Part IB — Complex Methods

Based on lectures by R. E. Hunt

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

Definition of an analytic function. Cauchy-Riemann equations. Analytic functions as

conformal mappings; examples. Application to the solutions of Laplace’s equation in

various domains. Discussion of log z and z

a

. [5]

Contour integration and Cauchy’s Theorem

[Proofs of theorems in this section will not be examined in this course.]

Contours, contour integrals. Cauchy’s theorem and Cauchy’s integral formula. Taylor

and Laurent series. Zeros, poles and essential singularities. [3]

Residue calculus

Residue theorem, calculus of residues. Jordan’s lemma. Evaluation of definite integrals

by contour integration. [4]

Fourier and Laplace transforms

Laplace transform: definition and basic properties; inversion theorem (proof not

required); convolution theorem. Examples of inversion of Fourier and Laplace transforms

by contour integration. Applications to differential equations. [4]

Contents

0 Introduction

1 Analytic functions

1.1 The complex plane and the Riemann sphere

1.2 Complex differentiation

1.3 Harmonic functions

1.4 Multi-valued functions

1.5 M¨obius map

1.6 Conformal maps

1.7 Solving Laplace’s equation using conformal maps

2 Contour integration and Cauchy’s theorem

2.1 Contour and integrals

2.2 Cauchy’s theorem

2.3 Contour deformation

2.4 Cauchy’s integral formula

3 Laurent series and singularities

3.1 Taylor and Laurent series

3.2 Zeros

3.3 Classification of singularities

3.4 Residues

4 The calculus of residues

4.1 The residue theorem

4.2 Applications of the residue theorem

4.3

Further applications of the residue theorem using rectangular

contours

4.4 Jordan’s lemma

5 Transform theory

5.1 Fourier transforms

5.2 Laplace transform

5.3 Elementary properties of the Laplace transform

5.4 The inverse Laplace transform

5.5 Solution of differential equations using the Laplace transform

5.6 The convolution theorem for Laplace transforms

0 Introduction

In Part IA, we learnt quite a lot about differentiating and integrating real

functions. Differentiation was fine, but integration was tedious. Integrals were

very difficult to evaluate.

In this course, we will study differentiating and integrating complex functions.

Here differentiation is nice, and integration is easy. We will show that complex

differentiable functions satisfy many things we hoped were true — a complex

differentiable function is automatically infinitely differentiable. Moreover, an

everywhere differentiable function must be constant if it is bounded.

On the integration side, we will show that integrals of complex functions can

be performed by computing things known as residues, which are much easier

to compute. We are not actually interested in performing complex integrals.

Instead, we will take some difficult real integrals, and pretend they are complex

ones.

This is a methods course. By this, we mean we will not focus too much on

proofs. We will at best just skim over the proofs. Instead, we focus on doing

things. We will not waste time proving things people have proved 300 years ago.

If you like proofs, you can go to the IB Complex Analysis course, or look them

up in relevant books.

1 Analytic functions

1.1 The complex plane and the Riemann sphere

We begin with a review of complex numbers. Any complex number

z ∈ C

can

be written in the form

x

+

iy

, where

x

=

Re z

,

y

=

Im z

are real numbers. We

can also write it as re

iθ

, where

Definition

(Modulus and argument)

.

The modulus and argument of a complex

number z = x + iy are given by

r = |z| =

p

x

2

+ y

2

, θ = arg z,

where x = r cos θ, y = r sin θ.

The argument is defined only up to multiples of 2

π

. So we define the following:

Definition

(Principal value of argument)

.

The principal value of the argument

is the value of θ in the range (−π, π].

We might be tempted to write down the formula

θ = tan

−1

y

x

,

but this does not always give the right answer — it is correct only if

x >

0. If

x ≤ 0, then it might be out by ±π (e.g. consider z = 1 + i and z = −1 − i).

Definition

(Open set)

.

An open set

D

is one which does not include its boundary.

More technically,

D ⊆ C

is open if for all

z

0

∈ D

, there is some

δ >

0 such that

the disc |z − z

0

| < δ is contained in D.

Definition

(Neighbourhood)

.

A neighbourhood of a point

z ∈ C

is an open set

containing z.

The extended complex plane

Often, the complex plane

C

itself is not enough. We want to consider the point

∞ as well. This forms the extended complex plane.

Definition

(The extended complex plane)

.

The extended complex plane is

C

∗

=

C ∪{∞}

. We can reach the “point at infinity” by going off in any direction

in the plane, and all are equivalent. In particular, there is no concept of

−∞

.

All infinities are the same. Operations with ∞ are done in the obvious way.

Sometimes, we do write down things like

−∞

. This does not refer to a

different point. Instead, this indicates a limiting process. We mean we are

approaching this infinity from the direction of the negative real axis. However,

we still end up in the same place.

Conceptually, we can visualize this using the Riemann sphere, which is a

sphere resting on the complex plane with its “South Pole” S at z = 0.

S

N

P

z

For any point

z ∈ C

, drawing a line through the “North Pole”

N

of the sphere

to

z

, and noting where this intersects the sphere. This specifies an equivalent

point

P

on the sphere. Then

∞

is equivalent to the North Pole of the sphere

itself. So the extended complex plane is mapped bijectively to the sphere.

This is a useful way to visualize things, but is not as useful when we actually

want to do computations. To investigate properties of

∞

, we use the substitution

ζ

=

1

z

. A function

f

(

z

) is said to have a particular property at

∞

if

f

(

1

ζ

) has

that same property at

ζ

= 0. This vague notion will be made precise when we

have specific examples to play with.

