part ib - complex methods · part ib | complex methods based on lectures by r. e. hunt notes taken...

Part IB — Complex Methods

Based on lectures by R. E. HuntNotes taken by Dexter Chua

Lent 2016

These notes are not endorsed by the lecturers, and I have modified them (oftensignificantly) after lectures. They are nowhere near accurate representations of what

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

Analytic functionsDefinition of an analytic function. Cauchy-Riemann equations. Analytic functions asconformal mappings; examples. Application to the solutions of Laplace’s equation invarious domains. Discussion of log z and za. [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. Taylorand Laurent series. Zeros, poles and essential singularities. [3]

Residue calculusResidue theorem, calculus of residues. Jordan’s lemma. Evaluation of definite integralsby 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]

1

Contents IB Complex Methods

Contents

0 Introduction 3

1 Analytic functions 41.1 The complex plane and the Riemann sphere . . . . . . . . . . . . 41.2 Complex differentiation . . . . . . . . . . . . . . . . . . . . . . . 51.3 Harmonic functions . . . . . . . . . . . . . . . . . . . . . . . . . . 91.4 Multi-valued functions . . . . . . . . . . . . . . . . . . . . . . . . 101.5 Mobius map . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.6 Conformal maps . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.7 Solving Laplace’s equation using conformal maps . . . . . . . . . 20

2 Contour integration and Cauchy’s theorem 232.1 Contour and integrals . . . . . . . . . . . . . . . . . . . . . . . . 232.2 Cauchy’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 262.3 Contour deformation . . . . . . . . . . . . . . . . . . . . . . . . . 272.4 Cauchy’s integral formula . . . . . . . . . . . . . . . . . . . . . . 28

3 Laurent series and singularities 313.1 Taylor and Laurent series . . . . . . . . . . . . . . . . . . . . . . 313.2 Zeros . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343.3 Classification of singularities . . . . . . . . . . . . . . . . . . . . . 353.4 Residues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

4 The calculus of residues 414.1 The residue theorem . . . . . . . . . . . . . . . . . . . . . . . . . 414.2 Applications of the residue theorem . . . . . . . . . . . . . . . . . 424.3 Further applications of the residue theorem using rectangular

contours . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 494.4 Jordan’s lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

5 Transform theory 555.1 Fourier transforms . . . . . . . . . . . . . . . . . . . . . . . . . . 555.2 Laplace transform . . . . . . . . . . . . . . . . . . . . . . . . . . 585.3 Elementary properties of the Laplace transform . . . . . . . . . . 605.4 The inverse Laplace transform . . . . . . . . . . . . . . . . . . . 615.5 Solution of differential equations using the Laplace transform . . 645.6 The convolution theorem for Laplace transforms . . . . . . . . . 65

2

0 Introduction IB Complex Methods

0 Introduction

In Part IA, we learnt quite a lot about differentiating and integrating realfunctions. Differentiation was fine, but integration was tedious. Integrals werevery 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 complexdifferentiable functions satisfy many things we hoped were true — a complexdifferentiable function is automatically infinitely differentiable. Moreover, aneverywhere differentiable function must be constant if it is bounded.

On the integration side, we will show that integrals of complex functions canbe performed by computing things known as residues, which are much easierto compute. We are not actually interested in performing complex integrals.Instead, we will take some difficult real integrals, and pretend they are complexones.

This is a methods course. By this, we mean we will not focus too much onproofs. We will at best just skim over the proofs. Instead, we focus on doingthings. 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 themup in relevant books.

3

1 Analytic functions IB Complex Methods

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 canbe written in the form x+ iy, where x = Re z, y = Im z are real numbers. Wecan also write it as reiθ, where

Definition (Modulus and argument). The modulus and argument of a complexnumber z = x+ iy are given by

r = |z| =√x2 + y2, θ = arg z,

where x = r cos θ, y = r sin θ.

The argument is defined only up to multiples of 2π. So we define the following:

Definition (Principal value of argument). The principal value of the argumentis the value of θ in the range (−π, π].

We might be tempted to write down the formula

θ = tan−1(yx

),

but this does not always give the right answer — it is correct only if x > 0. Ifx ≤ 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 z0 ∈ D, there is some δ > 0 such thatthe disc |z − z0| < δ is contained in D.

Definition (Neighbourhood). A neighbourhood of a point z ∈ C is an open setcontaining 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 isC∗ = C∪{∞}. We can reach the “point at infinity” by going off in any directionin 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 adifferent point. Instead, this indicates a limiting process. We mean we areapproaching 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 asphere resting on the complex plane with its “South Pole” S at z = 0.

4


S

N

P

z

For any point z ∈ C, drawing a line through the “North Pole” N of the sphereto z, and noting where this intersects the sphere. This specifies an equivalentpoint P on the sphere. Then ∞ is equivalent to the North Pole of the sphereitself. 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 actuallywant to do computations. To investigate properties of∞, we use the substitutionζ = 1

z . A function f(z) is said to have a particular property at ∞ if f( 1ζ ) has

that same property at ζ = 0. This vague notion will be made precise when wehave specific examples to play with.

1.2 Complex differentiation

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

f ′(x) = limδx→0

f(x+ δx)− f(x)

δx.

It is implicit that the limit must be the same whichever direction we approachfrom. 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 theorigin.

This is obvious and we already know that, but for complex differentiation,this issue is much more important, since there are many more directions. Wenow extend the definition of differentiation to complex number:

Definition (Complex differentiable function). A complex differentiable functionf : C→ C is differentiable at z if

f ′(z) = limδz→0

f(z + δz)− f(z)

δz

exists (and is therefore independent of the direction of approach — but nowthere are infinitely many possible directions).

This is the same definition as that for a real function. Often, we are notinterested in functions that are differentiable at a point, since this might allowsome rather exotic functions we do not want to consider. Instead, we want thefunction to be differentiable near the point.

Definition (Analytic function). We say f is analytic at a point z if thereexists a neighbourhood of z throughout which f ′ exists. The terms regular andholomorphic are also used.

5


Definition (Entire function). A complex function is entire if it is analyticthroughout 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. Thisis very very false for real functions. There are real functions differentiableN times, but no more (e.g. by taking a non-differentiable function andintegrating it N times).

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

There are many more interesting properties, but these are sufficient to show usthat 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 todetermine whether certain complex functions are differentiable? For example isthe function f(x+ iy) = cosx+ i sin y differentiable? In general, given a complexfunction

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

where z = x+iy are u, v are real functions, is there an easy criterion to determinewhether f is differentiable?

