→ U

that sends θ 7→ e

iθ

. Then

f(z) dz =

(

2πi n = −1

0 otherwise

To show this, we have

f(z) dz =

2π

inθ

iθ

dθ

= i

2π

i(n+1)θ

dθ.

−

1, then the integrand is constantly 1, and hence gives 2

πi

. Otherwise, the

integrand is a non-trivial exponential which is made of trigonometric functions,

and when integrated over 2π gives zero.

Example. Take γ to be the contour

R−R

We parametrize the path in segments by

: [−R, R] → C γ

: [0, 1] → C

t 7→ t t 7→ Re

iπt

Consider the function f(z) = z

. Then the integral is

f(z) dz =

−R

dt +

2πit

iπRe

iπt

+ R

iπ

3πit

+ R

iπ



3πit

3πi



= 0

We worked this out explicitly, but we have just wasted our time, since this is

just an instance of the fundamental theorem of calculus!

Definition

(Antiderivative)

Let

U ⊆ C

and

U → C

be continuous. An

antiderivative of

is a holomorphic function

U → C

such that

(

) =

(

Then the fundamental theorem of calculus tells us:

Theorem

(Fundamental theorem of calculus)

Let

U → C

be continuous

with antiderivative F . If γ : [a, b] → U is piecewise C

-smooth, then

f(z) dz = F (γ(b)) − F (γ(a)).

In particular, the integral depends only on the end points, and not the path

itself. Moreover, if γ is closed, then the integral vanishes.

Proof. We have

f(z) dz =

f(γ(t))γ

(t) dt =

(F ◦ γ)

(t) dt.

Then the result follows from the usual fundamental theorem of calculus, applied

to the real and imaginary parts separately.

Example.

This allows us to understand the first example we had. We had the

function f (z) = z

integrated along the path φ(t) = e

(for 0 ≤ t ≤ 2π).

If n 6= −1, then

f =



n+1

n + 1



has a well-defined antiderivative, and the integral vanishes. On the other

hand, if n = −1, then

f(z) =

(log z),

where

log

can only be defined on a slit plane. It is not defined on the whole unit

circle. So we cannot apply the fundamental theorem of calculus.

Reversing the argument around, since

−1

does not vanish, this implies

there is not a continuous branch of log on any set U containing the unit circle.

2.2 Cauchy’s theorem

A question we might ask ourselves is when the anti-derivative exists. A necessary

condition, as we have seen, is that the integral around any closed curve has to

vanish. This is also sufficient.

Proposition.

Let

U ⊆ C

be a domain (i.e. path-connected non-empty open

set), and f : U → C be continuous. Moreover, suppose

f(z) dz = 0

for any closed piecewise C

-smooth path γ in U. Then f has an antiderivative.

This is more-or-less the same proof we gave in IA Vector Calculus that a real

function is a gradient if and only if the integral about any closed path vanishes.

Proof.

Pick our favorite

∈ U

. For

w ∈ U

, we choose a path

: [0

→ U

such that γ

(0) = a

and γ

(1) = w.

We first go through some topological nonsense to show we can pick

such

that this is piecewise

. We already know a continuous path

: [0

→ U

from

exists, by definition of path connectedness. Since

is open, for

all

in the image of

, there is some

(

)

0 such that

(

x, ε

(

))

⊆ U

. Since

the image of

is compact, it is covered by finitely many such balls. Then it is

trivial to pick a piecewise straight path living inside the union of these balls,

which is clearly piecewise smooth.

We thus define

F (w) =

f(z) dz.

Note that this

(

) is independent of the choice of

, by our hypothesis on

— given another choice

˜γ

, we can form the new path

∗

(

−˜γ

), namely the

path obtained by concatenating γ

with −˜γ

˜γ

This is a closed piecewise C

-smooth curve. So

∗(−˜γ

)

f(z) dz = 0.

The left hand side is

f(z) dz +

−˜γ

f(z) dz =

f(z) dz −

˜γ

f(z) dz.

So the two integrals agree.

Now we need to check that

is complex differentiable. Since

is open, we

can pick

θ >

0 such that

(

;

)

⊆ U

. Let

be the radial path in

(

w, ε

) from

W to w + h, with |h| < ε.

w + h

Now note that γ

∗ δ

is a path from a

to w + h. So

F (w + h) =

∗δ

f(z) dz

= F (w) +

f(z) dz

= F (w) + hf(w) +

(f(z) − f(w)) dz.

Thus, we know



F (w + h) − F (w)

− f(w)



≤

|h|



f(z) − f(w) dz



≤

|h|

length(δ

) sup

|f(z) − f(w)|

= sup

|f(z) − f(w)|.

Since

is continuous, as

h →

0, we know

(

)

−f

(

)

→

0. So

is differentiable

with derivative f.

To construct the anti-derivative, we assumed

(

) d

= 0. But we didn’t

really need that much. To simplify matters, we can just consider curves consisting

of straight line segments. To do so, we need to make sure we really can draw

line segments between two points.

You might think — aha! We should work with convex spaces. No. We do not

need such a strong condition. Instead, all we need is that we have a distinguished

point a

such that there is a line segment from a

to any other point.

Definition

(Star-shaped domain)

A star-shaped domain or star domain is a

domain

such that there is some

∈ U

such that the line segment [

, w

]

⊆ U

for all w ∈ U .

This is weaker than requiring

to be convex, which says any line segment

between any two points in U, lies in U.

In general, we have the implications

U is a disc ⇒ U is convex ⇒ U is star-shaped ⇒ U is path-connected,

and none of the implications reverse.

In the proof, we also needed to construct a small straight line segment

However, this is a non-issue. By the openness of

, we can pick an open ball

B(w, ε) ⊆ U , and we can certainly construct the straight line in this ball.

Finally, we get to the integration part. Suppose we picked all our

to be the

fixed straight line segment from

. Then for antiderivative to be differentiable,

we needed

∗δ

f(z) dz =

w+h

f(z) dz.

In other words, we needed to the integral along the path

∗ δ

∗

(

−γ

w+h

) to

vanish. This is a rather simple kind of paths. It is just (the boundary of) a

triangle, consisting of three line segments.

Definition

(Triangle)

A triangle in a domain

is what it ought to be — the

Euclidean convex hull of 3 points in

, lying wholly in

. We write its boundary

as ∂T , which we view as an oriented piecewise C

path, i.e. a contour.

good

bad

very bad

Our earlier result on constructing antiderivative then shows:

Proposition. If U is a star domain, and f : U → C is continuous, and if

∂T

f(z) dz = 0

for all triangles T ⊆ U, then f has an antiderivative on U.

Proof. As before, taking γ

= [a

, w] ⊆ U if U is star-shaped about a

This is in some sense a weaker proposition — while our hypothesis only

requires the integral to vanish over triangles, and not arbitrary closed loops, we

are restricted to star domains only.

But well, this is technically a weakening, but how is it useful? Surely if we

can somehow prove that the integral of a particular function vanishes over all

triangles, then we can easily modify the proof such that it works for all possible

curves.

Turns out, it so happens that for triangles, we can fiddle around with some

geometry to prove the following result:

Theorem

(Cauchy’s theorem for a triangle)

Let

be a domain, and let

f : U → C be holomorphic. If T ⊆ U is a triangle, then

∂T

f(z) dz = 0.

So for holomorphic functions, the hypothesis of the previous theorem auto-

matically holds.

We immediately get the following corollary, which is what we will end up

using most of the time.

Corollary

(Convex Cauchy)

is a convex or star-shaped domain, and

U → C

is holomorphic, then for any closed piecewise

paths

γ ∈ U

, we

must have

f(z) dz = 0.

Proof of corollary.

is holomorphic, then Cauchy’s theorem says the integral

over any triangle vanishes. If

is star shaped, our proposition says

has an

antiderivative. Then the fundamental theorem of calculus tells us the integral

around any closed path vanishes.

Hence, all we need to do is to prove that fact about triangles.

Proof of Cauchy’s theorem for a triangle. Fix a triangle T. Let

η =



∂T

f(z) dz



, ` = length(∂T ).

The idea is to show to bound

, for every

ε >

0, and hence we must have

η = 0. To do so, we subdivide our triangles.

Before we start, it helps to motivate the idea of subdividing a bit. By

subdividing the triangle further and further, we are focusing on a smaller and

smaller region of the complex plane. This allows us to study how the integral

behaves locally. This is helpful since we are given that

is holomorphic, and

holomorphicity is a local property.

We start with T = T

We then add more lines to get

, T

(it doesn’t really matter which is

which).

We orient the middle triangle by the anti-clockwise direction. Then we have

∂T

f(z) dz =

a,b,c,d

∂T

f(z) dz,

since each internal edge occurs twice, with opposite orientation.

For this to be possible, if



∂T

f(z) dz



, then there must be some

subscript in {a, b, c, d} such that



∂T

f(z) dz



≥

We call this T

= T

. Then we notice ∂T

has length

length(∂T

) =

Iterating this, we obtain triangles

⊇ T

⊇ ···

such that



∂T

f(z) dz



≥

, length(∂T

) =

Now we are given a nested sequence of closed sets. By IB Metric and Topological

Spaces (or IB Analysis II), there is some z

∈

i≥0

Now fix an

ε >

0. Since

is holomorphic at

, we can find a

δ >

0 such that

|f(w) − f(z

) − (w − z

)| ≤ ε|w − z

whenever

|w − z

| < δ

. Since the diameters of the triangles are shrinking each

time, we can pick an

such that

⊆ B

(

, ε

). We’re almost there. We just

need to do one last thing that is slightly sneaky. Note that

∂T

1 dz = 0 =

∂T

z dz,

since these functions certainly do have anti-derivatives on

. Therefore, noting

that f (z

) and f

) are just constants, we have



∂T

f(z) dz



∂T

(f(z) − f(z

) − (z − z

)) dz



≤

∂T

|f(z) − f(z

) − (z − z

)| dz

≤ length(∂T

)ε sup

z∈∂T

|z − z

≤ ε length(∂T

)

where the last line comes from the fact that

∈ T

, and the distance between

any two points in the triangle cannot be greater than the perimeter of the

triangle. Substituting our formulas for these in, we have

≤

ε.

η ≤ `

ε.

Since ` is fixed and ε was arbitrary, it follows that we must have η = 0.

Is this the best we can do? Can we formulate this for an arbitrary domain,

and not just star-shaped ones? It is obviously not true if the domain is not

simply connected, e.g. for

(

) =

defined on

C \ {

}

. However, it turns out

Cauchy’s theorem holds as long as the domain is simply connected, as we will

show in a later part of the course. However, this is not surprising given the

Riemann mapping theorem, since any simply connected domain is conformally

equivalent to the unit disk, which is star-shaped (and in fact convex).

We can generalize our result when

U → C

is continuous on the whole

, and holomorphic except on finitely many points. In this case, the same

conclusion holds —

f(z) dz = 0 for all piecewise smooth closed γ.

Why is this? In the proof, it was sufficient to focus on showing

∂T

(

) d

= 0

for a triangle

T ⊆ U

. Consider the simple case where we only have a single point

of non-holomorphicity a ∈ T . The idea is again to subdivide.