1.2 Complex differentiation

Recall the definition of differentiation for a real function f(x):

f

0

(x) = lim

δx→0

f(x + δx) − f(x)

δx

.

It is implicit that the limit must be the same whichever direction we approach

from. For example, consider

|x|

at

x

= 0. If we approach from the right, i.e.

δx →

0

+

, then the limit is +1, whereas from the left, i.e.

δx →

0

−

, the limit is

−

1. Because these limits are different, we say that

|x|

is not differentiable at

the origin.

This is obvious and we already know that, but for complex differentiation,

this issue is much more important, since there are many more directions. We

now extend the definition of differentiation to complex number:

Definition

(Complex differentiable function)

.

A complex differentiable function

f : C → C is differentiable at z if

f

0

(z) = lim

δz→0

f(z + δz) − f (z)

δz

exists (and is therefore independent of the direction of approach — but now

there are infinitely many possible directions).

This is the same definition as that for a real function. Often, we are not

interested in functions that are differentiable at a point, since this might allow

some rather exotic functions we do not want to consider. Instead, we want the

function to be differentiable near the point.

Definition

(Analytic function)

.

We say

f

is analytic at a point

z

if there

exists a neighbourhood of

z

throughout which

f

0

exists. The terms regular and

holomorphic are also used.

Definition

(Entire function)

.

A complex function is entire if it is analytic

throughout C.

The property of analyticity is in fact a surprisingly strong one! For example,

two consequences are:

(i)

If a function is analytic, then it is differentiable infinitely many times. This

is very very false for real functions. There are real functions differentiable

N

times, but no more (e.g. by taking a non-differentiable function and

integrating it N times).

(ii) A bounded entire function must be a constant.

There are many more interesting properties, but these are sufficient to show us

that complex differentiation is very different from real differentiation.

The Cauchy-Riemann equations

We already know well how to differentiate real functions. Can we use this to

determine whether certain complex functions are differentiable? For example is

the function

f

(

x

+

iy

) =

cos x

+

i sin y

differentiable? In general, given a complex

function

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

where

z

=

x

+

iy

are

u, v

are real functions, is there an easy criterion to determine

whether f is differentiable?

We suppose that

f

is differentiable at

z

. We may take

δz

in any direction

we like. First, we take it to be real, with δz = δx. Then

f

0

(z) = lim

δx→0

f(z + δx) − f (z)

δx

= lim

δx→0

u(x + δx, y) + iv(x + δx, y) − (u(x, y) + iv(x, y))

δx

=

∂u

∂x

+ i

∂v

∂x

.

What this says is something entirely obvious — since we are allowed to take the

limit in any direction, we can take it in the

x

direction, and we get the corre-

sponding partial derivative. This is a completely uninteresting point. Instead,

let’s do the really fascinating thing of taking the limit in the y direction!

Let δz = iδy. Then we can compute

f

0

(z) = lim

δy→0

f(z + iδy) − f(z)

iδy

= lim

δy→0

u(x, y + δy) + iv(x, y + δy) − (u(x, y) + iv(x, y))

iδy

=

∂v

∂y

− i

∂u

∂y

.

By the definition of differentiability, the two results for

f

0

(

z

) must agree! So we

must have

∂u

∂x

+ i

∂v

∂x

=

∂v

∂y

− i

∂u

∂y

.

Taking the real and imaginary components, we get

Proposition

(Cauchy-Riemann equations)

.

If

f

=

u

+

iv

is differentiable, then

∂u

∂x

=

∂v

∂y

,

∂u

∂y

= −

∂v

∂x

.

Is the converse true? If these equations hold, does it follow that

f

is differ-

entiable? This is not always true. This holds only if

u

and

v

themselves are

differentiable, which is a stronger condition that the partial derivatives exist, as

you may have learnt from IB Analysis II. In particular, this holds if the partial

derivatives u

x

, u

y

, v

x

, v

y

are continuous (which implies differentiability). So

Proposition.

Given a complex function

f

=

u

+

iv

, if

u

and

v

are real differen-

tiable at a point z and

∂u

∂x

=

∂v

∂y

,

∂u

∂y

= −

∂v

∂x

,

then f is differentiable at z.

We will not prove this — proofs are for IB Complex Analysis.

Examples of analytic functions

Example.

(i) f

(

z

) =

z

is entire, i.e. differentiable everywhere. Here

u

=

x, v

=

y

. Then

the Cauchy-Riemann equations are satisfied everywhere, since

∂u

∂x

=

∂v

∂y

= 1,

∂u

∂y

= −

∂v

∂x

= 0,

and these are clearly continuous. Alternatively, we can prove this directly

from the definition.

(ii) f(z) = e

z

= e

x

(cos y + i sin y) is entire since

∂u

∂x

= e

x

cos y =

∂v

∂y

,

∂u

∂y

= −e

x

sin y = −

∂v

∂x

.

The derivative is

f

0

(z) =

∂u

∂x

+ i

∂v

∂x

= e

x

cos y + ie

x

sin y = e

z

,

as expected.

(iii) f

(

z

) =

z

n

for

n ∈ N

is entire. This is less straightforward to check. Writing

z = r(cos θ + i sin θ), we obtain

u = r

n

cos nθ, v = r

n

sin nθ.

We can check the Cauchy-Riemann equation using the chain rule, writing

r

=

p

x

2

= y

2

and

tan θ

=

y

x

. This takes quite a while, and it’s not worth

your time. But if you really do so, you will find the derivative to be

nz

n−1

.

(iv)

Any rational function, i.e.

f

(

z

) =

P (z)

Q(z)

where

P, Q

are polynomials, is

analytic except at points where

Q

(

z

) = 0 (where it is not even defined).