We suppose that f is differentiable at z. We may take δz in any direction welike. First, we take it to be real, with δz = δx. Then

f ′(z) = limδx→0

f(z + δx)− f(z)

δx

= limδx→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 thelimit 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 ′(z) = limδy→0

f(z + iδy)− f(z)

iδy

= limδy→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 ′(z) must agree! So wemust have

∂u

∂x+ i

∂v

∂x=∂v

∂y− i∂u

∂y.

Taking the real and imaginary components, we get

6


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 aredifferentiable, which is a stronger condition that the partial derivatives exist, asyou may have learnt from IB Analysis II. In particular, this holds if the partialderivatives ux, uy, vx, vy 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. Thenthe 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 directlyfrom the definition.

(ii) f(z) = ez = ex(cos y + i sin y) is entire since

∂u

∂x= ex cos y =

∂v

∂y,

∂u

∂y= −ex sin y = −∂v

∂x.

The derivative is

f ′(z) =∂u

∂x+ i

∂v

∂x= ex cos y + iex sin y = ez,

as expected.

(iii) f(z) = zn for n ∈ N is entire. This is less straightforward to check. Writingz = r(cos θ + i sin θ), we obtain

u = rn cosnθ, v = rn sinnθ.

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

√x2 = y2 and tan θ = y

x . This takes quite a while, and it’s not worthyour time. But if you really do so, you will find the derivative to be nzn−1.

7


(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

z2 + 1

is analytic except at ±i.

(v) Many standard functions can be extended naturally to complex functionsand obey the usual rules for their derivatives. For example,

– sin z = eiz−e−iz2i is differentiable with derivative cos z = eiz+e−iz

2 . Wecan also write

sin z = sin(x+ iy)

= sinx cos iy + cosx sin iy

= sinx cosh y + i cosx sinh y,

which is sometimes convenient.

– Similarly cos z, sinh z, cosh z etc. differentiate to what we expect themto differentiate to.

– log z = log |z|+ i arg z has derivative 1z .

– The product rule, quotient rule and chain rule hold in exactly thesame 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 =√x2 + y2, v = 0. This is thus nowhere

analytic.

(iii) The complex conjugate f(z) = z = z∗ = x− iy has u = x, v = −y. So theCauchy-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|2z also has to be analytic, which is not true.

(iv) We have to be a bit more careful with f(z) = |z|2 = x2 + y2. TheCauchy-Riemann equations are satisfied only at the origin. So f is onlydifferentiable at z = 0. However, it is not analytic since there is noneighbourhood of 0 throughout which f is differentiable.

8


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, ifu(x, y) = x2 − y2, then v must satisfy

∂v

∂y=∂u

∂x= 2x.

So we must have v = 2xy + g(x) for some function g(x). The other Cauchy-Riemann equation gives

−2y =∂u

∂y= −∂v

∂x= −2y − g′(x).

This tells us g′(x) = 0. So g must be a genuine constant, say α. The correspondinganalytic function whose real part is u is therefore

f(z) = x2 − y2 + 2ixy + iα = (x+ iy)2 + iα = z2 + iα.

Note that in an exam, if we were asked to find the analytic function f with realpart u (where u is given), then we must express it in terms of z, and not x andy, 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 wecan compute

∂2u

∂x2=

∂

∂x

(∂u

∂x

)=

∂

∂x

(∂v

∂y

)=

∂

∂y

(∂v

∂x

)=

∂

∂y

(−∂u∂y

)= −∂

2u

∂y2.

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

∇2u =∂2u

∂x2+∂2u

∂y2= 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.

9


1.4 Multi-valued functions

For z = riθ, we define log z = log r + iθ. There are infinitely many values oflog z, for every choice of θ. For example,

log i =πi

2or

5πi

2or − 3πi

2or · · · .

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, randomlyrequire θ to be in, say, (0, 2π], or else we will have some continuity problems, aswe will later see.

Branch points

Consider the three curves shown in the diagram.

C3

C1

C2

In C1, we could always choose θ to be always in the range(0, π2

), and then log z

would be continuous and single-valued going round C1.On C2, we could choose θ ∈

(π2 ,

3π2

)and log z would again be continuous

and single-valued.However, this doesn’t work for C3. Since this encircles the origin, there is no

such choice. Whatever we do, log z cannot be made continuous and single-valuedaround C3. It must either “jump” somewhere, or the value has to increase by2πi 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, sincethat is where all our problems occur.

Definition (Branch point). A branch point of a function is a point which isimpossible to encircle with a curve on which the function is both continuous andsingle-valued. The function is said to have a branch point singularity there.

10


Example.

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

(ii) log(z−1z+1

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

(iii) zα = rαeiαθ has a branch point at the origin as well for α 6∈ Z — considera circle of radius of r0 centered at 0, and wlog that we start at θ = 0 andgo once round anticlockwise. Just as before, θ must vary continuous toensure continuity of eiαθ. So as we get back almost to where we started,θ will approach 2π, and there will be a jump in θ from 2π back to 0. Sothere will be a jump in zα from rα0 e

2πiα to rα0 . So zα is not continuous ife2πiα 6= 1, i.e. α is not an integer.

(iv) log z also has a branch point at∞. Recall that to investigate the propertiesof a function f(z) at infinity, we investigate the property of f

(1z

)at zero.

If ζ = 1z , then log z = − log ζ, which has a branch point at ζ = 0. Similarly,

zα has a branch point at ∞ for α 6∈ Z.

(v) The function log(z−1z+1

)does not have a branch point at infinity, since if

ζ = 1z , then

log

(z − 1

z + 1

)= log

(1− ζ1 + ζ

).

For ζ close to zero, 1−ζ1+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−ζ1+ζ

)being discontinuous.

So we’ve identified the points where the functions have problems. How dowe deal with these problems?

Branch cuts

If we wish to make log z continuous and single valued, therefore, we must stopany curve from encircling the origin. We do this by introducing a branch cutfrom −∞ 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 θlying in the range (−π, π], and we have defined a branch of log z. This branchis single-valued and continuous on any curve C that does not cross the cut.

11


This branch is in fact analytic everywhere, with ddz log z = 1

z , except on thenon-positive real axis, where it is not even continuous.

Note that a branch cut is the squiggly line, while a branch is a particularchoice 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, atz = x+ i0+, we have θ = π. So log z = log |x|+ iπ. Just below it, at z = x+ i0−,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 otherbranch cuts or branches. Even with the same branch cut, we can still have adifferent branch — we can instead require θ to fall in (π, 3π]. Of course, we canalso pick other branch cuts, e.g. the non-negative imaginary axis. Any cut thatstops curves wrapping around the branch point will do.