We call the center triangle

. Along all other triangles in our subdivision, we

get

f(z) dz = 0, as these triangles lie in a region where f is holomorphic. So

∂T

f(z) dz =

∂T

f(z) dz.

Note now that we can make T

as small as we like. But



∂T

f(z) dz



≤ length(∂T

) sup

z∈∂T

|f(z)|.

Since

is continuous, it is bounded. As we take smaller and smaller subdivisions,

length(∂T

) → 0. So we must have

∂T

f(z) dz = 0.

From here, it’s straightforward to conclude the general case with many points

of non-holomorphicity — we can divide the triangle in a way such that each

small triangle contains one bad point.

2.3 The Cauchy integral formula

Our next amazing result will be Cauchy’s integral formula. This formula allows

us to find the value of

inside a ball

(

, r

) just given the values of

on the

boundary ∂B(z

, r).

Theorem

(Cauchy integral formula)

Let

be a domain, and

U → C

holomorphic. Suppose there is some

B(z

; r) ⊆ U

for some

and

r >

0. Then

for all z ∈ B(z

; r), we have

f(z) =

2πi

∂B(z

;r)

f(w)

w − z

dw.

Recall that we previously computed

∂B(0,1)

= 2

πi

. This is indeed a

special case of the Cauchy integral formula. We will provide two proofs. The

first proof relies on the above generalization of Cauchy’s theorem.

Proof.

Since

is open, there is some

δ >

0 such that

B(z

; r + δ) ⊆ U

. We

define g : B(z

; r + δ) → C by

g(w) =

(

f(w)−f (z)

w−z

w 6= z

(z) w = z

where we have fixed

z ∈ B

(

;

) as in the statement of the theorem. Now

note that

is holomorphic as a function of

w ∈ B

(

, r

), except perhaps at

. But since

is holomorphic, by definition

is continuous everywhere on

B(z

, r + δ). So the previous result says

∂B(z

;r)

g(w) dw = 0.

This is exactly saying that

∂B(z

;r)

f(w)

w − z

dw =

∂B(z

;r)

f(z)

w − z

dw.

We now rewrite

w − z

1 −



z−z

w−z



∞

n=0

(z − z

)

(w − z

)

n+1

Note that this sum converges uniformly on ∂B(z

; r) since



z − z

w − z



< 1

for w on this circle.

By uniform convergence, we can exchange summation and integration. So

∂B(z

;r)

f(w)

w − z

dw =

∞

n=0

∂B(z

,r)

f(z)

(z − z

)

(w − z

)

n+1

dw.

We note that

(

)(

z −z

)

is just a constant, and that we have previously proven

∂B(z

;r)

(w − z

)

dw =

(

2πi k = −1

0 k 6= −1

So the right hand side is just 2πif(z). So done.

Corollary

(Local maximum principle)

Let

(

z, r

)

→ C

be holomorphic.

Suppose

(

)

| ≤ |f

(

)

for all

w ∈ B

(

;

). Then

is constant. In other words,

a non-constant function cannot achieve an interior local maximum.

Proof. Let 0 < ρ < r. Applying the Cauchy integral formula, we get

|f(z)| =



2πi

∂B(z;ρ)

f(w)

w − z



Setting w = z + ρe

2πiθ

, we get



f(z + ρe

2πiθ

) dθ



≤ sup

|z−w|=ρ

|f(w)|

≤ f(z).

So we must have equality throughout. When we proved the supremum bound

for the integral, we showed equality can happen only if the integrand is constant.

(

)

is constant on the circle

|z − w|

, and is equal to

(

). Since this

is true for all

ρ ∈

, r

), it follows that

|f|

is constant on

(

;

). Then the

Cauchy–Riemann equations then entail that

must be constant, as you have

shown in example sheet 1.

Going back to the Cauchy integral formula, recall that we had

B(z

; r) ⊆ U

f : U → C holomorphic, and we want to show

f(z) =

2πi

∂B(z

;r)

f(w)

w − z

dw.

When we proved it last time, we remember we know how to integrate things of

the form

(w−z

)

, and manipulated the formula such that we get the integral is

made of things like this.

The second strategy is to change the contour of integration instead of changing

the integrand. If we can change it so that the integral is performed over a circle

around z instead of z

, then we know what to do.

Proof.

(of Cauchy integral formula again) Given

ε >

0, we pick

δ >

0 such that

B(z, δ) ⊆ B

(

, r

), and such that whenever

|w − z| < δ

, then

(

)

− f

(

)

| < ε

This is possible since

is uniformly continuous on the neighbourhood of

. We

now cut our region apart:

We know

f(w)

w−z

is holomorphic on sufficiently small open neighbourhoods of

the half-contours indicated. The area enclosed by the contours might not be

star-shaped, but we can definitely divide it once more so that it is. Hence the

integral of

f(w)

w−z

around the half-contour vanishes by Cauchy’s theorem. Adding

these together, we get

∂B(z

,r)

f(w)

w − z

dw =

∂B(z,δ)

f(w)

w − z

dw,

where the balls are both oriented anticlockwise. Now we have



f(z) −

2πi

∂B(z

,r)

f(w)

w − z



f(z) −

2πi

∂B(z,δ)

f(w)

w − z



Now we once again use the fact that

∂B(z,δ)

w − z

dz = 2πi

to show this is equal to



2πi

∂B(z,δ)

f(z) − f(w)

w − z



≤

2π

· 2πδ ·

· ε = ε.

Taking ε → 0, we see that the Cauchy integral formula holds.

Note that the subdivision we did above was something we can do in general.

Definition

(Elementary deformation)

Given a pair of

-smooth (or piecewise

smooth) closed paths

φ, ψ

: [0

→ U

, we say

is an elementary deformation of

if there exists convex open sets

, ··· , C

⊆ U

and a division of the interval

0 =

< x

< ··· < x

= 1 such that on [

i−1

, x

], both

(

) and

(

) belong

to C

φ(x

i−1

)

φ(x

)

φ(x

i−1

)

φ(x

)

Then there are straight lines

(

)

→ ψ

(

) lying inside

. If

is holomor-

phic on U, considering the shaded square, we find

f(z) dz =

f(z) dz

when φ and ψ are convex deformations.

We now explore some classical consequences of the Cauchy Integral formula.

The next is Liouville’s theorem, as promised.

Theorem

(Liouville’s theorem)

Let

C → C

be an entire function (i.e.

holomorphic everywhere). If f is bounded, then f is constant.

This, for example, means there is no interesting holomorphic period functions

like sin and cos that are bounded everywhere.

Proof.

Suppose

(

)

| ≤ M

for all

z ∈ C

. We fix

, z

∈ C

, and estimate

|f(z

) − f(z

)| with the integral formula.

Let R > max{2|z

|, 2|z

|}. By the integral formula, we know

|f(z

) − f(z

)| =



2πi

∂B(0,R)



f(w)

w − z

−

f(w)

w − z





2πi

∂B(0,R)

f(w)(z

− z

)

(w − z

)(w − z

)



≤

2π

· 2πR ·

M|z

− z

(R/2)

4M|z

− z

Note that we get the bound on the denominator since

|w|

implies

|w−z

| >

by our choice of

. Letting

R → ∞

, we know we must have

(

) =

(

). So

is constant.

Corollary

(Fundamental theorem of algebra)

A non-constant complex polyno-

mial has a root in C.

Proof. Let

P (z) = a

+ a

n−1

+ ··· + a

where

= 0 and

n >

0. So

is non-constant. Thus, as

|z| → ∞

(

)

| → ∞

In particular, there is some R such that for |z| > R, we have |P (z)| ≥ 1.

Now suppose for contradiction that

does not have a root in

. Then

consider

f(z) =

P (z)

which is then an entire function, since it is a rational function. On

B(0, R)

, we

know

is certainly continuous, and hence bounded. Outside this ball, we get

(

)

| ≤

1. So

(

) is constant, by Liouville’s theorem. But

is non-constant.

This is absurd. Hence the result follows.

There are many many ways we can prove the fundamental theorem of algebra.

However, none of them belong wholely to algebra. They all involve some analysis

or topology, as you might encounter in the IID Algebraic Topology and IID

Riemann Surface courses.

This is not surprising since the construction of

, and hence

, is intrinsically

analytic — we get from

by requiring it to have additive inverses;

by requiring multiplicative inverses;

by requiring the root to

+ 1 = 0.

These are all algebraic. However, to get from

, we are requiring something

about convergence in

. This is not algebraic. It requires a particular of metric

on Q. If we pick a different metric, then you get a different completion, as you

may have seen in IB Metric and Topological Spaces. Hence the construction of

R is actually analytic, and not purely algebraic.

2.4 Taylor’s theorem

When we first met Taylor series, we were happy, since we can express anything

as a power series. However, we soon realized this is just a fantasy — the Taylor

series of a real function need not be equal to the function itself. For example,

the function

(

) =

−x

−2

has vanishing Taylor series at 0, but does not vanish

in any neighbourhood of 0. What we do have is Taylor’s theorem, which gives

you an expression for what the remainder is if we truncate our series, but is

otherwise completely useless.

In the world of complex analysis, we are happy once again. Every holomorphic

function can be given by its Taylor series.

Theorem

(Taylor’s theorem)

Let

(

a, r

)

→ C

be holomorphic. Then

has

a convergent power series representation

f(z) =

∞

n=0

(z − a)

on B(a, r). Moreover,

(n)

(a)

2πi

∂B(a,ρ)

f(z)

(z − a)

n+1

for any 0 < ρ < r.

Note that the very statement of the theorem already implies any holomorphic

function has to be infinitely differentiable. This is a good world.

Proof. We’ll use Cauchy’s integral formula. If |w − a| < ρ < r, then

f(w) =

2πi

∂B(a,ρ)

f(z)

z − w

dz.

Now (cf. the first proof of the Cauchy integral formula), we note that

z − w

(z − a)



1 −

w−a

z−a



n=0

(w − a)

(z − a)

n+1

This series is uniformly convergent everywhere on the

disk, including its

boundary. By uniform convergence, we can exchange integration and summation

to get

f(w) =

∞

n=0

2πi

∂B(a,ρ)

f(z)

(z − a)

n+1

(w − a)

∞

n=0

(w − a)

Since

does not depend on

, this is a genuine power series representation,

and this is valid on any disk B(a, ρ) ⊆ B(a, r).

Then the formula for

in terms of the derivative comes for free since that’s

the formula for the derivative of a power series.

This tells us every holomorphic function behaves like a power series. In

particular, we do not get weird things like

−x

−2

that have a trivial Taylor

series expansion, but is itself non-trivial. Similarly, we know that there are no

“bump functions” on

that are non-zero only on a compact set (since power

series don’t behave like that). Of course, we already knew that from Liouville’s

theorem.

Corollary.

(

a, r

)

→ C

is holomorphic on a disc, then

is infinitely

differentiable on the disc.

Proof.

Complex power series are infinitely differentiable (and

had better be

infinitely differentiable for us to write down the formula for

in terms of

(n)

This justifies our claim from the very beginning that

(

) and

(

) are

harmonic functions if f is holomorphic.

Corollary.