For instance,

f(z) =

z

z

2

+ 1

is analytic except at ±i.

(v)

Many standard functions can be extended naturally to complex functions

and obey the usual rules for their derivatives. For example,

– sin z

=

e

iz

−e

−iz

2i

is differentiable with derivative

cos z

=

e

iz

+e

−iz

2

. We

can also write

sin z = sin(x + iy)

= sin x cos iy + cos x sin iy

= sin x cosh y + i cos x sinh y,

which is sometimes convenient.

–

Similarly

cos z, sinh z, cosh z

etc. differentiate to what we expect them

to differentiate to.

– log z = log |z| + i arg z has derivative

1

z

.

–

The product rule, quotient rule and chain rule hold in exactly the

same way, which allows us to prove (iii) and (iv) easily.

Examples of non-analytic functions

Example.

(i) Let f(z) = Re z. This has u = x, v = 0. But

∂u

∂x

= 1 6= 0 =

∂v

∂y

.

So Re z is nowhere analytic.

(ii)

Consider

f

(

z

) =

|z|

. This has

u

=

p

x

2

+ y

2

, v

= 0. This is thus nowhere

analytic.

(iii)

The complex conjugate

f

(

z

) =

¯z

=

z

∗

=

x − iy

has

u

=

x, v

=

−y

. So the

Cauchy-Riemann equations don’t hold. Hence this is nowhere analytic.

We could have deduced (ii) from this — if

|z|

were analytic, then so would

|z|

2

, and hence ¯z =

|z|

2

z

also has to be analytic, which is not true.

(iv)

We have to be a bit more careful with

f

(

z

) =

|z|

2

=

x

2

+

y

2

. The

Cauchy-Riemann equations are satisfied only at the origin. So

f

is only

differentiable at

z

= 0. However, it is not analytic since there is no

neighbourhood of 0 throughout which f is differentiable.

1.3 Harmonic functions

This is the last easy section of the course.

Definition

(Harmonic conjugates)

.

Two functions

u, v

satisfying the Cauchy-

Riemann equations are called harmonic conjugates.

If we know one, then we can find the other up to a constant. For example, if

u(x, y) = x

2

− y

2

, then v must satisfy

∂v

∂y

=

∂u

∂x

= 2x.

So we must have

v

= 2

xy

+

g

(

x

) for some function

g

(

x

). The other Cauchy-

Riemann equation gives

−2y =

∂u

∂y

= −

∂v

∂x

= −2y − g

0

(x).

This tells us

g

0

(

x

) = 0. So

g

must be a genuine constant, say

α

. The corresponding

analytic function whose real part is u is therefore

f(z) = x

2

− y

2

+ 2ixy + iα = (x + iy)

2

+ iα = z

2

+ iα.

Note that in an exam, if we were asked to find the analytic function

f

with real

part

u

(where

u

is given), then we must express it in terms of

z

, and not

x

and

y, or else it is not clear this is indeed analytic.

On the other hand, if we are given that

f

(

z

) =

u

+

iv

is analytic, then we

can compute

∂

2

u

∂x

2

=

∂

∂x

∂u

∂x

=

∂

∂x

∂v

∂y

=

∂

∂y

∂v

∂x

=

∂

∂y

−

∂u

∂y

= −

∂

2

u

∂y

2

.

So u satisfies Laplace’s equation in two dimensions, i.e.

∇

2

u =

∂

2

u

∂x

2

+

∂

2

u

∂y

2

= 0.

Similarly, so does v.

Definition

(Harmonic function)

.

A function satisfying Laplace’s equation equa-

tion in an open set is said to be harmonic.

Thus we have shown the following:

Proposition.

The real and imaginary parts of any analytic function are har-

monic.

1.4 Multi-valued functions

For

z

=

r

iθ

, we define

log z

=

log r

+

iθ

. There are infinitely many values of

log z, for every choice of θ. For example,

log i =

πi

2

or

5πi

2

or −

3πi

2

or ··· .

This is fine, right? Functions can be multi-valued. Nothing’s wrong.

Well, when we write down an expression, it’d better be well-defined. So we

really should find some way to deal with this.

This section is really more subtle than it sounds like. It turns out it is non-

trivial to deal with these multi-valued functions. We can’t just, say, randomly

require

θ

to be in, say,

(0, 2π]

, or else we will have some continuity problems, as

we will later see.

Branch points

Consider the three curves shown in the diagram.

C

3

C

1

C

2

In

C

1

, we could always choose

θ

to be always in the range

0,

π

2

, and then

log z

would be continuous and single-valued going round C

1

.

On

C

2

, we could choose

θ ∈

π

2

,

3π

2

and

log z

would again be continuous

and single-valued.

However, this doesn’t work for

C

3

. Since this encircles the origin, there is no

such choice. Whatever we do,

log z

cannot be made continuous and single-valued

around

C

3

. It must either “jump” somewhere, or the value has to increase by

2πi every time we go round the circle, i.e. the function is multi-valued.

We now define what a branch point is. In this case, it is the origin, since

that is where all our problems occur.

Definition

(Branch point)

.

A branch point of a function is a point which is

impossible to encircle with a curve on which the function is both continuous and

single-valued. The function is said to have a branch point singularity there.

Example.

(i) log(z − a) has a branch point at z = a.

(ii) log

z−1

z+1

= log(z − 1) − log(z + 1) has two branch points at ±1.