Here we can choose θ ∈(− 3π

2 ,π2

]. 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 validcut, since it stops curves from encircling the origin.

Exactly the same considerations (and possible branch cuts) apply for zα (forα 6∈ Z).

In practice, whenever a problem requires the use of a branch, it is importantto 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 requiredbranch at a single point not on the cut. The values everywhere else are

12


then defined uniquely by continuity. For example, we have log z with abranch cut along R≤0 and log 1 = 0. Of course, we could have definedlog 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, Riemannimagined different branches as separate copies of C, all stacked on top of eachother but each one joined to the next at the branch cut. This structure is aRiemann 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 thiscorresponds to a different branch of our function.

We will not say any more about this — there is a whole Part II coursedevoted to these, uncreatively named IID Riemann Surfaces.

Multiple branch cuts

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

f(z) = (z(z − 1))13 ,

there are two branch points, at 0 and 1. So we need two branch cuts. A possibilityis shown below. Then no curve can wrap around either 0 or 1.

13


10

z

r r1

θ θ1

For any z, we write z = reiθ and z− 1 = r1eiθ1 with θ ∈ (−π, π] and θ1 ∈ [0, 2π),

and definef(z) = 3

√rr1e

i(θ+θ1)/3.

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

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 = reiθ and z − 1 = r1eiθ1 , we can write this as

f(z) = log(z − 1)− log(z + 1)

= log(r1/r) + i(θ1 − 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 “windaround∞” (e.g. moving around a circle of arbitrarily large radius). Yet∞ is nota branch point, and we don’t have to make this unnecessary restriction. Instead,we can use the following branch cut:

1−1

z

r r1

θ θ1

Drawing this branch cut is not hard. However, picking the values of θ, θ1 ismore tricky. What we really want to pick is θ, θ1 ∈ [0, 2π). This might not lookintuitive 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 ofθ passes through the branch cut, while the value of θ1 varies smoothly. So the

14


value of f(z) jumps. This is expected since we have a branch cut there. If wepass through the negative real axis on the left of the branch cut, then nothinghappens, since θ = θ1 = π are not at a point of discontinuity.

The interesting part is when we pass through the positive real axis on theright of branch cut. When we do this, both θ and θ1 jump by 2π. However, thisdoes not induce a discontinuity in f(z), since f(z) depends on the differenceθ1 − θ, which has not experienced a jump.

1.5 Mobius map

We are now going to consider a special class of maps, namely the Mobius maps, asdefined 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 niceconformal mappings in the next section.

We know from general theory that the Mobius map

z 7→ w =az + b

cz + d

with ad − bc 6= 0 is analytic except at z = −dc . It is useful to consider it as amap from C∗ → C∗ = C ∪ {∞}, with

−dc7→ ∞, ∞ 7→ a

c.

It is then a bijective map between C∗ and itself, with the inverse being

w 7→ −dw + b

cw − a,

another Mobius map. These are all analytic everywhere when considered as amap C∗ → C∗.

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

The key property of Mobius maps is the following:

Proposition. Mobius maps take circlines to circlines.

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

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

|z − z1| = λ|z − z2|,

where z1, z2 ∈ 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. Substitutingz in terms of w, we get∣∣∣∣−dw + b

cw − a− z1

∣∣∣∣ = λ

∣∣∣∣−dw + b

cw − a− z2

∣∣∣∣ .

15


Rearranging this gives

|(cz1 + d)w − (az1 + b)| = λ|(cz2 + d)w − (az2 + b)|. (∗)

A bit more rearranging gives∣∣∣∣w − az1 + b

cz1 + d

∣∣∣∣ = λ

∣∣∣∣cz2 + d

cz1 + d

∣∣∣∣ ∣∣∣∣w − az2 + b

cz2 + d

∣∣∣∣ .This is another circle of Apollonius.

Note that the proof fails if either cz1 + d = 0 or cz2 + d = 0, but then (∗)trivially represents a circle.

Geometrically, it is clear that choosing three distinct points in C∗ uniquelyspecifies a circline (if one of the points is ∞, then we have specified the straightline through the other two points).

Also,

Proposition. Given six points α, β, γ, α′, β′, γ′ ∈ C∗, we can find a Mobiusmap which sends α 7→ α′, β 7→ β′ and γ → γ′.

Proof. Define the Mobius map

f1(z) =β − γβ − α

z − αz − γ

.

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

f2(z) =β′ − γ′

β′ − α′z − α′

z − γ′.

This clearly sends α′ → 0, β′ → 1 and γ′ → ∞. Then f−12 ◦ f1 is the required

mapping. It is a Mobius map since Mobius maps form a group.

Therefore, we can therefore find a Mobius map taking any given circline toany other, which is convenient.

1.6 Conformal maps

Sometimes, we might be asked to solve a problem on some complicated subspaceU ⊆ C. For example, we might need to solve Laplace’s equation subject tosome boundary conditions. In such cases, it is often convenient to transformour space U into some nicer space V , such as the open disk. To do so, we willneed a complex function f that sends U to V . For this function to preserve ourproperties such that the solution on V can be transferred back to a solution onU , we would of course want f to be differentiable. Moreover, we would like it tohave non-vanishing derivative, so that it is at least locally invertible.

Definition (Conformal map). A conformal map f : U → V , where U, V areopen 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 callthese conformal equivalences.

Unfortunately, after many hundred years, we still haven’t managed to agreeon what being conformal means. An alternative definition is that a conformal

16


map is one that preserves the angle (in both magnitude and orientation) betweenintersecting curves.

We shall show that our definition implies this is true; the converse is alsotrue, but the proof is omitted. So the two definitions are equivalent.

Proposition. A conformal map preserves the angles between intersecting curves.

Proof. Suppose z1(t) is a curve in C, parameterised by t ∈ R, which passesthrough a point z0 when t = t1. Suppose that its tangent there, z′1(t1), has awell-defined direction, i.e. is non-zero, and the curve makes an angle φ = arg z′1(t1)to the x-axis at z0.

Consider the image of the curve, Z1(t) = f(z1(t)). Its tangent direction att = t1 is

Z ′1(t1) = z′1(t1)f ′(z1(t1)) = z′1(t0)f ′(z0),

and therefore makes an angle with the x-axis of

arg(Z ′1(t1)) = arg(z′1(t1)f ′(z0)) = φ+ arg f ′(z0),

noting that arg f ′(z0) exists since f is conformal, and hence f ′(z0) 6= 0.In other words, the tangent direction has been rotated by arg f ′(z0), and this

is independent of the curve we started with.Now if z2(t) is another curve passing through z0. Then its tangent direction

will also be rotated by arg f ′(z0). The result then follows.