U → C

is a complex-valued function, then

holomorphic at

p ∈ U

if and only if

u, v

satisfy the Cauchy–Riemann equations,

and that u

, u

, v

are continuous in a neighbourhood of p.

Proof.

, u

, v

exist and are continuous in an open neighbourhood of

then

and

are differentiable as functions

→ R

, and then we proved

that the Cauchy–Riemann equations imply differentiability at each point in the

neighbourhood of p. So f is differentiable at a neighbourhood of p.

On the other hand, if

is holomorphic, then it is infinitely differentiable. In

particular,

(

) is also holomorphic. So

, u

, v

are differentiable, hence

continuous.

We also get the following (partial) converse to Cauchy’s theorem.

Corollary

(Morera’s theorem)

Let

U ⊆ C

be a domain. Let

U → C

continuous such that

f(z) dz = 0

for all piecewise-C

closed curves γ ∈ U. Then f is holomorphic on U.

Proof.

We have previously shown that the condition implies that

has an

antiderivative

U → C

, i.e.

is a holomorphic function such that

But F is infinitely differentiable. So f must be holomorphic.

Recall that Cauchy’s theorem required

to be sufficiently nice, e.g. being

star-shaped or just simply-connected. However, Morera’s theorem does not. It

just requires that

is a domain. This is since holomorphicity is a local property,

while vanishing on closed curves is a global result. Cauchy’s theorem gets us

from a local property to a global property, and hence we need to assume more

about what the “globe” looks like. On the other hand, passing from a global

property to a local one does not. Hence we have this asymmetry.

Corollary.

Let

U ⊆ C

be a domain,

;

U → C

be a holomorphic function. If

→ f uniformly, then f is in fact holomorphic, and

(z) = lim

(z).

Proof. Given a piecewise C

path γ, uniformity of convergence says

(z) dz →

f(z) dz

uniformly. Since

being holomorphic is a local condition, so we fix

p ∈ U

and

work in some small, convex disc

(

p, ε

)

⊆ U

. Then for any curve

inside this

disk, we have

(z) dz = 0.

Hence we also have

(

) d

= 0. Since this is true for all curves, we conclude

is holomorphic inside

(

p, ε

) by Morera’s theorem. Since

was arbitrary, we

know f is holomorphic.

We know the derivative of the limit is the limit of the derivative since we can

express f

(a) in terms of the integral of

f(z)

(z−a)

, as in Taylor’s theorem.

There is a lot of passing between knowledge of integrals and knowledge of

holomorphicity all the time, as we can see in these few results. These few sections

are in some sense the heart of the course, where we start from Cauchy’s theorem

and Cauchy’s integral formula, and derive all the other amazing consequences.

2.5 Zeroes

Recall that for a polynomial

(

), we can talk about the order of its zero at

by looking at the largest power of (

z −a

) dividing

. A priori, it is not

clear how we can do this for general functions. However, given that everything

is a Taylor series, we know how to do this for holomorphic functions.

Definition

(Order of zero)

Let

(

a, r

)

→ C

be holomorphic. Then we

know we can write

f(z) =

∞

n=0

(z − a)

as a convergent power series. Then either all

= 0, in which case

= 0 on

(

a, r

), or there is a least

such that

= 0 (

is just the smallest

such

that f

(n)

(a) 6= 0).

If N > 0, then we say f has a zero of order N.

If f has a zero of order N at a, then we can write

f(z) = (z − a)

g(z)

on B(a, r), where g(a) = c

6= 0.

Often, it is not the actual order that is too important. Instead, it is the

ability to factor f in this way. One of the applications is the following:

Lemma

(Principle of isolated zeroes)

Let

(

a, r

)

→ C

be holomorphic and

not identically zero. Then there exists some 0

< ρ < r

such that

(

)

= 0 in the

punctured neighbourhood B(a, ρ) \ {a}.

Proof. If f (a) 6= 0, then the result is obvious by continuity of f.

The other option is not too different. If

has a zero of order

, then

we can write

(

) = (

z − a

)

(

) with

(

)

= 0. By continuity of

does

not vanish on some small neighbourhood of

, say

(

a, ρ

). Then

(

) does not

vanish on B(a, ρ) \ {a}.

A consequence is that given two holomorphic functions on the same domain,

if they agree on sufficiently many points, then they must in fact be equal.

Corollary

(Identity theorem)

Let

U ⊆ C

be a domain, and

f, g

U → C

holomorphic. Let

{z ∈ U

(

) =

(

)

}

. Suppose

contains a non-isolated

point, i.e. there exists some

w ∈ S

such that for all

ε >

S ∩ B

(

w, ε

)

{w}

Then f = g on U .

Proof.

Consider the function

(

) =

(

)

− g

(

). Then the hypothesis says

(

)

has a non-isolated zero at

, i.e. there is no non-punctured neighbourhood of

on which

is non-zero. By the previous lemma, this means there is some

ρ >

such that h = 0 on B(w, ρ) ⊆ U.

Now we do some topological trickery. We let

= {a ∈ U : h = 0 on some neighbourhood B(a, ρ) of a in U},

= {a ∈ U : there exists n ≥ 0 such that h

(n)

6= 0}.

Clearly,

∩U

∅

, and the existence of Taylor expansions shows

∪U

Moreover,

is open by definition, and

is open since

(n)

(

) is continuous

near any given

a ∈ U

. Since

is (path) connected, such a decomposition can

happen if one of

and

is empty. But

w ∈ U

. So in fact

, i.e.

vanishes on the whole of U. So f = g.

In particular, if two holomorphic functions agree on some small open subset

of the domain, then they must in fact be identical. This is a very strong result,

and is very false for real functions. Hence, to specify, say, an entire function, all

we need to do is to specify it on an arbitrarily small domain we like.

Definition

(Analytic continuiation)

Let

⊆ U ⊆ C

be domains, and

→

be holomorphic. An analytic continuation of

is a holomorphic function

h : U → C such that h|

= f, i.e. h(z) = f(z) for all z ∈ U

By the identity theorem, we know the analytic continuation is unique if it

exists.

Thus, given any holomorphic function

U → C

, it is natural to ask how

far we can extend the domain, i.e. what is the largest

⊇ U

such that there is

an analytic continuation of f to U

There is no general method that does this for us. However, one useful trick

is to try to write our function

in a different way so that it is clear how we can

extend it to elsewhere.

Example. Consider the function

f(z) =

n≥0

= 1 + z + z

+ ···

defined on B(0, 1).

By itself, this series diverges for

outside

1). However, we know well

that this function is just

f(z) =

1 − z

This alternative representation makes sense on the whole of

except at

= 1.

So we see that

has an analytic continuation to

C \ {

}

. There is clearly no

extension to the whole of C, since it blows up near z = 1.

Example. Alternatively, consider

f(z) =

n≥0

Then this again converges on

1). You will show in example sheet 2 that

there is no analytic continuation of f to any larger domain.

Example. The Riemann zeta function

ζ(z) =

∞

n=1

−z

defines a holomorphic function on

(

)

} ⊆ C

. Indeed, we have

−z

Re(z)

, and we know

−t

converges for

t ∈ R

, and in fact does so

uniformly on any compact domain. So the corollary of Morera’s theorem tells us

that ζ(z) is holomorphic on Re(z) > 1.

We know this cannot converge as

z →

1, since we approach the harmonic

series which diverges. However, it turns out

(

) has an analytic continuation to

C \ {1}. We will not prove this.

At least formally, using the fundamental theorem of arithmetic, we can

expand n as a product of its prime factors, and write

ζ(z) =

primes p

(1 + p

−z

+ p

−2z

+ ···) =

primes p

1 − p

−z

If there were finitely many primes, then this would be a well-defined function

on all of

, since this is a finite product. Hence, the fact that this blows up at

z = 1 implies that there are infinitely many primes.

2.6 Singularities

The next thing to study is singularities of holomorphic functions. These are

places where the function is not defined. There are many ways a function can

be ill-defined. For example, if we write

f(z) =

1 − z

then on the face of it, this function is not defined at

= 1. However, elsewhere,

is just the constant function 1, and we might as well define

(1) = 1. Then we

get a holomorphic function. These are rather silly singularities, and are singular

solely because we were not bothered to define f there.

Some singularities are more interesting, in that they are genuinely singular.

For example, the function

f(z) =

1 − z

is actually singular at

= 1, since

is unbounded near the point. It turns out

these are the only possibilities.

Proposition

(Removal of singularities)

Let

be a domain and

∈ U

. If

U \{z

} → C

is holomorphic, and

is bounded near

, then there exists an

a such that f(z) → a as z → z

Furthermore, if we define

g(z) =

(

f(z) z ∈ U \{z

}

a z = z

then g is holomorphic on U.

Proof. Define a new function h : U → C by

h(z) =

(

(z − z

)

f(z) z 6= z

0 z = z

Then since

is holomorphic away from

, we know

is also holomorphic away

from z

Also, we know

is bounded near

. So suppose

(

)

| < M

in some

neighbourhood of z

. Then we have



h(z) − h(z

)

z − z



≤ |z − z

|M.

So in fact

is also differentiable at

, and

(

) =

(

) = 0. So near

has a Taylor series

h(z) =

n≥0

(z − z

)

Since we are told that a

= a

= 0, we can define a g(z) by

g(z) =

n≥0

n+2

(z − z

)

defined on some ball

(

, ρ

), where the Taylor series for

is defined. By

construction, on the punctured ball

(

, ρ

)

\{z

}

, we get

(

) =

(

). Moreover,

g(z) → a

as z → z

. So f(z) → a

as z → z

Since g is a power series, it is holomorphic. So the result follows.

This tells us the only way for a function to fail to be holomorphic at an

isolated point is that it blows up near the point. This won’t happen because

fails to be continuous in some weird ways.

However, we are not yet done with our classification. There are many ways

in which things can blow up. We can further classify these into two cases — the

case where

(

)

| → ∞

z → z

, and the case where

(

)

does not converge as

z → z

. It happens that the first case is almost just as boring as the removable

ones.

Proposition.

Let

be a domain,

∈ U

and

U \{z

} → C

be holomorphic.

Suppose

(

)

| → ∞

z → z

. Then there is a unique

k ∈ Z

≥1

and a unique

holomorphic function g : U → C such that g(z

) 6= 0, and

f(z) =

g(z)

(z − z

)

Proof.

We shall construct

near

in some small neighbourhood, and then

apply analytic continuation to the whole of

. The idea is that since

(

) blows

up nicely as z → z

, we know

f(z)

behaves sensibly near z

We pick some

δ >

0 such that

(

)

| ≥

1 for all

z ∈ B

(

;

)

\ {z

}

. In

particular, f (z) is non-zero on B(z

; δ) \ {z

}. So we can define

h(z) =

(

f(z)

z ∈ B(z

; δ) \ {z

}

0 z = z

Since

f(z)

| ≤

1 on

(

;

)

\{z

}

, by the removal of singularities,

is holomorphic

(

, δ

). Since

vanishes at the

, it has a unique definite order at

, i.e.

there is a unique integer

k ≥

1 such that

has a zero of order

. In other

words,

h(z) = (z − z

)

`(z),

for some holomorphic ` : B(z

; δ) → C and `(z

) 6= 0.