(iii) z

α

=

r

α

e

iαθ

has a branch point at the origin as well for

α 6∈ Z

— consider

a circle of radius of

r

0

centered at 0, and wlog that we start at

θ

= 0 and

go once round anticlockwise. Just as before,

θ

must vary continuous to

ensure continuity of

e

iαθ

. So as we get back almost to where we started,

θ

will approach 2

π

, and there will be a jump in

θ

from 2

π

back to 0. So

there will be a jump in

z

α

from

r

α

0

e

2πiα

to

r

α

0

. So

z

α

is not continuous if

e

2πiα

6= 1, i.e. α is not an integer.

(iv) log z

also has a branch point at

∞

. Recall that to investigate the properties

of a function

f

(

z

) at infinity, we investigate the property of

f

1

z

at zero.

If

ζ

=

1

z

, then

log z

=

−log ζ

, which has a branch point at

ζ

= 0. Similarly,

z

α

has a branch point at ∞ for α 6∈ Z.

(v)

The function

log

z−1

z+1

does not have a branch point at infinity, since if

ζ =

1

z

, then

log

z − 1

z + 1

= log

1 − ζ

1 + ζ

.

For

ζ

close to zero,

1−ζ

1+ζ

remains close to 1, and therefore well away from

the branch point of

log

at the origin. So we can encircle

ζ

= 0 without

log

1−ζ

1+ζ

being discontinuous.

So we’ve identified the points where the functions have problems. How do

we deal with these problems?

Branch cuts

If we wish to make

log z

continuous and single valued, therefore, we must stop

any curve from encircling the origin. We do this by introducing a branch cut

from −∞ on the real axis to the origin. No curve is allowed to cross this cut.

z

θ

Once we’ve decided where our branch cut is, we can use it to fix on values of

θ

lying in the range (

−π, π

], and we have defined a branch of

log z

. This branch

is single-valued and continuous on any curve

C

that does not cross the cut.

This branch is in fact analytic everywhere, with

d

dz

log z

=

1

z

, except on the

non-positive real axis, where it is not even continuous.

Note that a branch cut is the squiggly line, while a branch is a particular

choice of the value of log z.

The cut described above is the canonical (i.e. standard) branch cut for

log z

.

The resulting value of log z is called the principal value of the logarithm.

What are the values of

log z

just above and just below the branch cut?

Consider a point on the negative real axis,

z

=

x <

0. Just above the cut, at

z

=

x

+

i

0

+

, we have

θ

=

π

. So

log z

=

log |x|

+

iπ

. Just below it, at

z

=

x

+

i

0

−

,

we have log z = log |x| − iπ. Hence we have a discontinuity of 2πi.

We have picked an arbitrary branch cut and branch. We can pick other

branch cuts or branches. Even with the same branch cut, we can still have a

different branch — we can instead require

θ

to fall in (

π,

3

π

]. Of course, we can

also pick other branch cuts, e.g. the non-negative imaginary axis. Any cut that

stops curves wrapping around the branch point will do.

Here we can choose θ ∈

−

3π

2

,

π

2

. We can also pick a branch cut like this:

The exact choice of

θ

is more difficult to write down, but this is an equally valid

cut, since it stops curves from encircling the origin.

Exactly the same considerations (and possible branch cuts) apply for

z

α

(for

α 6∈ Z).

In practice, whenever a problem requires the use of a branch, it is important

to specify it clearly. This can be done in two ways:

(i) Define the function and parameter range explicitly, e.g.

log z = log |z| + i arg z, arg z ∈ (−π, π].

(ii)

Specify the location of the branch cut and give the value of the required

branch at a single point not on the cut. The values everywhere else are

then defined uniquely by continuity. For example, we have

log z

with a

branch cut along

R

≤0

and

log

1 = 0. Of course, we could have defined

log 1 = 2πi as well, and this would correspond to picking arg z ∈ (π, 3π].

Either way can be used, but it must be done properly.

Riemann surfaces*

Instead of this brutal way of introducing a cut and forbidding crossing, Riemann

imagined different branches as separate copies of

C

, all stacked on top of each

other but each one joined to the next at the branch cut. This structure is a

Riemann surface.

C

C

C

C

C

The idea is that traditionally, we are not allowed to cross branch cuts. Here,

when we cross a branch cut, we will move to a different copy of

C

, and this

corresponds to a different branch of our function.

We will not say any more about this — there is a whole Part II course

devoted to these, uncreatively named IID Riemann Surfaces.

Multiple branch cuts

When there is more than one branch point, we may need more than one branch

cut. For

f(z) = (z(z − 1))

1

3

,

there are two branch points, at 0 and 1. So we need two branch cuts. A possibility

is shown below. Then no curve can wrap around either 0 or 1.

10

z

r

r

1

θ

θ

1

For any

z

, we write

z

=

re

iθ

and

z −

1 =

r

1

e

iθ

1

with

θ ∈

(

−π, π

] and

θ

1

∈

[0

,

2

π

),

and define

f(z) =

3

√

rr

1

e

i(θ+θ

1

)/3

.

This is continuous so long as we don’t cross either branch cut. This is all and

simple.

However, sometimes, we need fewer branch cuts than we might think. Con-

sider instead the function

f(z) = log

z − 1

z + 1

.

Writing z + 1 = re

iθ

and z − 1 = r

1

e

iθ

1

, we can write this as

f(z) = log(z − 1) − log(z + 1)

= log(r

1

/r) + i(θ

1

− θ).

This has branch points at

±

1. We can, of course, pick our branch cut as above.

However, notice that these two cuts also make it impossible for

z

to “wind

around

∞

” (e.g. moving around a circle of arbitrarily large radius). Yet

∞

is not

a branch point, and we don’t have to make this unnecessary restriction. Instead,

we can use the following branch cut:

1−1

z

r

r

1

θ

θ

1

Drawing this branch cut is not hard. However, picking the values of

θ, θ

1

is

more tricky. What we really want to pick is

θ, θ

1

∈

[0

,

2

π

). This might not look

intuitive at first, but we will shortly see why this is the right choice.