Often, the easiest way to find the image set of a conformal map acting on aset 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 pointof 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. Itrotates by arg a, enlarges by |a|, and translates by b. This is conformaleverywhere.

(ii) The map f(z) = z2 is a conformal map from

U ={z : 0 < |z| < 1, 0 < arg z <

π

2

}to

V = {w : 0 < |w| < 1, 0 < argw < π}.

1

Uf

1

V

17


Note that the right angles between the boundary curves at z = 1 and i arepreserved, because f is conformal there; but the right angle at z = 0 is notpreserved because f is not conformal there (f ′(0) = 0). Fortunately, thisdoes 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 ={w : −π

4< argw ≤ π

4

}.

U

V

We need to halve the angle. We saw that z 7→ z2 doubles then angle, so wemight try z

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

not lie in U , since z12 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 isdefined by reiθ 7→

√reiθ/2, where θ ∈

(−π2 ,

3π2

]. This produces the wedge

{z′ : π4 < arg z′ < 3π4 }. This isn’t exactly the wedge we want. So we need

to rotate it through −π2 . So the final map is

f(z) = −iz 12 .

(iv) ez takes rectangles conformally to sectors of annuli:

Uiy1

iy2

x1 x2

ex1 ex2

y1 y2

V

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

18


(v) Mobius maps (which are conformal equivalence except at the point that issent to ∞) are very useful in taking circles, or parts of them to straightlines, or vice versa.

Consider f(z) = z−1z+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 Mobius maps take circlines to circlines, the image of ∂U is theimaginary axis. Since f(0) = −1, we see that the image of U is theleft-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-handhalf plane.

In fact, this particular map f(z) = z−1z+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. Inparticular, this agrees with what we had above — it sends the completecircle to the left hand half plane.

(vi) Consider the map f(z) = 1z . This is just another Mobius map! Hence

everything we know about Mobius maps apply to this. In particular, it isuseful for acting on vertical and horizontal lines. Details are left for thefirst example sheet.

19


In practice, complicated conformal maps are usually built up from individualbuilding 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 ofconformal maps is conformal, by the chain rule.

Example. Suppose we want to map the upper half-disc |z| < 1, Im z > 0 to thefull disc |z| < 1. We might want to just do z 7→ z2. 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 f1(z) = z−1z+1 to take the half-disc to the second quadrant.

(ii) We now recall that f1 also takes the right-hand half plane to the disc. So wesquare and rotate to get the right-hand half plane. We apply f2(z) = iz2.

(iii) We apply f3(z) = f1(z) again to obtain the disc.

Then the desired conformal map is f3 ◦ f2 ◦ f1, you can, theoretically, expandthis out and get an explicit expression, but that would be a waste of time.

z 7→ z−1z+1

z 7→ iz2 z 7→ z−1z+1

1.7 Solving Laplace’s equation using conformal maps

As we have mentioned, conformal maps are useful for transferring problems froma complicated domain to a simple domain. For example, we can use it to solveLaplace’s equation, since solutions to Laplace’s equations are given by real andimaginary parts of holomorphic functions.

More concretely, the following algorithm can be used to solve Laplace’sEquation ∇2φ(x, y) = 0 on a tricky domain U ⊆ R2 with given Dirichletboundary conditions on ∂U . We now pretend R2 is actually C, and identifysubsets of R2 with subsets of C in the obvious manner.

(i) Find a conformal map f : U → V , where U is now considered a subset ofC, and V is a “nice” domain of our choice. Our aim is to find a harmonicfunction Φ in V that satisfies the same boundary conditions as φ.

20


(ii) Map the boundary conditions on ∂U directly to the equivalent points on∂V .

(iii) Now solve ∇2Φ = 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, usethe Cauchy-Riemann equation, and expand the mess.

Alternatively, we perform magic. Note that since Φ is harmonic, it is thereal part of some complex analytic function F (z) = Φ(x, y) + iΨ(x, y), wherez = 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 directlyjust by looking at it, using what you’ve learnt from IB Methods. However, wewill do it with this complex methods magic.

Example. We want to find a bounded solution of ∇2φ = 0 on the first quadrantof R2 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 isactually 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 < π2 .

U1

0

z 7→ log z V

iπ2

00

1

Recall that we said log maps an annulus to a rectangle. This is indeed the casehere — U is an annulus with zero inner radius and infinite outer radius; V is aninfinitely long rectangle.

Now, we must now solve ∇2Φ = 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) takespositive real axis of ∂V to the line Im z = 0, and the positive imaginary axis toIm z = π

2 .By inspection, the solution is

Φ(x, y) =2

πy.

21


Hence,

Φ(x, y) = Φ(Re log z, Im log z)

=2

πIm log z

=2

πtan−1

(yx

).

Notice this is just the argument θ.

22

2 Contour integration and Cauchy’s theorem IB Complex Methods

2 Contour integration and Cauchy’s theorem

In the remaining of the course, we will spend all our time studying integrationof complex functions, and see what we can do with it. At first, you might thinkthis is just an obvious generalization of integration of real functions. This is nottrue. Complex integrals have many many nice properties, and it turns out thereare some really convenient tricks for evaluating complex integrals. In fact, wewill 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 thereis just one possible way we can get from 0 to 1 along the real line. However, inthe complex plane, there are many paths we can take to get from a point toanother. Integrating along different paths may produce different results. So wehave 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 ofnotation, often use the symbol γ to denote both the map and its image, namelythe 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 γ1 and γ2 with γ1(1) = γ2(0), γ1 + γ2 denotes the twocontours joined end-to-end. Formally,

(γ1 + γ2)(t) =

{γ1(2t) t < 1

2

γ2(2t− 1) t ≥ 12

.

Definition (Contour integral). The contour integral∫γf(z) dz is defined to be

the usual real integral ∫γ

f(z) dz =

∫ 1

0

f(γ(t))γ′(t) dt.

Alternatively, and equivalently, dissect [0, 1] into 0 = t0 < t1 < · · · < tn = 1,and let zn = γ(tn) for n = 0, · · · , N . We define

δtn = tn+1 − tn, δzn = zn+1 − zn.