Now by continuity of

, there is some 0

< ε < δ

such that

(

)

= 0 for all

z ∈ B(z

, ε). Now define g : B(z

; ε) → C by

g(z) =

`(z)

Then g is holomorphic on this disc.

By construction, at least away from z

, we have

g(z) =

`(z)

h(z)

· (z −z

)

= (z − z

)

f(z).

was initially defined on

(

;

)

→ C

, but now this expression certainly makes

sense on all of

. So

admits an analytic continuation from

(

;

) to

. So

done.

We can start giving these singularities different names. We start by formally

defining what it means to be a singularity.

Definition

(Isolated singularity)

Given a domain

and

∈ U

, and

U \ {z

} → C holomorphic, we say z

is an isolated singularity of f .

Definition

(Removable singularity)

A singularity

is a removable singu-

larity if f is bounded near z

Definition

(Pole)

A singularity

is a pole of order

(

)

| → ∞

z → z

and one can write

f(z) =

g(z)

(z − z

)

with g : U → C, g(z

) 6= 0.

Definition

(Isolated essential singularity)

An isolated singularity is an isolated

essential singularity if it is neither removable nor a pole.

It is easy to give examples of removable singularities and poles. So let’s look

at some essential singularities.

Example. z 7→ e

1/z

has an isolated essential singularity at z = 0.

Note that if

(

, ε

)

\{z

} → C

has a pole of order

, then

naturally

defines a map

f : B(z

; ε) → CP

= C ∪ {∞}, the Riemann sphere, by

f(z) =

(

∞ z = z

f(z) z 6= z

This is then a “continuous” function. So a singularity is just a point that gets

mapped to the point ∞.

As was emphasized in IA Groups, the point at infinity is not a special point

in the Riemann sphere. Similarly, poles are also not really singularities from the

viewpoint of the Riemann sphere. It’s just that we are looking at it in a wrong

way. Indeed, if we change coordinates on the Riemann sphere so that we label

each point

w ∈ CP

instead, then

just maps

to 0 under the new

coordinate system. In particular, at the point

, we find that

is holomorphic

and has an innocent zero of order k.

Since poles are not bad, we might as well allow them.

Definition

(Meromorphic function)

is a domain and

S ⊆ U

is a finite or

discrete set, a function

U \ S → C

which is holomorphic and has (at worst)

poles on S is said to be meromorphic on U.

The requirement that

is discrete is so that each pole in

is actually an

isolated singularity.

Example.

A rational function

P (z)

Q(z)

, where

P, Q

are polynomials, is holomorphic

C \ {z

(

) = 0

}

, and meromorphic on

. More is true — it is in fact

holomorphic as a function CP

→ CP

These ideas are developed more in depth in the IID Riemann Surfaces course.

As an aside, if we want to get an interesting holomorphic function with

domain

, its image must contain the point

∞

, or else its image will be

a compact subset of

(since

is compact), thus bounded, and therefore

constant by Liouville’s theorem.

At this point, we really should give essential singularities their fair share of

attention. Not only are they bad. They are bad spectacularly.

Theorem

(Casorati-Weierstrass theorem)

Let

be a domain,

∈ U

, and

suppose

U \ {z

} → C

has an essential singularity at

. Then for all

w ∈ C

there is a sequence z

→ z

such that f (z

) → w.

In other words, on any punctured neighbourhood

(

;

)

\ {z

}

, the image

of f is dense in C.

This is not actually too hard to proof.

Proof. See example sheet 2.

If you think that was bad, actually essential singularities are worse than that.

The theorem only tells us the image is dense, but not that we will hit every

point. It is in fact not true that every point will get hit. For example

can

never be zero. However, this is the worst we can get

Theorem

(Picard’s theorem)

has an isolated essential singularity at

, then

there is some

b ∈ C

such that on each punctured neighbourhood

(

;

)

\ {z

}

the image of f contains C \ {b}.

The proof is beyond this course.

2.7 Laurent series

If f is holomorphic at z

, then we have a local power series expansion

f(z) =

∞

n=0

(z − z

)

near

. If

is singular at

(and the singularity is not removable), then there

is no hope we can get a Taylor series, since the existence of a Taylor series would

imply f is holomorphic at z = z

However, it turns out we can get a series expansion if we allow ourselves to

have negative powers of z.

Theorem (Laurent series). Let 0 ≤ r < R < ∞, and let

A = {z ∈ C : r < |z − a| < R}

denote an annulus on C.

Suppose

A → C

is holomorphic. Then

has a (unique) convergent series

expansion

f(z) =

∞

n=−∞

(z − a)

where

2πi

∂B(a,ρ)

f(z)

(z − a)

n+1

for

r < ρ < R

. Moreover, the series converges uniformly on compact subsets of

the annulus.

The Laurent series provides another way of classifying singularities. In the

case where r = 0, we just have

f(z) =

∞

−∞

(z − a)

on B(a, R) \ {a}, then we have the following possible scenarios:

(i) c

= 0 for all

n <

0. Then

is bounded near

, and hence this is a

removable singularity.

(ii)

Only finitely many negative coefficients are non-zero, i.e. there is a

k ≥

such that

= 0 for all

n < −k

and

−k

= 0. Then

has a pole of order

k at a.

(iii)

There are infinitely many non-zero negative coefficients. Then we have an

isolated essential singularity.

So our classification of singularities fit nicely with the Laurent series expansion.

We can interpret the Laurent series as follows — we can write

f(z) = f

(z) + f

out

(z),

where

consists of the terms with positive power and

out

consists of those with

negative power. Then

is the part that is holomorphic on the disk

|z −a| < R

while

out

(

) is the part that is holomorphic on

|z − a| > r

. These two combine

to give an expression holomorphic on

r < |z − a| < R

. This is just a nice way

of thinking about it, and we will not use this anywhere. So we will not give a

detailed proof of this interpretation.

Proof.

The proof looks very much like the blend of the two proofs we’ve given for

the Cauchy integral formula. In one of them, we took a power series expansion

of the integrand, and in the second, we changed our contour by cutting it up.

This is like a mix of the two.

Let w ∈ A. We let r < ρ

< |w − a| < ρ

< R.

˜γ

We let ˜γ be the contour containing w, and

˜γ be the other contour.

Now we apply the Cauchy integral formula to say

f(w) =

2πi

˜γ

f(z)

z − w

and

0 =

2πi

˜γ

f(z)

z − w

dz.

So we get

f(w) =

2πi

∂B(a,ρ

)

f(z)

z − w

dz −

2πi

∂B(a,ρ

)

f(z)

z − w

dz.

As in the first proof of the Cauchy integral formula, we make the following

expansions: for the first integral, we have w − a < z − a. So

z − w

z − a

1 −

w−a

z−a

∞

n=0

(w − a)

(z − a)

n+1

which is uniformly convergent on z ∈ ∂B(a, ρ

For the second integral, we have w − a > z − a. So

−1

z − w

w − a

1 −

z−a

w−a

∞

m=1

(z − a)

m−1

(w − a)

which is uniformly convergent for z ∈ ∂B(a, ρ

By uniform convergence, we can swap summation and integration. So we get

f(w) =

∞

n=0

2πi

∂B(a,ρ

)

f(z)

(z − a)

n+1

(w − a)

∞

m=1

2πi

∂B(a,ρ

)

f(z)

(z − a)

−m+1

(w − a)

−m

Now we substitute n = −m in the second sum, and get

f(w) =

∞

n=−∞

˜c

(w − a)

for the integrals

˜c

. However, some of the coefficients are integrals around the

circle, while the others are around the

circle. This is not a problem. For

any

r < ρ < R

, these circles are convex deformations of

|z − a|

inside the

annulus A. So

∂B(a,ρ)

f(z)

(z − a)

n+1

is independent of ρ as long as ρ ∈ (r, R). So we get the result stated.

Definition

(Principal part)

(

a, r

)

\ {a} → C

is holomorphic and if

has Laurent series

f(z) =

∞

n=−∞

(z − a)

then the principal part of f at a is

principal

−1

n=−∞

(z − a)

f − f

principal

is holomorphic near

, and

principal

carries the information

of what kind of singularity f has at a.

When we talked about Taylor series, if

(

a, r

)

→ C

is holomorphic

with Taylor series

(

) =

∞

n=0

(

z − a

)

, then we had two possible ways of

expressing the coefficients of c

. We had

2πi

∂B(a,ρ)

f(z)

(z − a)

n+1

dz =

(n)

(a)

In particular, the second expansion makes it obvious the Taylor series is uniquely

determined by f .

For the Laurent series, we cannot expect to have a simple expression of the

coefficients in terms of the derivatives of the function, for the very reason that

is not even defined, let alone differentiable, at

. So is the Laurent series unique?

Lemma. Let f : A → C be holomorphic, A = {r < |z −a| < R}, with

f(z) =

∞

n=−∞

(z − a)

Then the coefficients c

are uniquely determined by f .

Proof. Suppose also that

f(z) =

∞

n=−∞

(z − a)

Using our formula for c

, we know

2πic

∂B(a,ρ)

f(z)

(z − a)

k+1

∂B(a,ρ)

(z − a)

n−k−1

∂B(a,ρ)

(z − a)

n−k−1

= 2πib

So c

= b

While we do have uniqueness, we still don’t know how to find a Laurent

series. For a Taylor series, we can just keep differentiating and then get the

coefficients. For Laurent series, the above integral is often almost impossible to

evaluate. So the technique to compute a Laurent series is blind guesswork.

Example. We know

sin z = z −

− ···

defines a holomorphic function, with a radius of convergence of

∞

. Now consider

cosec z =

sin z

which is holomorphic except for

kπ

, with

k ∈ Z

. So

cosec z

has a Laurent

series near z = 0. Using

sin z = z



1 −

+ O(z

)



we get

cosec z =



1 +

+ O(z

)



From this, we can read off that the Laurent series has

= 0 for all

n ≤ −

−1

= 1,

. If we want, we can go further, but we already see that

cosec

has a simple pole at z = 0.

By periodicity, cosec has a simple pole at all other singularities.

Example. Consider instead

sin





−

3!z

5!z

− ··· .

We see this is holomorphic on

∗

, with

= 0 for infinitely many

n <

0. So this

has an isolated essential singularity.

Example.

Consider

cosec





. This has singularities at

kπ

for

k ∈ N

{

, ···}

. So it is not holomorphic at any punctured neighbourhood

, r

)

{

}

of zero. So this has a non-isolated singularity at zero, and there is no Laurent

series in a neighbourhood of zero.

We’ve already done most of the theory. In the remaining of the course, we

will use these techniques to do stuff. We will spend most of our time trying to

evaluate integrals, but before that, we will have a quick look on how we can use

Laurent series to evalue some series.

Example (Series summation). We claim that

f(z) =

∞

n=−∞

(z − n)

is holomorphic on C \ Z, and moreover if we let

f(z) =

sin

(πz)

We will reserve the name

for the original series, and refer to the function

z 7→

sin

(πz)

as g instead, until we have proven that they are the same.

Our strategy is as follows — we first show that

(

) converges and is holo-

morphic, which is not hard, given the Weierstrass

-test and Morera’s theorem.