Suppose that we are unlawful and cross the branch cut. Then the value of

θ

passes through the branch cut, while the value of

θ

1

varies smoothly. So the

value of

f

(

z

) jumps. This is expected since we have a branch cut there. If we

pass through the negative real axis on the left of the branch cut, then nothing

happens, since θ = θ

1

= π are not at a point of discontinuity.

The interesting part is when we pass through the positive real axis on the

right of branch cut. When we do this, both

θ

and

θ

1

jump by 2

π

. However, this

does not induce a discontinuity in

f

(

z

), since

f

(

z

) depends on the difference

θ

1

− θ, which has not experienced a jump.

1.5 M¨obius map

We are now going to consider a special class of maps, namely the M¨obius maps, as

defined in IA Groups. While these maps have many many different applications,

the most important thing we are going to use it for is to define some nice

conformal mappings in the next section.

We know from general theory that the M¨obius map

z 7→ w =

az + b

cz + d

with

ad − bc 6

= 0 is analytic except at

z

=

−

d

c

. It is useful to consider it as a

map from C

∗

→ C

∗

= C ∪ {∞}, with

−

d

c

7→ ∞, ∞ 7→

a

c

.

It is then a bijective map between C

∗

and itself, with the inverse being

w 7→

−dw + b

cw − a

,

another M¨obius map. These are all analytic everywhere when considered as a

map C

∗

→ C

∗

.

Definition (Circline). A circline is either a circle or a line.

The key property of M¨obius maps is the following:

Proposition. M¨obius maps take circlines to circlines.

Note that if we start with a circle, we might get a circle or a line; if we start

with a line, we might get a circle or a line.

Proof. Any circline can be expressed as a circle of Apollonius,

|z − z

1

| = λ|z − z

2

|,

where z

1

, z

2

∈ C and λ ∈ R

+

.

This was proved in the first example sheet of IA Vectors and Matrices. The

case

λ

= 1 corresponds to a line, while

λ 6

= 1 corresponds to a circle. Substituting

z in terms of w, we get

−dw + b

cw − a

− z

1

= λ

−dw + b

cw − a

− z

2

.

Rearranging this gives

|(cz

1

+ d)w − (az

1

+ b)| = λ|(cz

2

+ d)w − (az

2

+ b)|. (∗)

A bit more rearranging gives

w −

az

1

+ b

cz

1

+ d

= λ

cz

2

+ d

cz

1

+ d

w −

az

2

+ b

cz

2

+ d

.

This is another circle of Apollonius.

Note that the proof fails if either

cz

1

+

d

= 0 or

cz

2

+

d

= 0, but then (

∗

)

trivially represents a circle.

Geometrically, it is clear that choosing three distinct points in

C

∗

uniquely

specifies a circline (if one of the points is

∞

, then we have specified the straight

line through the other two points).

Also,

Proposition.

Given six points

α, β, γ, α

0

, β

0

, γ

0

∈ C

∗

, we can find a M¨obius

map which sends α 7→ α

0

, β 7→ β

0

and γ → γ

0

.

Proof. Define the M¨obius map

f

1

(z) =

β − γ

β − α

z − α

z − γ

.

By direct inspection, this sends α → 0, β → 1 and γ → ∞. Again, we let

f

2

(z) =

β

0

− γ

0

β

0

− α

0

z − α

0

z − γ

0

.

This clearly sends

α

0

→

0

, β

0

→

1 and

γ

0

→ ∞

. Then

f

−1

2

◦ f

1

is the required

mapping. It is a M¨obius map since M¨obius maps form a group.

Therefore, we can therefore find a M¨obius map taking any given circline to

any other, which is convenient.

1.6 Conformal maps

Sometimes, we might be asked to solve a problem on some complicated subspace

U ⊆ C

. For example, we might need to solve Laplace’s equation subject to

some boundary conditions. In such cases, it is often convenient to transform

our space

U

into some nicer space

V

, such as the open disk. To do so, we will

need a complex function

f

that sends

U

to

V

. For this function to preserve our

properties such that the solution on

V

can be transferred back to a solution on

U

, we would of course want

f

to be differentiable. Moreover, we would like it to

have non-vanishing derivative, so that it is at least locally invertible.

Definition

(Conformal map)

.

A conformal map

f

:

U → V

, where

U, V

are

open subsets of C, is one which is analytic with non-zero derivative.

In reality, we would often want the map to be a bijection. We sometimes call

these conformal equivalences.

Unfortunately, after many hundred years, we still haven’t managed to agree

on what being conformal means. An alternative definition is that a conformal

map is one that preserves the angle (in both magnitude and orientation) between

intersecting curves.

We shall show that our definition implies this is true; the converse is also

true, but the proof is omitted. So the two definitions are equivalent.

Proposition.

A conformal map preserves the angles between intersecting curves.

Proof.

Suppose

z

1

(

t

) is a curve in

C

, parameterised by

t ∈ R

, which passes

through a point

z

0

when

t

=

t

1

. Suppose that its tangent there,

z

0

1

(

t

1

), has a

well-defined direction, i.e. is non-zero, and the curve makes an angle

φ

=

arg z

0

1

(

t

1

)

to the x-axis at z

0

.