Then ∫γ

f(z) dz = lim∆→0

N−1∑n=0

f(zn)δzn,

23


where∆ = max

n=0,··· ,N−1δtn,

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

I1 =

∫γ1

dz

z, I2 =

∫γ2

dz

z,

where the paths are given by

γ1

γ2

θ1−1 0

In both cases, we integrate from z = −1 to +1 around a unit circle: γ1 above,γ2 below the real axis. Substitute z = eiθ, dz = ieiθ dθ. Then we get

I1 =

∫ 0

π

ieiθ dθ

eiθ= −iπ

I2 =

∫ 0

−π

ieiθ dθ

eiθ= 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 γ1 + γ2 for the path obtained by joining γ1 and γ2. We have∫γ1+γ2

f(z) dz =

∫γ1

f(z) dz +

∫γ2

f(z) dz

Compare this with the equivalent result on the real line:∫ c

a

f(x) dx =

∫ b

a

f(x) dx+

∫ c

b

f(x) dx.

24


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

f(z) dz = −∫γ

f(z) dz.

Compare this with the real result∫ b

a

f(x) dx = −∫ a

b

f(x) dx.

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

f ′(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 atevery 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, thisdoes not imply the integrals along different paths are the same, since weneed to pick different branches of log for different paths, and things becomemessy.

(iv) Integration by substitution and by parts work exactly as for integrals onthe real line.

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

f(z) dz

∣∣∣∣ ≤ LM.

This is since ∣∣∣∣∫γ

f(z) dz

∣∣∣∣ ≤ ∫γ

|f(z)|| dz| ≤M∫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;∮γf(z) dz means the same thing in any case, so long as we go all the way round

(∮

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 previousresult. More technically, the positive sense is the direction that keeps the interiorof 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 isthe 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”.

25


Simply connected domain

Definition (Simply connected domain). A domain D (an open subset of C) issimply connected if it is connected and every closed curve in D encloses onlypoints 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 whicha 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 thethings we do will be based upon this single important result.

Theorem (Cauchy’s theorem). If f(z) is analytic in a simply-connected domainD, then for every simple closed contour γ in D, we have∮

γ

f(z) dz = 0.

This is quite a powerful statement, and will allow us to do a lot! On the otherhand, 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 acatch), and follows from the Cauchy-Riemann equations and Green’s theorem.Recall that Green’s theorem says∮

∂S

(P dx+Q dy) =

∫∫S

(∂Q

∂x− ∂P

∂y

)dx dy.

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

f(z) dz =

∮γ

(u+ iv)(dx+ i dy)

=

∮γ

(u dx− v dy) + i

∮γ

(v dx+ u dy)

=

∫∫S

(−∂v∂x− ∂u

∂y

)dx dy + i

∫∫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.

26


Actually, this proof requires u and v to have continuous partial derivativesin S, otherwise Green’s theorem does not apply. We shall see later that in factf is differentiable infinitely many time, so actually u and v do have continuouspartial derivatives. However, our proof of that will utilize Cauchy’s theorem! Sowe are trapped.

Thus a completely different proof (and a very elegant one!) is required if wedo not wish to make assumptions about u and v. However, we shall not worryabout this in this course since it is easy to verify that the functions we use dohave 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 contoursalong regions where f is defined without changing the value of the integral.

Proposition. Suppose that γ1 and γ2 are contours from a to b, and that f isanalytic on the contours and between the contours. Then∫

γ1

f(z) dz =

∫γ2

f(z) dz.

a

b

γ2

γ1

Proof. Suppose first that γ1 and γ2 do not cross. Then γ1− γ2 is a simple closedcontour. So ∮

γ1−γ2f(z) dz = 0

by Cauchy’s theorem. Then the result follows.If γ1 and γ2 do cross, then dissect them at each crossing point, and apply

the previous result to each section.

So we conclude that if f has no singularities, then∫ baf(z) dz does not depend

on the chosen contour.This result of path independence, and indeed Cauchy’s theorem itself, becomes

less surprising if we think of∫f(z) dz as a path integral in R2, because

f(z) dz = (u+ iv)(dz + i dy) = (u+ iv) dx+ (−v + iu) dy

is an exact differential, since

∂

∂y(u+ iv) =

∂

∂x(−v + iu)

27


from the Cauchy-Riemann equations.The same idea of “moving the contour” applies to closed contours. Suppose

that γ1 is a closed contour that can be continuously deformed to another one,γ2, inside it; and suppose f has no singularities in the region between them.

γ2γ1

××

×

We can instead consider the following contour γ:

γ

××

×

By Cauchy’s theorem, we know∮γf(z) dz = 0 since f(z) is analytic throughout

the region enclosed by γ. Now we let the distance between the two “cross-cuts”tend to zero: those contributions cancel and, in the limit, we have∮

γ1−γ2f(z) dz = 0.

hence we know ∮γ1

f(z) dz =

∮γ2

f(z) dz = 0.

2.4 Cauchy’s integral formula

Theorem (Cauchy’s integral formula). Suppose that f(z) is analytic in a domainD and that z0 ∈ D. Then

f(z0) =1

2πi

∮γ

f(z)

z − z0dz

for any simple closed contour γ in D encircling z0 anticlockwise.

This result is going to be very important in a brief moment, for proving onething. Afterwards, it will be mostly useless.

28


Proof. (non-examinable) We let γε be a circle of radius ε about z0, within γ.

z0 γε

γ

Since f(z)z−z0 is analytic except when z = z0, we know∮

γ

f(z)

z − z0dz =

∮γε

f(z)

z − z0dz.

We now evaluate the right integral directly. Substituting z = z0 + εeiθ, we get∮γε

f(z)

z − z0dz =

∫ 2π

0

f(z0 + εeiθ)

εeiθiεeiθ dθ

= i

∫ 2

0

π(f(z0) +O(ε)) dθ

→ 2πif(z0)

as we take the limit ε→ 0. The result then follows.

So, if we know f on γ, then we know it at all points within γ. While thisseems magical, it is less surprising if we look at it in another way. We can writef = u + iv, where u and v are harmonic functions, i.e. they satisfy Laplace’sequation. Then if we know the values of u and v on γ, then what we essentiallyhave is Laplace’s equation with Dirichlet boundary conditions! Then the factthat this tells us everything about f within the boundary is just the statementthat Laplace’s equation with Dirichlet boundary conditions has a unique solution!

The difference between this and what we’ve got in IA Vector Calculus is thatCauchy’s integral formula gives an explicit formula for the value of f(z0), whilein IA Vector Calculus, we just know there is one solution, whatever that mightbe.

Note that this does not hold if z0 does not lie on or inside γ, since Cauchy’stheorem just gives

1

2πi

∮γ

f(z)

z − z0dz = 0.

Now, we can differentiate Cauchy’s integral formula with respect to z0, andobtain

f ′(z0) =1

2πi

∮γ

f(z)

(z − z0)2dz.

We have just taken the differentiation inside the integral sign. This is valid sinceit’s Complex Methods and we don’t care the integrand, both before and after,is a continuous function of both z and z0.