To show that indeed we have

(

) =

(

), we first show that they have equal

principal part, so that

(

)

− g

(

) is entire. We then show it is zero by proving

f − g

is bounded, hence constant, and that

(

)

− g

(

)

→

0 as

z → ∞

(in some

appropriate direction).

For any fixed

w ∈ C \ Z

, we can compare it with

and apply the

Weierstrass

-test. We pick

r >

0 such that

|w − n| >

for all

n ∈ Z

. Then

for all z ∈ B(w; r), we have

|z − n| ≥ max{r, n − |w|− r}.

Hence

|z − n|

≤ min



(n − |w| − r)



= M

By comparison to

, we know

converges. So by the Weierstrass

M-test, we know our series converges uniformly on B(w, r).

By our results around Morera’s theorem, we see that

is a uniform limit of

holomorphic functions

n=−N

(z−n)

, and hence holomorphic.

Since

was arbitrary, we know

is holomorphic on

C \ Z

. Note that we

do not say the sum converges uniformly on

C \ Z

. It’s just that for any point

w ∈ C \ Z

, there is a small neighbourhood of

on which the sum is uniformly

convergent, and this is sufficient to apply the result of Morera’s.

For the second part, note that

is periodic, since

(

+ 1) =

(

). Also, at

has a double pole, since

(

) =

+ holomorphic stuff near

= 0. So

has

a double pole at each

k ∈ Z

. Note that

sin

(πz)

also has a double pole at each

k ∈ Z.

Now, consider the principal parts of our functions — at

k ∈ Z

(

) has

principal part

(z−k)

. Looking at our previous Laurent series for cosec(z), if

g(z) =



sin πz



then

lim

z→0

(

) = 1. So

(

) must have the same principal part at 0 and

hence at k for all k ∈ Z.

Thus

(

) =

(

)

−g

(

) is holomorphic on

C \Z

. However, since its principal

part vanishes at the integers, it has at worst a removable singularity. Removing

the singularity, we know h(z) is entire.

Since we want to prove f(z) = g(z), we need to show h(z) = 0.

We first show it is boundedness. We know

and

are both periodic with

period 1. So it suffices to focus attention on the strip

−

≤ x = Re(z) ≤

To show this is bounded on the rectangle, it suffices to show that

(

)

→

y → ±∞

, by continuity. To do so, we show that

and

both vanish as

y → ∞.

So we set z = x + iy, with |x| ≤

. Then we have

|g(z)| ≤

4π

πy

− e

−πy

→ 0

as y → ∞. Exactly analogously,

|f(z)| ≤

n∈Z

|x + iy −n|

≤

+ 2

∞

n=1

(n −

)

+ y

→ 0

y → ∞

. So

is bounded on the strip, and tends to 0 as

y → ∞

, and is hence

constant by Liouville’s theorem. But if

h →

0 as

y → ∞

, then the constant

better be zero. So we get

h(z) = 0.

3 Residue calculus

3.1 Winding numbers

Recall that the type of the singularity of a point depends on the coefficients in

the Laurent series, and these coefficients play an important role in determining

the behaviour of the functions. Among all the infinitely many coefficients, it

turns out the coefficient of

−1

is the most important one, as we will soon see.

We call this the residue of f .

Definition

(Residue)

Let

(

a, r

)

\ {a} → C

be holomorphic, with Laurent

series

f(z) =

∞

n=−∞

(z − a)

Then the residue of f at a is

Res(f, a) = Res

(a) = c

−1

Note that if ρ < r, then by definition of the Laurent coefficients, we know

∂B(a,ρ)

f(z) dz = 2πic

−1

So we can alternatively write the residue as

Res

(a) =

2πi

∂B(a,ρ)

f(z) dz.

This gives us a formulation of the residue without reference to the Laurent series.

Deforming paths if necessary, it is not too far-fetching to imagine that for

any simple curve γ around the singularity a, we have

f(z) dz = 2πi Res(f, a).

Moreover, if the path actually encircles two singularities

and

, then deforming

the path, we would expect to have

f(z) dz = 2πi(Res(f, a) + Res(f, b)),

and this generalizes to multiple singularities in the obvious way.

If this were true, then it would be very helpful, since this turns integration

into addition, which is (hopefully) much easier!

Indeed, we will soon prove that this result holds. However, we first get rid of

the technical restriction that we only work with simple (i.e. non-self intersecting)

curves. This is completely not needed. We are actually not really worried in

the curve intersecting itself. The reason why we’ve always talked about simple

closed curves is that we want to avoid the curve going around the same point

many times.

There is a simple workaround to this problem — we consider arbitrary curves,

and then count how many times we are looping around the point. If we are

looping around it twice, then we count its contribution twice!

Indeed, suppose we have the following curve around a singularity:

We see that the curve loops around

twice. Also, by the additivity of the

integral, we can break this curve into two closed contours. So we have

2πi

f(z) dz = 2 Res

(a).

So what we want to do now is to define properly what it means for a curve to

loop around a point n times. This will be called the winding number.

There are many ways we can define the winding number. The definition we

will pick is based on the following observation — suppose, for convenience, that

the point in question is the origin. As we move along a simple closed curve around

0, our argument will change. If we keep track of our argument continuously,

then we will find that when we return to starting point, the argument would

have increased by 2

. If we have a curve that winds around the point twice,

then our argument will increase by 4π.

What we do is exactly the above — given a path, find a continuous function

that gives the “argument” of the path, and then define the winding number to

be the difference between the argument at the start and end points, divided by

2π.

For this to make sense, there are two important things to prove. First, we

need to show that there is indeed a continuous “argument” function of the curve,

in a sense made precise in the lemma below. Then we need to show the winding

number is well-defined, but that is easier.

Lemma.

Let

: [

a, b

]

→ C

be a continuous closed curve, and pick a point

w ∈ C \ image

(

). Then there are continuous functions

: [

a, b

]

→ R >

0 and

θ : [a, b] → R such that

γ(t) = w + r(t)e

iθ(t)

Of course, at each point

, we can find

and

such that the above holds.

The key point of the lemma is that we can do so continuously.

Proof.

Clearly

(

) =

|γ

(

)

− w|

exists and is continuous, since it is the composi-

tion of continuous functions. Note that this is never zero since

(

) is never

The actual content is in defining θ.

To define

(

), we for simplicity assume

= 0. Furthermore, by considering

instead the function

γ(t)

r(t)

, which is continuous and well-defined since

is never

zero, we can assume |γ(t)| = 1 for all t.

Recall that the principal branch of

log

, and hence of the argument

(

log

takes values in (−π, π) and is defined on C \ R

≤0

(

) always lied in, say, the right-hand half plane, we would have no problem

defining θ consistently, since we can just let

θ(t) = arg(γ(t))

for

arg

the principal branch. There is nothing special about the right-hand half

plane. Similarly, if γ lies in the region as shaded below:

i.e. we have

γ(t) ∈

z : Re



iα



> 0

for a fixed α, we can define

θ(t) = α + arg



γ(t)

iα



Since γ : [a, b] → C is continuous, it is uniformly continuous, and we can find a

subdivision

a = a

< a

< ··· < a

= b,

such that if

s, t ∈

[

i−1

, a

], then

|γ

(

)

− γ

(

)

| <

√

, and hence

(

) and

(

)

belong to such a half-plane.

So we define θ

: [a

j−1

, a

] → R such that

γ(t) = e

iθ

(t)

for t ∈ [a

j−1

, a

], and 1 ≤ j ≤ n − 1.

On each region [

j−1

, a

], this gives a continuous argument function. We

cannot immediately extend this to the whole of [

a, b

], since it is entirely possible

that

(

) =

j+1

(

). However, we do know that

(

) are both values of the

argument of

(

). So they must differ by an integer multiple of 2

, say 2

nπ

Then we can just replace

j+1

−

nπ

, which is an equally valid argument

function, and then the two functions will agree at a

Hence, for

j >

1, we can successively re-define

such that the resulting map

θ is continuous. Then we are done.

We can thus use this to define the winding number.

Definition

(Winding number)

Given a continuous path

: [

a, b

]

→ C

such

that γ(a) = γ(b) and w 6∈ image(γ), the winding number of γ about w is

θ(b) − θ(a)

2π

where

: [

a, b

]

→ R

is a continuous function as above. This is denoted by

(

γ, w

)

or n

(W ).

I and n stand for index and number respectively.

Note that we always have

(

γ, w

)

∈ Z

, since

(

) and

(

) are arguments

of the same number. More importantly,

(

γ, w

) is well-defined — suppose

(

) =

(

)

iθ

(t)

(

)

iθ

(t)

for continuous functions

, θ

: [

a, b

]

→ R

. Then

− θ

: [

a, b

]

→ R

is continuous, but takes values in the discrete set 2

πZ

. So it

must in fact be constant, and thus θ

(b) − θ

(a) = θ

(b) − θ

(a).

So far, what we’ve done is something that is true for arbitrary continuous

closed curve. However, if we focus on piecewise

-smooth closed path, then we

get an alternative expression:

Lemma.

Suppose

: [

a, b

]

→ C

is a piecewise

-smooth closed path, and

w 6∈ image(γ). Then

I(γ, w) =

2πi

z − w

dz.

Proof. Let γ(t) − w = r(t)e

iθ(t)

, with now r and θ piecewise C

-smooth. Then

z − w

dz =

(t)

γ(t) − w



(t)

r(t)

+ iθ

(t)



= [ln r(t) + iθ(t)]

= i(θ(b) − θ(a))

= 2πiI(γ, w).

So done.

In some books, this integral expression is taken as the definition of the

winding number. While this is elegant in complex analysis, it is not clear a priori

that this is an integer, and only works for piecewise

-smooth closed curves,

not arbitrary continuous closed curves.

On the other hand, what is evident from this expression is that

(

γ, w

) is

continuous as a function of

w ∈ C \ image

(

), since it is even holomorphic as a

function of

. Since

(

;

) is integer valued,

(

) must be locally constant on

path components of C \ image(γ).

We can quickly verify that this is a sensible definition, in that the winding

number around a point “outside” the curve is zero. More precisely, since

image

(

)

is compact, all points of sufficiently large modulus in

belong to one component

C \ image

(

). This is indeed the only path component of

C \ image

(

) that is

unbounded.

To find the winding number about a point in this unbounded component, note

that

(

;

) is consistent on this component, and so we can consider arbitrarily

larger w. By the integral formula,

|I(γ, w)| ≤

2π

length(γ) max

z∈γ

|w − z|

→ 0

w → ∞

. So it does vanish outside the curve. Of course, inside the other path

components, we can still have some interesting values of the winding number.

3.2 Homotopy of closed curves

The last ingredient we need before we can get to the residue theorem is the

idea of homotopy. Recall we had this weird, ugly definition of elementary

deformation of curves — given

φ, ψ

: [

a, b

]

→ U

, which are closed, we say

is an

elementary deformation or convex deformation of

if there exists a decomposition

< x

< ··· < x

and convex open sets

, ··· , C

⊆ U

such that for

i−1

≤ t ≤ x

, we have φ(t) and ψ(t) in C

It was a rather unnatural definition, since we have to make reference to

this arbitrarily constructed dissection of [

a, b

] and convex sets

. Moreover,

this definition fails to be transitive (e.g. on

R \ {

}

, rotating a circle about the

center by, say,

is elementary, but rotating by

is not). Yet, this definition

was cooked up just so that it immediately follows that elementary deformations

preserve integrals of holomorphic functions around the loop.