Consider the image of the curve,

Z

1

(

t

) =

f

(

z

1

(

t

)). Its tangent direction at

t = t

1

is

Z

0

1

(t

1

) = z

0

1

(t

1

)f

0

(z

1

(t

1

)) = z

0

1

(t

0

)f

0

(z

0

),

and therefore makes an angle with the x-axis of

arg(Z

0

1

(t

1

)) = arg(z

0

1

(t

1

)f

0

(z

0

)) = φ + arg f

0

(z

0

),

noting that arg f

0

(z

0

) exists since f is conformal, and hence f

0

(z

0

) 6= 0.

In other words, the tangent direction has been rotated by

arg f

0

(

z

0

), and this

is independent of the curve we started with.

Now if

z

2

(

t

) is another curve passing through

z

0

. Then its tangent direction

will also be rotated by arg f

0

(z

0

). The result then follows.

Often, the easiest way to find the image set of a conformal map acting on a

set

U

is first to find the image of its boundary,

∂U

, which will form the boundary

∂V

of

V

; but, since this does not reveal which side of

∂V V

lies on, take a point

of your choice within U, whose image will lie within V .

Example.

(i)

The map

z 7→ az

+

b

, where

a, b ∈ C

and

a 6

= 0, is a conformal map. It

rotates by

arg a

, enlarges by

|a|

, and translates by

b

. This is conformal

everywhere.

(ii) The map f(z) = z

2

is a conformal map from

U =

n

z : 0 < |z| < 1, 0 < arg z <

π

2

o

to

V = {w : 0 < |w| < 1, 0 < arg w < π}.

1

U

f

1

V

Note that the right angles between the boundary curves at

z

= 1 and

i

are

preserved, because

f

is conformal there; but the right angle at

z

= 0 is not

preserved because

f

is not conformal there (

f

0

(0) = 0). Fortunately, this

does not matter, because U is an open set and does not contain 0.

(iii) How could we conformally map the left-hand half-plane

U = {z : Re z < 0}

to a wedge

V =

n

w : −

π

4

< arg w ≤

π

4

o

.

U

V

We need to halve the angle. We saw that

z 7→ z

2

doubles then angle, so we

might try

z

1

2

, for which we need to choose a branch. The branch cut must

not lie in

U

, since

z

1

2

is not analytic on the branch cut. In particular, the

principal branch does not work.

So we choose a cut along the negative imaginary axis, and the function is

defined by

re

iθ

7→

√

re

iθ/2

, where

θ ∈

−

π

2

,

3π

2

. This produces the wedge

{z

0

:

π

4

< arg z

0

<

3π

4

}

. This isn’t exactly the wedge we want. So we need

to rotate it through −

π

2

. So the final map is

f(z) = −iz

1

2

.

(iv) e

z

takes rectangles conformally to sectors of annuli:

U

iy

1

iy

2

x

1

x

2

e

x

1

e

x

2

y

1

y

2

V

With an appropriate choice of branch, log z does the reverse.

(v)

M¨obius maps (which are conformal equivalence except at the point that is

sent to

∞

) are very useful in taking circles, or parts of them to straight

lines, or vice versa.

Consider

f

(

z

) =

z−1

z+1

acting on the unit disk

U

=

{z

:

|z| <

1

}

. The

boundary of

U

is a circle. The three points

−

1

, i

and +1 lie on this circle,

and are mapped to ∞, i and 0 respectively.

Since M¨obius maps take circlines to circlines, the image of

∂U

is the

imaginary axis. Since

f

(0) =

−

1, we see that the image of

U

is the

left-hand half plane.

U

V

We can derive this alternatively by noting

w =

z − 1

z + 1

⇔ z = −

w + 1

w − 1

.

So

|z| < 1 ⇔ |w + 1| < |w − 1|,

i.e.

w

is closer to

−

1 than it is to +1, which describes precisely the left-hand

half plane.

In fact, this particular map

f

(

z

) =

z−1

z+1

can be deployed more generally on

quadrants, because it permutes 8 divisions on the complex plane as follows:

23

67

14

58

The map sends 1

7→

2

7→

3

7→

4

7→

1 and 5

7→

6

7→

7

7→

8

7→

5. In

particular, this agrees with what we had above — it sends the complete

circle to the left hand half plane.

(vi)

Consider the map

f

(

z

) =

1

z

. This is just another M¨obius map! Hence

everything we know about M¨obius maps apply to this. In particular, it is

useful for acting on vertical and horizontal lines. Details are left for the

first example sheet.

In practice, complicated conformal maps are usually built up from individual

building blocks, each a simple conformal map. The required map is the com-

position of these. For this to work, we have to note that the composition of

conformal maps is conformal, by the chain rule.

Example.

Suppose we want to map the upper half-disc

|z| <

1,

Im z >

0 to the

full disc

|z| <

1. We might want to just do

z 7→ z

2

. However, this does not work,

since the image does not include the non-negative real axis, say

z

=

1

2

. Instead,

we need to do something more complicated. We will do this in several steps:

(i) We apply f

1

(z) =

z−1

z+1

to take the half-disc to the second quadrant.

(ii)

We now recall that

f

1

also takes the right-hand half plane to the disc. So we

square and rotate to get the right-hand half plane. We apply

f

2

(

z

) =

iz

2

.

(iii) We apply f

3

(z) = f

1

(z) again to obtain the disc.

Then the desired conformal map is

f

3

◦ f

2

◦ f

1

, you can, theoretically, expand

this out and get an explicit expression, but that would be a waste of time.

z 7→

z−1

z+1

z 7→ iz

2

z 7→

z−1

z+1

1.7 Solving Laplace’s equation using conformal maps

As we have mentioned, conformal maps are useful for transferring problems from

a complicated domain to a simple domain. For example, we can use it to solve

Laplace’s equation, since solutions to Laplace’s equations are given by real and

imaginary parts of holomorphic functions.