29


We see that the integrand is still differentiable. So we can differentiate itagain, and obtain

f (n)(z0) =n!

2πi

∮γ

f(z)

(z − z0)n+1dz.

Hence at any point z0 where f is analytic, all its derivatives exist, and we havejust found a formula for them. So it is differentiable infinitely many times asadvertised.

A classic example of Cauchy’s integral formula is Liouville’s theorem.

Theorem (Liouville’s theorem*). Any bounded entire function is a constant.

Proof. (non-examinable) Suppose that |f(z)| ≤M for all z, and consider a circleof radius r centered at an arbitrary point z0 ∈ C. Then

f ′(z0) =1

2πi

∮|z−z0|=r

f(z)

(z − z0)2dz.

Hence we know1

2πi≤ 1

2π· 2πr · M

r2→ 0

as r →∞. So f ′(z0) = 0 for all z0 ∈ C. So f is constant.

30

3 Laurent series and singularities IB Complex Methods

3 Laurent series and singularities

3.1 Taylor and Laurent series

If f is analytic at z0, then it has a Taylor series

f(z) =

∞∑n=0

an(z − z0)n

in a neighbourhood of z0. We will prove this as a special case of the comingproposition. Exactly which neighbourhood it applies in depends on the function.Of course, we know the coefficients are given by

an =f (n)(z0)

n!,

but this doesn’t matter. All the standard Taylor series from real analysis applyin C as well. For example,

ez =

∞∑n=0

zn

n!,

and this converges for all z. Also, we have

(1− z)−1 =

∞∑n=0

zn.

This converges for |z| < 1.But if f has a singularity at z0, we cannot expect such a Taylor series, since

it would imply f is non-singular at z0. However, it turns out we can get a seriesexpansion if we allow ourselves to have negative powers of z.

Proposition (Laurent series). If f is analytic in an annulus R1 < |z−z0| < R2,then it has a Laurent series

f(z) =

∞∑n=−∞

an(z − z0)n.

This is convergent within the annulus. Moreover, the convergence is uniformwithin compact subsets of the annulus.

Proof. (non-examinable) We wlog z0 = 0. Given a z in the annulus, we pickr1, r2 such that

R1 < r1 < |z| < r2 < R2,

and we let γ1 and γ2 be the contours |z| = r1, |z| = r2 traversed anticlockwiserespectively. We choose γ to be the contour shown in the diagram below.

z0

r1r2

z

γ

31


We now apply Cauchy’s integral formula (after a change of notation):

f(z) =1

2πi

∮γ

f(ζ)

ζ − zdζ.

We let the distance between the cross-cuts tend to zero. Then we get

f(z) =1

2πi

∮γ2

f(ζ)

ζ − zdz − 1

2πi

∮γ1

f(ζ)

ζ − zdζ,

We have to subtract the second integral because it is traversed in the oppositedirection. We do the integrals one by one. We have∮

γ1

f(ζ)

ζ − z= −1

z

∮γ1

f(ζ)

1− ζz

dζ

Taking the Taylor series of 11− ζz

, which is valid since |ζ| = r1 < |z| on γ1, we

obtain

= −1

z

∮γ1

f(ζ)

∞∑m=0

(ζ

z

)mdζ

= −∞∑m=0

z−m−1

∮γ1

f(ζ)ζm dζ.

This is valid since∮ ∑

=∑∮

by uniform convergence. So we can deduce

− 1

2πi

∮γ1

f(ζ)

ζ − zdζ =

−1∑n=−∞

anzn,

where

an =1

2πi

∮γ1

f(ζ)ζ−n−1 dζ

for n < 0.Similarly, we obtain

1

2πi

∮γ2

f(ζ)

ζ − zdζ =

∞∑n=0

anzn,

for the same definition of an, except n ≥ 0, by expanding

1

ζ − z=

1

ζ

1

1− zζ

=

∞∑n=0

zn

ζn+1.

This is again valid since |ζ| = r2 > |z| on γ2. Putting these result together, weobtain the Laurent series. The general result then follows by translating theorigin by z0.

We will not prove uniform convergence — go to IB Complex Analysis.

32


It can be shown that Laurent series are unique. Then not only is there aunique Laurent series for each annulus, but if we pick two different annuli onwhich f is analytic (assuming they overlap), they must have the same coefficient.This is since we can restrict the two series to their common intersection, andthen uniqueness requires the coefficients to be must be the same.

Note that the Taylor series is just a special case of Laurent series, and if f isholomorphic at z0, then our explicit formula for the coefficients plus Cauchy’stheorem tells us we have an = 0 for n < 0.

Note, however, that we needed the Taylor series of 11−z in order to prove

Taylor’s theorem.

Example. Consider ez

z3 . What is its Laurent series about z0 = 0? We alreadyhave a Taylor series for ez, and all we need to do is to divide it by z3. So

ez

z3=

∞∑n=0

zn−3

n!=

∞∑n=−3

zn

(n+ 3)!.

Example. What is the Laurent series for e1/z about z0? Again, we just use theTaylor series for ez. We have

e1z = 1 +

1

z+

1

2!z2+

1

3!z3+ · · · .

So the coefficients are

an =1

(−n)!for n ≤ 0.

Example. This is a little bit more tricky. Consider

f(z) =1

z − a,

where a ∈ C. Then f is analytic in |z| < |a|. So it has a Taylor series aboutz0 = 0 given by

1

z − a= −1

a

(1− z

a

)−1

= −∞∑n=0

1

an+1zn.

What about in |z| > |a|? This is an annulus, that goes from |a| to infinity. So ithas a Laurent series. We can find it by

1

z − a=

1

z

(1− a

z

)−1

=

∞∑m=0

am

zm+1=

−1∑n=−∞

a−n−1zn.

Example. Now consider

f(z) =ez

z2 − 1.

This has a singularity at z0 = 1 but is analytic in an annulus 0 < |z − z0| < 2(the 2 comes from the other singularity at z = −1). How do we find its Laurent

33


series? This is a standard trick that turns out to be useful — we write everythingin terms of ζ = z − z0. So

f(z) =eζez0

ζ(ζ + 2)

= ez02ζeζ(

1 +1

2ζ

)−1

=e

2ζ

(1 + ζ +

1

2!ζ2 + · · ·

)(1− 1

2ζ + · · ·

)=

e

2ζ

(1 +

1

2ζ + · · ·

)=e

2

(1

z − z0+

1

2+ · · ·

).

This is now a Laurent series, with a−1 = 12e, a0 = 1

4e etc.

Example. This doesn’t seem to work for f(z) = z−1/2. The reason is that the

required branch cut of z−12 would pass through any annulus about z = 0. So we

cannot find an annulus on which f is analytic.