The idea now is to define a more general and natural notion of deforming a

curve, known as “homotopy”. We will then show that each homotopy can be

given by a sequence of elementary deformations. So homotopies also preserve

integrals of holomorphic functions.

Definition

(Homotopy of closed curves)

Let

U ⊆ C

be a domain, and let

: [

a, b

]

→ U

and

: [

a, b

]

→ U

be piecewise

-smooth closed paths. A

homotopy from φ : ψ is a continuous map F : [0, 1] × [a, b] → U such that

F (0, t) = φ(t), F (1, t) = ψ(t),

and moreover, for all

s ∈

, t

], the map

t 7→ F

(

s, t

) viewed as a map [

a, b

]

→ U

is closed and piecewise C

-smooth.

We can imagine this as a process of “continuously deforming” the path

ψ, with a path F (s, ·) at each point in time s ∈ [0, 1].

Proposition.

Let

φ, ψ

: [

a, b

]

→ U

be homotopic (piecewise

) closed paths in

a domain

. Then there exists some

, φ

, ··· , φ

such that each

is piecewise

closed and

i+1

is obtained from

by elementary deformation.

Proof.

This is an exercise in uniform continuity. We let

: [0

[

a, b

]

→ U

be a homotopy from

. Since

image

(

) is compact and

is open, there

is some

ε >

0 such that

(

s, t

)

, ε

)

⊆ U

for all (

s, t

)

∈

[

a, b

] (for each

s, t

, pick the maximum

s,t

0 such that

(

s, t

)

, ε

s,t

)

⊆ U

. Then

s,t

varies

continuously with

s, t

, hence attains its minimum on the compact set [0

[

a, b

Then picking ε to be the minimum works).

Since

is uniformly continuous, there is some

such that

(

s, t

)

−

(

, t

)

k < δ

implies |F (s, t) − F (s

, t

)| < ε.

Now we pick n ∈ N such that

1+(b−a)

< δ, and let

= a + (b − a)

(t) = F



, t



= B



, x



, ε



Then

is clearly convex. These definitions are cooked up precisely so that if

s ∈



i−1



and t ∈ [x

j−1

, x

], then F (s, t) ∈ C

. So the result follows.

Corollary.

Let

be a domain,

U → C

be holomorphic, and

, γ

homotopic piecewise C

-smooth closed curves in U. Then

f(z) dz =

f(z) dz.

This means the integral around any path depends only on the homotopy

class of the path, and not the actual path itself.

We can now use this to “upgrade” our Cauchy’s theorem to allow arbitrary

simply connected domains. The theorem will become immediate if we adopt the

following alternative definition of a simply connected domain:

Definition

(Simply connected domain)

A domain

is simply connected if

every C

smooth closed path is homotopic to a constant path.

This is in fact equivalent to our earlier definition that every continuous

map

→ U

can be extended to a continuous map

→ U

. This is almost

immediately obvious, except that our old definition only required the map to be

continuous, while the new definition only works with piecewise

paths. We

will need something that allows us to approximate any continuous curve with a

piecewise

-smooth one, but we shall not do that here. Instead, we will just

forget about the old definition and stick to the new one.

Rewriting history, we get the following corollary:

Corollary

(Cauchy’s theorem for simply connected domains)

Let

be a simply

connected domain, and let

U → C

be holomorphic. If

is any piecewise

-smooth closed curve in U, then

f(z) dz = 0.

We will sometimes refer to this theorem as “simply-connected Cauchy”, but

we are not in any way suggesting that Cauchy himself is simply connected.

Proof.

By definition of simply-connected,

is homotopic to the constant path,

and it is easy to see the integral along a constant path is zero.

3.3 Cauchy’s residue theorem

We finally get to Cauchy’s residue theorem. This in some sense a mix of all the

results we’ve previously had. Simply-connected Cauchy tells us the integral of a

holomorphic

around a closed curve depends only on its homotopy class, i.e. we

can deform curves by homotopy and this preserves the integral. This means the

value of the integral really only depends on the “holes” enclosed by the curve.

We also had the Cauchy integral formula. This says if

(

a, r

)

→ C

holomorphic, w ∈ B(a, ρ) and ρ < r, then

f(w) =

2πi

∂B(a,ρ)

f(z)

z − w

dz.

Note that

(

) also happens to be the residue of the function

f(z)

z−w

. So this

really says if

has a simple pole at

inside the region bounded by a simple

closed curve γ, then

2π

g(z) dz = Res(g, a).

The Cauchy’s residue theorem says the result holds for any type of singularities,

and any number of singularities.

Theorem

(Cauchy’s residue theorem)

Let

be a simply connected domain,

and

, ··· , z

} ⊆ U

. Let

U \ {z

, ··· , z

} → C

be holomorphic. Let

: [

a, b

]

→ U

be a piecewise

-smooth closed curve such that

image

(

)

for all i. Then

2πi

f(z) dz =

j=1

I(γ, z

) Res(f; z

The Cauchy integral formula and simply-connected Cauchy are special cases

of this.

Proof. At each z

, f has a Laurent expansion

f(z) =

n∈Z

(i)

(z − z

)

valid in some neighbourhood of z

. Let g

(z) be the principal part, namely

(z) =

−1

n=−∞

(i)

(z − z

)

From the proof of the Laurent series, we know

(

) gives a holomorphic function

on U \ {z

We now consider

f −g

−g

−···−g

, which is holomorphic on

U \{z

, ··· , z

}

and has a removable singularity at each z

. So

(f − g

− ··· − g

)(z) dz = 0,

by simply-connected Cauchy. Hence we know

f(z) dz =

j=1

(z) dz.

For each

, we use uniform convergence of the series

n≤−1

(j)

(

z − z

)

compact subsets of U \ {z

}, and hence on γ, to write

(z) dz =

n≤−1

(j)

(z − z

)

dz.

However, for

n 6

−

1, the function (

z − z

)

has an antiderivative, and hence

the integral around γ vanishes. So this is equal to

(j)

−1

z − z

dz.

But

(j)

−1

is by definition the residue of

, and the integral is just the integral

definition of the winding number (up to a factor of 2πi). So we get

f(z) dz = 2πi

j=1

Res(f; z

)I(γ, z

So done.

3.4 Overview

We’ve done most of the theory we need. In the remaining of the time, we are

going to use these tools to do something useful. In particular, we will use the

residue theorem heavily to compute integrals.

But before that, we shall stop and look at what we have done so far.

Our first real interesting result was Cauchy’s theorem for a triangle, which

had a rather weird hypothesis — if

U → C

is holomorphic and ∆

⊆ U

is at

triangle, then

∂∆

f(z) dz = 0.

To prove this, we dissected our triangle into smaller and smaller triangles, and

then the result followed how the numbers and bounds magically fit in together.

To accompany this, we had another theorem that used triangles. Suppose

is a star domain and f : U → C is continuous. Then if

∂∆

f(z) dz = 0

for all triangles, then there is a holomorphic

with

(

) =

(

). Here we

defined F by

F (z) =

f(z) dz,

where

is the “center” of the star, and we integrate along straight lines. The

triangle condition ensures this is well-defined.

These are the parts where we used some geometric insight — in the first

case we thought of subdividing, and in the second we decided to integrate along

paths.

These two awkward theorems about triangles fit in perfectly into the convex

Cauchy theorem, via the fundamental theorem of calculus. This tells us that if

f : U → C is holomorphic and U is convex, then

f(z) dz = 0

for all closed γ ⊆ U.

We then noticed this allows us to deform paths nicely and still preserve the

integral. We called these nice deformations elementary deformations, and then

used it to obtain the Cauchy integral formula, namely

f(w) =

2πi

∂B(a,ρ)

f(z)

z − w

for f : B(a, r) → C, ρ < r and w ∈ B(a, ρ).

This formula led us to some classical theorems like the Liouville theorem and

the maximum principle. We also used the power series trick to prove Taylor’s

theorem, saying any holomorphic function is locally equal to some power series,

which we call the Taylor series. In particular, this shows that holomorphic

functions are infinitely differentiable, since all power series are.

We then notice that for

a convex domain, if

U → C

is continuous and

f(z) dz = 0

for all curves

, then

has an antiderivative. Since

is the derivative of its

antiderivative (by definition), it is then (infinitely) differentiable. So a function

is holomorphic on a simply connected domain if and only if the integral along

any closed curve vanishes. Since the latter property is easily shown to be

conserved by uniform limits, we know the uniform limit of holomorphic functions

is holomorphic.

Then we figured out that we can use the same power series expansion trick

to deal with functions with singularities. It’s just that we had to include

negative powers of

. Adding in the ideas of winding numbers and homotopies,

we got the residue theorem. We showed that if

is simply connected and

f : U \ {z

, ··· , z

} → C is holomorphic, then

2πi

f(z) dz =

Res(f, z

)I(γ, z

This will further lead us to Rouch´e’s theorem and the argument principle, to be

done later.

Throughout the course, there weren’t too many ideas used. Everything was

built upon the two “geometric” theorems of Cauchy’s theorem for triangles and

the antiderivative theorem. Afterwards, we repeatedly used the idea of deforming

and cutting paths, as well as the power series expansion of

z−w

, and that’s it.

3.5 Applications of the residue theorem

This section is more accurately described as “Integrals, integrals, integrals”. Our

main objective is to evaluate real integrals, but to do so, we will pretend they

are complex integrals, and apply the residue theorem.

Before that, we first come up with some tools to compute residues, since we

will have to do that quite a lot.

Lemma.

Let

U \ {a} → C

be holomorphic with a pole at

, i.e

meromorphic on U.

(i) If the pole is simple, then

Res(f, a) = lim

z→a

(z − a)f(z).

(ii) If near a, we can write

f(z) =

g(z)

h(z)

where

(

)

= 0 and

has a simple zero at

, and

g, h

are holomorphic on

B(a, ε) \ {a}, then

Res(f, a) =

g(a)

(a)

(iii) If

f(z) =

g(z)

(z − a)

near a, with g(a) 6= 0 and g is holomorphic, then

Res(f, a) =

(k−1)

(a)

(k − 1)!

Proof.

(i) By definition, if f has a simple pole at a, then

f(z) =

−1

(z − a)

+ c

(z − a) + ··· ,

and by definition c

−1

= Res(f, a). Then the result is obvious.

(ii) This is basically L’Hˆopital’s rule. By the previous part, we have

Res(f; a) = lim

z→a

(z − a)

g(z)

h(z)

= g(a) lim

z→a

z − a

h(z) − h(a)

g(a)

(a)

(iii)

We know the residue

Res

(

;

) is the coefficient of (

z −a

)

k−1

in the Taylor

series of g at a, which is exactly

(k−1)!

(k−1)

(a).

Example. We want to compute the integral

∞

1 + x

dx.