More concretely, the following algorithm can be used to solve Laplace’s

Equation

∇

2

φ

(

x, y

) = 0 on a tricky domain

U ⊆ R

2

with given Dirichlet

boundary conditions on

∂U

. We now pretend

R

2

is actually

C

, and identify

subsets of R

2

with subsets of C in the obvious manner.

(i)

Find a conformal map

f

:

U → V

, where

U

is now considered a subset of

C

, and

V

is a “nice” domain of our choice. Our aim is to find a harmonic

function Φ in V that satisfies the same boundary conditions as φ.

(ii)

Map the boundary conditions on

∂U

directly to the equivalent points on

∂V .

(iii) Now solve ∇

2

Φ = 0 in V with the new boundary conditions.

(iv) The required harmonic function φ in U is then given by

φ(x, y) = Φ(Re(f(x + iy)), Im f(x + iy)).

To prove this works, we can take the

∇

2

of this expression, write

f

=

u

+

iv

, use

the Cauchy-Riemann equation, and expand the mess.

Alternatively, we perform magic. Note that since Φ is harmonic, it is the

real part of some complex analytic function

F

(

z

) = Φ(

x, y

) +

i

Ψ(

x, y

), where

z

=

x

+

iy

. Now

F

(

f

(

z

)) is analytic, as it is a composition of analytic functions.

So its real part, which is Φ(Re f, Im f), is harmonic.

Let’s do an example. In this case, you might be able to solve this directly

just by looking at it, using what you’ve learnt from IB Methods. However, we

will do it with this complex methods magic.

Example.

We want to find a bounded solution of

∇

2

φ

= 0 on the first quadrant

of R

2

subject to φ(x, 0) = 0 and φ(0, y) = 1 when, x, y > 0.

This is a bit silly, since our

U

is supposed to be a nasty region, but our

U

is

actually quite nice. Nevertheless, we still do this since this is a good example.

We choose f(z) = log z, which maps U to the strip 0 < Im z <

π

2

.

U

1

0

z 7→ log z

V

i

π

2

00

1

Recall that we said

log

maps an annulus to a rectangle. This is indeed the case

here —

U

is an annulus with zero inner radius and infinite outer radius;

V

is an

infinitely long rectangle.

Now, we must now solve ∇

2

Φ = 0 in V subject to

Φ(x, 0) = 0, Φ

x,

π

2

= 1

for all

x ∈ R

. Note that we have these boundary conditions since

f

(

z

) takes

positive real axis of

∂V

to the line

Im z

= 0, and the positive imaginary axis to

Im z =

π

2

.

By inspection, the solution is

Φ(x, y) =

2

π

y.

Hence,

Φ(x, y) = Φ(Re log z, Im log z)

=

2

π

Im log z

=

2

π

tan

−1

y

x

.

Notice this is just the argument θ.

2 Contour integration and Cauchy’s theorem

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

of complex functions, and see what we can do with it. At first, you might think

this is just an obvious generalization of integration of real functions. This is not

true. Complex integrals have many many nice properties, and it turns out there

are some really convenient tricks for evaluating complex integrals. In fact, we

will learn how to evaluate certain real integrals by pretending they are complex.

2.1 Contour and integrals

With real functions, we can just integrate a function, say, from 0 to 1, since there

is just one possible way we can get from 0 to 1 along the real line. However, in

the complex plane, there are many paths we can take to get from a point to

another. Integrating along different paths may produce different results. So we

have to carefully specify our path of integration.

Definition (Curve). A curve γ(t) is a (continuous) map γ : [0, 1] → C.

Definition (Closed curve). A closed curve is a curve γ such that γ(0) = γ(1).

Definition

(Simple curve)

.

A simple curve is one which does not intersect itself,

except at t = 0, 1 in the case of a closed curve.

Definition (Contour). A contour is a piecewise smooth curve.

Everything we do is going to be about contours. We shall, in an abuse of

notation, often use the symbol

γ

to denote both the map and its image, namely

the actual curve in C traversed in a particular direction.

Notation.

The contour

−γ

is the contour

γ

traversed in the opposite direction.

Formally, we say

(−γ)(t) = γ(1 − t).

Given two contours

γ

1

and

γ

2

with

γ

1

(1) =

γ

2

(0),

γ

1

+

γ

2

denotes the two

contours joined end-to-end. Formally,

(γ

1

+ γ

2

)(t) =

(

γ

1

(2t) t <

1

2

γ

2

(2t − 1) t ≥

1

2

.

Definition

(Contour integral)

.

The contour integral

R

γ

f

(

z

) d

z

is defined to be

the usual real integral

Z

γ

f(z) dz =

Z

1

0

f(γ(t))γ

0

(t) dt.

Alternatively, and equivalently, dissect [0

,

1] into 0 =

t

0

< t

1

< ··· < t

n

= 1,

and let z

n

= γ(t

n

) for n = 0, ··· , N. We define

δt

n

= t

n+1

− t

n

, δz

n

= z

n+1

− z

n

.

Then

Z

γ

f(z) dz = lim

∆→0

N−1

X

n=0

f(z

n

)δz

n

,

where

∆ = max

n=0,··· ,N −1

δt

n

,

and as ∆ → 0, N → ∞.

All this says is that the integral is what we expect it to be — an infinite sum.

The result of a contour integral between two points in

C

may depend on the

choice of contour.