The radius of convergence of a Laurent series is at least the size of the annulus,as we that’s how we have constructed it. So how large is it?

It turns out the radius of convergence of a Laurent series is always thedistance from z0 to the closest singularity of f(z), for we may always chooseR2 to be that distance, and obtain a Laurent series valid all the way to R2

(not inclusive). This Laurent series must be the same series as we would haveobtained with a smaller radius, because Laurent series are unique.

While this sounds just like some technicalities, this is actually quite useful.When deriving a Laurent series, we can make any assumptions we like aboutz − z0 being small. Then even if we have derived a Laurent series for a reallysmall neighbourhood, we automatically know the series is valid up to the otherpoint where f is singular.

3.2 Zeros

Recall that for a polynomial p(z), we can talk about the order of its zero atz = a by looking at the largest power of (z − a) dividing p. A priori, it is notclear how we can do this for general functions. However, given that everythingis a Taylor series, we know how to do this for holomorphic functions.

Definition (Zeros). The zeros of an analytic function f(Z) are the points z0

where f(z0) = 0. A zero is of order N if in its Taylor series∑∞n=0 an(z − z0)n,

the first non-zero coefficient is aN .Alternatively, it is of order N if 0 = f(z0) = f ′(z0) = · · · = f (N−1), but

f (N)(z0) 6= 0.

Definition (Simple zero). A zero of order one is called a simple zero.

Example. z3 + iz2 + z + i = (z − i)(z + i)2 has a simple zero at z = i and azero of order 2 at z = −i.

34


Example. sinh z has zeros where 12 (ez − e−z) = 0, i.e. e2z = 1, i.e. z = nπi,

where n ∈ Z. The zeros are all simple, since coshnπi = cosnπ 6= 0.

Example. Since sinh z has a simple zero at z = πi, we know sinh3 z has a zeroof order 3 there. This is since the first term of the Taylor series of sinh z aboutz = πi has order 1, and hence the first term of the Taylor series of sinh3 z hasorder 3.

We can also find the Taylor series about πi by writing ζ = z − πi:

sinh3 z = [sinh(ζ + πi)]3

= [− sinh ζ]3

= −(ζ +

1

3!+ · · ·

)3

= −ζ3 − 1

2ζ5 − · · ·

= −(z − πi)3 − 1

2(z − πi)5 − · · · .

3.3 Classification of singularities

The previous section was rather boring — you’ve probably seen all of that before.It is just there as a buildup for our study of singularities. These are in somesense the “opposites” of zeros.

Definition (Isolated singularity). Suppose that f has a singularity at z0 = z.If there is a neighbourhood of z0 within which f is analytic, except at z0 itself,then f has an isolated singularity at z0. If there is no such neighbourhood, thenf has an essential (non-isolated) singularity at z0.

Example. cosech z has isolated singularities at z = nπi, n ∈ Z, since sinh haszeroes at these points.

Example. cosech 1z has isolated singularities at z = 1

nπi , with n 6= 0, and anessential non-isolated singularity at z = 0 (since there are other arbitrarily closesingularities).

Example. cosech z also has an essential non-isolated singularity at z =∞, sincecosech 1

z has an essential non-isolated singularity at z = 0.

Example. log z has a non-isolated singularity at z = 0, because it is not analyticat any point on the branch cut. This is normally referred to as a branch pointsingularity.

If f has an isolated singularity, at z0, we can find an annulus 0 < |z− z0| < rwithin which f is analytic, and it therefore has a Laurent series. This gives us away to classify singularities:

(i) Check for a branch point singularity.

(ii) Check for an essential (non-isolated) singularity.

(iii) Otherwise, consider the coefficients of the Laurent series∑∞n=−∞ an(z −

z0)n:

35


(a) If an = 0 for all n < 0, then f has a removable singularity at z0.

(b) If there is a N > 0 such that an = 0 for all n < −N but a−N 6= 0,then f has a pole of order N at z0 (for N = 1, 2, · · · , this is also calleda simple pole, double pole etc.).

(c) If there does not exist such an N , then f has an essential isolatedsingularity.

A removable singularity (one with Laurent series a0 +a1(z−z0)+ · · · ) is so calledbecause we can remove the singularity by redefining f(z0) = a0 = lim

z→z0f(z);

then f will become analytic at z0.Let’s look at some examples. In fact, we have 10 examples here.

Example.

(i) 1z−i has a simple pole at z = i. This is since its Laurent series is, err, 1

z−i .

(ii) cos zz has a singularity at the origin. This has Laurent series

cos z

z= z−1 − 1

2z +

1

24z3 − · · · ,

and hence it has a simple pole.

(iii) Consider z2

(z−1)3(z−i)2 . This has a double pole at z = i and a triple pole at

z = 1. To show formally that, for instance, there is a double pole at z = i,

notice first that z2

(z−1)3 is analytic at z = i. So it has a Taylor series, say,

b0 + b1(z − i) + b2(z − i)2 + · · ·

for some bn. Moreover, since z2

(z−1)3 is non-zero at z = i, we have b0 6= 0.

Hencez2

(z − 1)3(z − i)2=

b0(z − i)2

+b1z − i

+ b2 + · · · .

So this has a double pole at z = i.

(iv) If g(z) has zero of order N at z = z0, then 1g(z) has a pole of order N there,

and vice versa. Hence cot z has a simple pole at the origin, because tan zhas a simple zero there. To prove the general statement, write

g(z) = (z − z0)NG(z)

for some G with G(z0) 6= 0. Then 1G(z) has a Taylor series about z0, and

then the result follows.

(v) z2 has a double pole at infinity, since 1ζ2 has a double pole at ζ = 0.

(vi) e1/z has an essential isolated singularity at z = 0 because all the an’s arenon-zero for n ≤ 0.

(vii) sin 1z also has an essential isolated singularity at z = 0 because (using the

standard Taylor series for sin) there are non-zero an’s for infinitely manynegative n.

36


(viii) f(z) = ez−1z has a removable singularity at z = 0, because its Laurent

series is

f(z) = 1 +1

2!z +

1

3!z2 + · · · .

By defining f(0) = 1, we would remove the singularity and obtain an entirefunction.

(ix) f(z) = sin zz is not defined at z = 0, but has a removable singularity there;

remove it by setting f(0) = 1.

(x) A rational function f(z) = P (z)Q(z) (where P,Q are polynomials) has a

singularity at any point z0 where Q has a zero. Assuming Q has a simplezero, if P (z0) = 0 as well, then the singularity is removable by redefining

f(z0) = P ′(z0)Q′(z0) (by L’Hopital’s rule).