We consider the following contour:

−R R

iπ/4

3iπ/4

××

We notice

1+x

has poles at

−

1, as indicated in the diagram. Note that

the two of the poles lie in the unbounded region. So I(γ, ·) = 0 for these.

We can write the integral as

1 + z

dz =

−R

1 + x

dx +

iRe

iθ

1 + R

4iθ

dθ.

The first term is something we care about, while the second is something we

despise. So we might want to get rid of it. We notice the integrand of the second

integral is O(R

−3

). Since we are integrating it over something of length R, the

whole thing tends to 0 as R → ∞.

We also know the left hand side is just

1 + z

dz = 2πi(Res(f, e

iπ/4

) + Res(f, e

3iπ/4

)).

So we just have to compute the residues. But our function is of the form given

by part (ii) of the lemma above. So we know

Res(f, e

iπ/4

) =



z=e

iπ/4

−3πi/4

and similarly at

i3π/4

. On the other hand, as

R → ∞

, the first integral on the

right is

∞

−∞

1+x

dx, which is, by evenness, twice of what we want. So

∞

1 + x

dx =

∞

−∞

1 + x

dx = −

2πi

iπ/4

+ e

3πi/4

) =

√

Hence our integral is

∞

1 + x

dx =

√

When computing contour integrals, there are two things we have to decide.

First, we need to pick a nice contour to integrate along. Secondly, as we will see

in the next example, we have to decide what function to integrate.

Example. Suppose we want to integrate

cos(x)

1 + x + x

dx.

We know

cos

, as a complex function, is everywhere holomorphic, and 1 +

have two simple zeroes, namely at the cube roots of unity. We pick the same

contour, and write ω = e

2πi/3

. Then we have

−R R

Life would be good if

cos

were bounded, for the integrand would then be

(

−2

and the circular integral vanishes. Unfortunately, at, say,

cos

(

) is large. So

instead, we consider

f(z) =

1 + z + z

Now, again by the previous lemma, we get

Res(f; ω) =

iω

2ω + 1

On the semicircle, we have



f(Re

iθ

)Re

iθ

dθ



≤

−sin θR

2iθ

+ Re

iθ

+ 1|

dθ,

which is O(R

−1

). So this vanishes as R → ∞.

The remaining is not quite the integral we want, but we can just take the

real part. We have

cos x

1 + x + x

dx = Re

f(z) dz

= Re lim

R→∞

f(z) dz

= Re(2πi Res(f, ω))

2π

√

−

√

3/2

cos

Another class of integrals that often come up are integrals of trigonometric

functions, where we are integrating along the unit circle.

Example. Consider the integral

π/2

1 + sin

(t)

dt.

We use the expression of sin in terms of the exponential function, namely

sin(t) =

− e

−it

So if we are on the unit circle, and z = e

, then

sin(t) =

z − z

−1

Moreover, we can check

= ie

dt =

Hence we get

π/2

1 + sin

(t)

dt =

2π

1 + sin

(t)

|z|=1

1 +

(z−z

−1

)

−4

|z|=1

− 6z

+ 1

dz.

The base is a quadratic in

, which we can solve. We find the roots to be 1

√

and −1 ±

√

× × ××

The residues at the point

√

2 −

1 and

−

√

+ 1 give

−

√

. So the integral we

want is

π/2

1 + sin

(t)

= 2πi

−

√

−

√

Most rational functions of trigonometric functions can be integrated around

|z| = 1 in this way, using the fact that

sin(kt) =

ikt

− e

−ikt

− z

−k

, cos(kt) =

ikt

+ e

−ikt

+ z

−k

We now develop a few lemmas that help us evaluate the contributions of certain

parts of contours, in order to simplify our work.

Lemma.

Let

(

a, r

)

\{a} → C

be holomorphic, and suppose

has a simple

pole at a. We let γ

: [α, β] → C be given by

t 7→ a + εe

Then

lim

ε→0

f(z) dz = (β − α) · i · Res(f, a).

Proof. We can write

f(z) =

z − a

+ g(z)

near

, where

Res

(

;

), and

(

a, δ

)

→ C

is holomorphic near

. We

take ε < δ. Then



g(z) dz



≤ (β −α) · ε sup

z∈γ

|g(z)|.

But

is bounded on

(

α, δ

). So this vanishes as

ε →

0. So the remaining

integral is

lim

ε→0

z − a

dz = c lim

ε→0

z − a

= c lim

ε→0

εe

· iεe

= i(β −α)c,

as required.

A lemma of a similar flavor allows us to consider integrals on expanding

semicircles.

Lemma

(Jordan’s lemma)

Let

be holomorphic on a neighbourhood of infinity

, i.e. on

{|z| > r}

for some

r >

0. Assume that

(

) is bounded in this

region. Then for α > 0, we have

f(z)e

iαz

dz → 0

R → ∞

, where

(

) =

for

t ∈

, π

] is the semicircle (which is not

closed).

−R R

In previous cases, we had

(

) =

(

−2

), and then we can bound the

integral simply as

(

−1

)

→

0. In this case, we only require

(

) =

(

−1

The drawback is that the case

(

) d

need not work — it is possible that

this does not vanish. However, if we have the extra help from e

iαx

, then we do

get that the integral vanishes.

Proof. By assumption, we have

|f(z)| ≤

|z|

for large |z| and some constant M > 0. We also have

iαz

| = e

−Rα sin t

. To avoid messing with

sin t

, we note that on (0

], the function

sin θ

decreasing, since

dθ



sin θ



θ cos θ −sin θ

≤ 0.

Then by consider the end points, we find

sin(t) ≥

for t ∈ [0,

]. This gives us the bound

iαz

| = e

−Rα sin t

≤

(

−Ra2t/π

0 ≤ t ≤

−Ra2t

/π

0 ≤ t

= π − t ≤

So we get



π/2

iRαe

f(Re

)Re



≤

2π

−2αRt/π

· M dt

(1 − e

αR

)

→ 0

as R → ∞.

The estimate for

π/2

f(z)e

iαz

is analogous.

Example. We want to show

∞

sin x

dx =

Note that

sin x

has a removable singularity at x = 0. So everything is fine.

Our first thought might be to use our usual semi-circular contour that looks

like this:

−R R

If we look at this and take the function

sin z

, then we get no control at

iR ∈ γ

So what we would like to do is to replace the sine with an exponential. If we let

f(z) =

then we now have the problem that

has a simple pole at 0. So we consider a

modified contour

−R

−ε

R,ε

Now if

R,ε

denotes the modified contour, then the singularity of

lies outside

the contour, and Cauchy’s theorem says

R,ε

f(z) dz = 0.

Considering the

-semicircle

, and using Jordan’s lemma with

= 1 and

as the function, we know

f(z) dz → 0

as R → ∞.

Considering the

-semicircle

, and using the first lemma, we get a contribu-

tion of

−iπ

, where the sign comes from the orientation. Rearranging, and using

the fact that the function is even, we get the desired result.

Example. Suppose we want to evaluate

∞

−∞

cosh x

dx,

where a ∈ (−1, 1) is a real constant.

To do this, note that the function

f(z) =

cosh z

has simple poles where



n +



iπ

for

n ∈ Z

. So if we did as we have done

above, then we would run into infinitely many singularities, which is not fun.

Instead, we note that

cos(x + iπ) = −cosh x.

Consider a rectangular contour

−R

vert

−

vert

πi

We now enclose only one singularity, namely ρ =

iπ

, where

Res(f, ρ) =

aρ

cosh

(ρ)

= ie

aπi/2

We first want to see what happens at the edges. We have

vert

f(z) dz =

a(R+iy)

cosh(R + iy)

i dy.

hence we can bound this as



vert

f(z) dz



≤



− e

−R



dy → 0 as R → ∞,

since

a <

1. We can do a similar bound for

vert

−

, where we use the fact that

a > −1.

Thus, letting R → ∞, we get

cosh x

dx +

−∞

+∞

aπi

cosh(x + iπ)

dx = 2πi(−ie

aπi/2

Using the fact that cosh(x + iπ) = −cos(x), we get

cosh x

dx =

2πe

aiπ/2

1 + e

aπi

= π sec



πa



Example. We provide a(nother) proof that

n≥1

Recall we just avoided having to encircle infinitely poles by picking a rectangular

contour. Here we do the opposite — we encircle infinitely many poles, and then

we can use this to evaluate the infinite sum of residues using contour integrals.

We consider the function

(

) =

π cot(πz)

, which is holomorphic on

except

for simple poles at Z \ {0}, and a triple pole at 0.

We can check that at n ∈ Z \ {0}, we can write

f(z) =

π cos(πz)

sin(πz)

where the second term has a simple zero at

, and the first is non-vanishing at

n 6= 0. Then we have compute

Res(f; n) =

π cos(πn)

Note that the reason why we have those funny

’s all around the place is so that

we can get this nice expression for the residue.

At z = 0, we get

cot(z) =



1 −

+ O(z

)



z −

+ O(z

)



−1

−

+ O(z

So we get

π cot(πz)

−

+ ···

So the residue is −

. Now we consider the following square contour:

× × × × × × × × × × ×

(N +

−(N +

N +

−(N +

)

Since we don’t want the contour itself to pass through singularities, we make

the square pass through ±



N +



. Then the residue theorem says

f(z) dz = 2πi

n=1

−

We can thus get the desired series if we can show that

f(z) dz → 0 as n → ∞.

We first note that



f(z) dz



≤ sup



π cot πz



4(2N + 1)

≤ sup

|cot πz|

4(2N + 1)π



N +



= sup

|cot πz|O(N

−1

So everything is good if we can show sup

|cot πz| is bounded as N → ∞.

On the vertical sides, we have

z = π



N +



+ iy,

and thus

|cot(πz)| = |tan(iπy)| = |tanh(πy)| ≤ 1,

while on the horizontal sides, we have

z = x ± i



N +



and

|cot(πz)| ≤

π(N+1/2)

+ e

−π(N+1/2)

π(N+1/2)

− e

−π(N+1/2)

= coth



N +



π.

While it is not clear at first sight that this is bounded, we notice

x 7→ coth x

decreasing and positive for x ≥ 0. So we win.

Example. Suppose we want to compute the integral

∞

log x

1 + x

dx.

The point is that to define

log z

, we have to cut the plane to avoid multi-

valuedness. In this case, we might choose to cut it along

iR ≤

0, giving a branch

log

, for which

arg

(

)

∈



−

3π



. We need to avoid running through zero. So

we might look at the following contour:

−R

ε ε

On the large semicircular arc of radius R, the integrand

|f(z)||dz| = O



R ·

log R



= O



log R



→ 0 as R → ∞.

On the small semicircular arc of radius ε, the integrand

|f(z)||dz| = O(ε log ε) → 0 as ε → 0.

Hence, as

ε →

0 and

R → ∞

, we are left with the integral along the negative

real axis. Along the negative real axis, we have

log z = log |z| + iπ.

So the residue theorem says

∞

log x

1 + x

dx +

∞

log |z| + iπ

1 + x

(−dx) = 2πi Res(f; i).

We can compute the residue as

Res(f, i) =

log i

iπ

So we find

∞

log x

1 + x

dx + iπ

∞

1 + x

dx =

iπ

Taking the real part of this, we obtain

∞

log x

1 + x

dx = 0.