Example. Consider

I

1

=

Z

γ

1

dz

z

, I

2

=

Z

γ

2

dz

z

,

where the paths are given by

γ

1

γ

2

θ

1−1 0

In both cases, we integrate from

z

=

−

1 to +1 around a unit circle:

γ

1

above,

γ

2

below the real axis. Substitute z = e

iθ

, dz = ie

iθ

dθ. Then we get

I

1

=

Z

0

π

ie

iθ

dθ

e

iθ

= −iπ

I

2

=

Z

0

−π

ie

iθ

dθ

e

iθ

= iπ.

So they can in fact differ.

Elementary properties of the integral

Contour integrals behave as we would expect them to.

Proposition.

(i) We write γ

1

+ γ

2

for the path obtained by joining γ

1

and γ

2

. We have

Z

γ

1

+γ

2

f(z) dz =

Z

γ

1

f(z) dz +

Z

γ

2

f(z) dz

Compare this with the equivalent result on the real line:

Z

c

a

f(x) dx =

Z

b

a

f(x) dx +

Z

c

b

f(x) dx.

(ii) Recall −γ is the path obtained from reversing γ. Then we have

Z

−γ

f(z) dz = −

Z

γ

f(z) dz.

Compare this with the real result

Z

b

a

f(x) dx = −

Z

a

b

f(x) dx.

(iii) If γ is a contour from a to b in C, then

Z

γ

f

0

(z) dz = f(b) − f(a).

This looks innocuous. This is just the fundamental theorem of calculus.

However, there is some subtlety. This requires

f

to be differentiable at

every point on

γ

. In particular, it must not cross a branch cut. For example,

our previous example had

log z

as the antiderivative of

1

z

. However, this

does not imply the integrals along different paths are the same, since we

need to pick different branches of

log

for different paths, and things become

messy.

(iv)

Integration by substitution and by parts work exactly as for integrals on

the real line.

(v) If γ has length L and |f (z)| is bounded by M on γ, then

Z

γ

f(z) dz

≤ LM.

This is since

Z

γ

f(z) dz

≤

Z

γ

|f(z)|| dz| ≤ M

Z

γ

|dz| = ML.

We will be using this result a lot later on.

We will not prove these. Again, if you like proofs, go to IB Complex Analysis.

Integrals on closed contours

If

γ

is a closed contour, then it doesn’t matter where we start from on

γ

;

H

γ

f

(

z

) d

z

means the same thing in any case, so long as we go all the way round

(

H

denotes an integral around a closed contour).

The usual direction of traversal is anticlockwise (the “positive sense”). If

we traverse

γ

in a negative sense (clockwise), then we get negative the previous

result. More technically, the positive sense is the direction that keeps the interior

of the contour on the left. This “more technical” definition might seem pointless

— if you can’t tell what anticlockwise is, then you probably can’t tell which is

the left. However, when we deal with more complicated structures in the future,

it turns out it is easier to define what is “on the left” than “anticlockwise”.

Simply connected domain

Definition

(Simply connected domain)

.

A domain

D

(an open subset of

C

) is

simply connected if it is connected and every closed curve in

D

encloses only

points which are also in D.

In other words, it does not have holes. For example, this is not simply-

connected:

These “holes” need not be big holes like this, but just individual points at which

a function under consider consideration is singular.

2.2 Cauchy’s theorem

We now come to the highlight of the course — Cauchy’s theorem. Most of the

things we do will be based upon this single important result.

Theorem

(Cauchy’s theorem)

.

If

f

(

z

) is analytic in a simply-connected domain

D, then for every simple closed contour γ in D, we have

I

γ

f(z) dz = 0.

This is quite a powerful statement, and will allow us to do a lot! On the other

hand, this tells us functions that are analytic everywhere are not too interesting.

Instead, we will later look at functions like

1

z

that have singularities.

Proof.

(non-examinable) The proof of this remarkable theorem is simple (with a

catch), and follows from the Cauchy-Riemann equations and Green’s theorem.

Recall that Green’s theorem says

I

∂S

(P dx + Q dy) =

ZZ

S

∂Q

∂x

−

∂P

∂y

dx dy.

Let u, v be the real and imaginary parts of f. Then

I

γ

f(z) dz =

I

γ

(u + iv)(dx + i dy)

=

I

γ

(u dx − v dy) + i

I

γ

(v dx + u dy)

=

ZZ

S

−

∂v

∂x

−

∂u

∂y

dx dy + i

ZZ

S

∂u

∂x

−

∂v

∂y

dx dy

But both integrands vanish by the Cauchy-Riemann equations, since

f

is differ-

entiable throughout S. So the result follows.

Actually, this proof requires

u

and

v

to have continuous partial derivatives

in

S

, otherwise Green’s theorem does not apply. We shall see later that in fact

f

is differentiable infinitely many time, so actually

u

and

v

do have continuous

partial derivatives. However, our proof of that will utilize Cauchy’s theorem! So

we are trapped.

Thus a completely different proof (and a very elegant one!) is required if we

do not wish to make assumptions about

u

and

v

. However, we shall not worry

about this in this course since it is easy to verify that the functions we use do

have continuous partial derivatives. And we are not doing Complex Analysis.

2.3 Contour deformation

One useful consequence of Cauchy’s theorem is that we can freely deform contours

along regions where f is defined without changing the value of the integral.

Proposition.

Suppose that

γ

1

and

γ

2

are contours from

a

to

b

, and that

f

is

analytic on the contours and between the contours. Then

Z

γ

1

f(z) dz =

Z

γ

2

f(z) dz.

a

b

γ

2

γ

1

Proof.

Suppose first that

γ

1

and

γ

2

do not cross. Then

γ

1

−γ

2

is a simple c