In this case, we had a branch cut, and we managed to avoid it by going

around our magic contour. Sometimes, it is helpful to run our integral along the

branch cut.

Example. We want to compute

∞

√

+ ax + b

dx,

where

a, b ∈ R

. To define

√

, we need to pick a branch cut. We pick it to lie

along the real line, and consider the keyhole contour

As usual this has a small circle of radius

around the origin, and a large circle

of radius R. Note that these both avoid the branch cut.

Again, on the R circle, we have

|f(z)||dz| = O



√



→ 0 as R → ∞.

On the ε-circle, we have

|f(z)||dz| = O(ε

3/2

) → 0 as ε → 0.

Viewing

√

log z

, on the two pieces of the contour along

≥0

log z

differs

by 2

πi

. So

√

changes sign. This cancels with the sign change arising from

going in the wrong direction. Therefore the residue theorem says

2πi

residues inside contour = 2

∞

√

+ ax + b

dx.

What the residues are depends on what the quadratic actually is, but we will

not go into details.

3.6 Rouch´es theorem

We now want to move away from computing integrals, and look at a different

application — Rouch´es theorem. Recall one of the first applications of complex

analysis is to use Liouville’s theorem to prove the fundamental theorem of algebra,

and show that every polynomial has a root. One might wonder — if we know

a bit more about the polynomial, can we say a bit more about how the roots

behave?

To do this, recall we said that if

(

;

)

→ C

is holomorphic, and

(

) = 0,

then f has a zero of order k if, locally,

f(z) = (z − a)

g(z),

with g holomorphic and g(a) 6= 0.

Analogously, if

(

a, r

)

\ {a} → C

is holomorphic, and

has at worst a

pole at a, we can again write

f(z) = (z − a)

g(z),

where now

k ∈ Z

may be negative. Since we like numbers to be positive, we say

the order of the zero/pole is |k|.

It turns out we can use integrals to help count poles and zeroes.

Theorem

(Argument principle)

Let

be a simply connected domain, and let

be meromorphic on

. Suppose in fact

has finitely many zeroes

, ··· , z

and finitely many poles

, ··· , w

. Let

be a piecewise-

closed curve such

that z

, w

6∈ image(γ) for all i, j. Then

I(f ◦ γ, 0) =

2πi

(z)

f(z)

dz =

i=1

ord(f; z

) −

j=1

ord(f, w

)I(γ, w

Note that the first equality comes from the fact that

I(f ◦ γ, 0) =

2πi

f◦γ

2πi

f(z)

2πi

(z)

f(z)

dz.

In particular, if

is a simple closed curve, then all the winding numbers of

about points

, w

lying in the region bound by

are all +1 (with the right

choice of orientation). Then

number of zeroes − number of poles =

2π

(change in argument of f along γ).

Proof. By the residue theorem, we have

2πi

(z)

f(z)

dz =

z∈U

Res



, z



I(γ, z),

where we sum over all zeroes and poles of

. Note that outside these zeroes and

poles, the function

(z)

f(z)

is holomorphic.

Now at each

, if

(

) = (

z − z

)

(

), with

(

)

= 0, then by direct

computation, we get

(z)

f(z)

z − z

(z)

g(z)

Since at

is holomorphic and non-zero, we know

(z)

g(z)

is holomorphic near

. So

Res



, z



= k = ord(f, z

Analogously, by the same proof, at the w

, we get

Res



, w



= −ord(f; w

So done.

This might be the right place to put the following remark — all the time, we

have assumed that a simple closed curve “bounds a region”, and then we talk

about which poles or zeroes are bounded by the curve. While this seems obvious,

it is not. This is given by the Jordan curve theorem, which is actually hard.

Instead of resorting to this theorem, we can instead define what it means to

bound a region in a more convenient way. One can say that for a domain

, a

closed curve γ ⊆ U bounds a domain D ⊆ U if

I(γ, z) =

(

+1 z ∈ D

0 z 6∈ D

for a particular choice of orientation on

. However, we shall not worry ourselves

with this.

The main application of the argument principle is Rouch´es theorem.

Corollary

(Rouch´es theorem)

Let

be a domain and

a closed curve which

bounds a domain in

(the key case is when

is simply connected and

is a

simple closed curve). Let

f, g

be holomorphic on

, and suppose

|f| > |g|

for all

z ∈ image

(

). Then

and

have the same number of zeroes in the domain

bound by γ, when counted with multiplicity.

Proof.

|f| > |g|

, then

and

cannot have zeroes on the curve

We let

h(z) =

f(z) + g(z)

f(z)

= 1 +

g(z)

f(z)

This is a natural thing to consider, since zeroes of

is zeroes of

, while

poles of h are zeroes of f . Note that by assumption, for all z ∈ γ, we have

h(z) ∈ B(1, 1) ⊆ {z : Re z > 0}.

Therefore

h◦γ

is a closed curve in the half-plane

Re z >

}

. So

(

h◦γ

; 0) = 0.

Then by the argument principle,

must have the same number of zeros as poles

, when counted with multiplicity (note that the winding numbers are all

+1).

Thus, as the zeroes of

are the zeroes of

, and the poles of

are the

poles of f, the result follows.

Example.

Consider the function

+ 6

+ 3. This has three roots (with

multiplicity) in {1 < |z| < 2}. To show this, note that on |z| = 2, we have

|z|

= 16 > 6|z| + 3 ≥ |6z + 3|.

So if we let

(

) =

and

(

) = 6

+ 3, then

and

have the same number

of roots in {|z| < 2}. Hence all four roots lie inside {|z| < 2}.

On the other hand, on |z| = 1, we have

|6z| = 6 > |z

+ 3|.

So 6

and

+ 6

+ 3 have the same number of roots in

{|z| <

}

. So there is

exactly one root in there, and the remaining three must lie in

{

< |z| <

}

(the

bounds above show that |z| cannot be exactly 1 or 2). So done.

Example. Let

P (x) = x

+ a

n−1

+ ··· + a

x + a

∈ Z[x],

and suppose a

6= 0. If

n−1

| > 1 + |a

n−2

| + ··· + |a

| + |a

then

is irreducible over

(and hence irreducible over

, by Gauss’ lemma

from IB Groups, Rings and Modules).

To show this, we let

f(z) = a

n−1

g(z) = z

+ a

n−2

+ ··· + a

z + a

Then our hypothesis tells us |f| > |g| on |z| = 1.

So f and P = f + g both have n − 1 roots in the open unit disc {|z| < 1}.

Now if we could factor

(

) =

(

)

(

), where

Q, R ∈ Z

[

], then at least

one of

Q, R

must have all its roots inside the unit disk. Say all roots of

are

inside the unit disk. But we assumed

= 0. So 0 is not a root of

. Hence it

is not a root of

. But the product of the roots

is a coefficient of

, hence an

integer strictly between 0 and 1. This is a contradiction.

The argument principle and Rouch´es theorem tell us how many roots we

have got. However, we do not know if they are distinct or not. This information

is given to us via the local degree theorem. Before we can state it, we have to

define the local degree.

Definition

(Local degree)

Let

(

a, r

)

→ C

be holomorphic and non-

constant. Then the local degree of

, written

deg

(

f, a

) is the order of the

zero of f(z) − f (a) at a.

If we take the Taylor expansion of

about

, then the local degree is the

degree of the first non-zero term after the constant term.

Lemma. The local degree is given by

deg(f, a) = I(f ◦γ, f(a)),

where

γ(t) = a + re

with 0 ≤ t ≤ 2π, for r > 0 sufficiently small.

Proof.

Note that by the identity theorem, we know that,

(

)

− f

(

) has an

isolated zero at

(since

is non-constant). So for sufficiently small

, the

function

(

)

−f

(

) does not vanish on

B(a, r) \{a}

. If we use this

, then

f ◦γ

never hits f (a), and the winding number is well-defined.

The result then follows directly from the argument principle.

Proposition

(Local degree theorem)

Let

(

a, r

)

→ C

be holomorphic and

non-constant. Then for

r >

0 sufficiently small, there is

ε >

0 such that for any

w ∈ B

(

)

, ε

)

\ {f

(

)

}

, the equation

(

) =

has exactly

deg

(

f, a

) distinct

solutions in B(a, r).

Proof.

We pick

r >

0 such that

(

)

−f

(

) and

(

) don’t vanish on

(

a, r

)

\{a}

We let

(

) =

. Then

(

)

6∈ image

(

f ◦ γ

). So there is some

ε >

0 such

that

B(f(a), ε) ∩ image(f ◦ γ) = ∅.

We now let

w ∈ B

(

)

, ε

). Then the number of zeros of

(

)

− w

(

a, r

) is

just

(

f ◦γ, w

), by the argument principle. This is just equal to

(

f ◦γ, f

(

)) =

deg(f, a), by the invariance of I(Γ, ∗) as we move ∗ in a component C \ Γ.

Now if

w 6

(

), since

(

)

= 0 on

(

a, r

)

\{a}

, all roots of

(

)

−w

must

be simple. So there are exactly deg(f; a) distinct zeros.

The local degree theorem says the equation

(

) =

has

deg

(

f, a

) roots for

sufficiently close to

(

). In particular, we know there are some roots. So

B(f(a), ε) is contained in the image of f . So we get the following result:

Corollary

(Open mapping theorem)

Let

be a domain and

U → C

holomorphic and non-constant, then

is an open map, i.e. for all open

V ⊆ U

we get that f(V ) is open.

Proof.

This is an immediate consequence of the local degree theorem. It suffices

to prove that for every

z ∈ U

and

r >

0 sufficiently small, we can find

ε >

such that

(

)

, ε

)

⊆ f

(

a, r

)). This is true by the local degree theorem.

Recall that Liouville’s theorem says every holomorphic

C → B

1) is

constant. However, for any other simply connected domain, we know there are

some interesting functions we can write down.

Corollary.

Let

U ⊆ C

be a simply connected domain, and

U 6

. Then there

is a non-constant holomorphic function U → B(0, 1).

This is a weak form of the Riemann mapping theorem, which says that there

is a conformal equivalence to

1). This just says there is a map that is not

boring.

Proof.

We let

q ∈ C \ U

, and let

(

) =

z −q

. So

U → C

is non-vanishing.

It is also clearly holomorphic and non-constant. By an exercise (possibly on the

example sheet), there is a holomorphic function

U → C

such that

(

) =

g(z)

for all

. In particular, our function

(

) =

z − q

U → C

∗

can be written as

φ(z) = h(z)

, for some function h : U → C

∗

(by letting h(z) = e

g(z)

We let

y ∈ h

(

), and then the open mapping theorem says there is some

r >

0 with

(

y, r

)

⊆ h

(

). But notice

is injective by observation, and that

(

) =

±h

(

) implies

(

) =

(

). So we deduce that

(

−y, r

)

∩ h

(

) =

∅

(note that since y 6= 0, we have B(y, r) ∩ B(−y, r) = ∅ for sufficiently small r).

Now define

f : z 7→

2(h(z) + y)

This is a holomorphic function f : U → B(0, 1), and is non-constant.

This shows the amazing difference between C and C \ {0}.