Near an essential isolated singularity of a function f(z), it can be shown thatf takes all possible complex values (except at most one) in any neighbourhood,

however small. For example, e1z takes all values except zero. We will not prove

this. Even in IB Complex Analysis.

3.4 Residues

So far, we’ve mostly been making lots of definitions. We haven’t actually usedthem to do much useful things. We are almost there. It turns out we caneasily evaluate integrals of analytic functions by looking at their Laurent series.Moreover, we don’t need the whole Laurent series. We just need one of thecoefficients.

Definition (Residue). The residue of a function f at an isolated singularity z0

is the coefficient a−1 in its Laurent expansion about z0. There is no standardnotation, but shall denote the residue by res

z=z0f(z).

Proposition. At a simple pole, the residue is given by

resz=z0

f(z) = limz→z0

(z − z0)f(z).

Proof. We can simply expand the right hand side to obtain

limz→z0

(z − z0)

(a−1

z − z0+ a0 + a1(z − z0) + · · ·

)= limz→z0

(a−1 + a0(z − z0) + · · · )

= a−1,

as required.

How about for more complicated poles? More generally, at a pole of orderN , the formula is a bit messier.

Proposition. At a pole of order N , the residue is given by

limz→z0

1

(N − 1)!

dN−1

dzN−1(z − z0)Nf(z).

37


This can be proved in a similar manner (see example sheet 2).In practice, a variety of techniques can be used to evaluate residues — no

single technique is optimal for all situations.

Example. Consider f(z) = ez

z3 . We can find the residue by directly computingthe Laurent series about z = 0:

ez

z3= z−3 + z−2 +

1

2z−1 +

1

3!+ · · · .

Hence the residue is 12 .

Alternatively, we can use the fact that f has a pole of order 3 at z = 0. Sowe can use the formula to obtain

resz=0

f(z) = limz→0

1

2!

d2

dz2(z3f(z)) = lim

z→0

1

2

d2

dz2ez =

1

2.

Example. Consider

g(z) =ez

z2 − 1.

This has a simple pole at z = 1. Recall we have found its Laurent series at z = 1to be

ez

z2 − 1=e

2

(1

z − 1+

1

2+ · · ·

).

So the residue is e2 .

Alternatively, we can use our magic formula to obtain

resz=1

g(z) = limz→1

(z − 1)ez

z2 − 1= limz→1

ez

z + 1=e

2.

Example. Consider h(z) = (z8−w8)−1, for any complex constant w. We knowthis has 8 simple poles at z = wenπi/4 for n = 0, · · · , 7. What is the residue atz = w?

We can try to compute this directly by

resz=w

h(z) = limz→w

z − w(z − w)(z − weiπ/4) · · · (z − we7πi/4)

=1

(w − weiπ/4) · · · (w − we7πi/4)

=1

w7

1

(1− eiπ/4) · · · (1− e7iπ/4)

Now we are quite stuck. We don’t know what to do with this. We can thinkreally hard about complex numbers and figure out what it should be, but this isdifficult. What we should do is to apply L’Hopital’s rule and obtain

resz=w

h(z) = limz→w

z − wz8 − w8

=1

8z7=

1

8w7.

Example. Consider the function (sinhπz)−1. This has a simple pole at z = nifor all integers n (because the zeros of sinh z are at nπi and are simple). Again,we can compute this by finding the Laurent expansion. However, it turns out itis easier to use our magic formula together with L’Hopital’s rule. We have

limz→ni

z − nisinhπz

= limz→ni

1

π coshπz=

1

π coshnπi=

1

π cosnπ=

(−1)n

π.

38


Example. Consider the function (sinh3 z)−1. This time, we find the residue bylooking at the Laurent series. We first look at sinh3 z. This has a zero of order3 at z = πi. Its Taylor series is

sinh3 z = −(z − πi)3 − 1

2(z − πi)5.

Therefore

1

sinh3 z= −(z − πi)−3

(1 +

1

2(z − πi)2 + · · ·

)−1

= −(z − πi)−3

(1− 1

2(z − πi)2 + · · ·

)= −(z − πi)−3 +

1

2(z − πi)−1 + · · ·

Therefore the residue is 12 .

So we can compute residues. But this seems a bit odd — why are weinterested in the coefficient of a−1? Out of the doubly infinite set of coefficients,why a−1?

The purpose of this definition is to aid in evaluating integrals∮f(z) dz,

where f is a function analytic within the anticlockwise simple closed contour γ,except for an isolated singularity z0. We let γr be a circle of radius r centeredon z0, lying within γ.

z0 γr

Now f has some Laurent series expansion∑∞n=−∞ an(z − z0)n about z0. Recall

that we are allowed to deform contours along regions where f is analytic. So∮γ

f(z) dz =

∮γr

f(z) dz

=

∮γr

∞∑n=−∞

an(z − z0)n dz

By uniform convergence, we can swap the integral and sum to obtain

=

∞∑n=−∞

an

∮γr

(z − z0)n dz

We know how to integrate around the circle:∮γr

(z − z0)n dz =

∫ 2π

0

rneinθireiθ dθ

= irn+1

∫ 2π

0

ei(n+1)θ dθ

=

{2πi n = −1rn+1

n+1

[ei(n+1)θ

]= 0, n 6= −1

.

39


Hence we have ∮γ

f(z) dz = 2πia−1 = 2πi resz=z0

f(z).

Theorem. γ be an anticlockwise simple closed contour, and let f be analyticwithin γ except for an isolated singularity z0. Then∮

γ

f(z) dz = 2πia−1 = 2πi resz=z0

f(z).

40

4 The calculus of residues IB Complex Methods

4 The calculus of residues

Nowadays, we use “calculus” to mean differentiation and integration. However,historically, the word “calculus” just means doing calculations. The word “calcu-lus” in the calculus of residues does not refer to differentiation and integration,even though in this case they are related, but this is just a coincidence.

4.1 The residue theorem

We are now going to massively generalize the last result we had in the previouschapter. We are going consider a function f with many singularities, and obtainan analogous formula.

Theorem (Residue theorem). Suppose f is analytic in a simply-connectedregion except at a finite number of isolated singularities z1, · · · , zn, and that asimple closed contour γ encircles the singularities anticlockwise. Then∮

γ

f(z) dz = 2πi

n∑k=1

resz=zk

f(z).

z1

z2

z3

γ

Note that we have already proved the case n = 1 in the previous section. Toprove this, we just need a difficult drawing.

Proof. Consider the following curve γ, consisting of small clockwise circlesγ1, · · · , γn around each singularity; cross cuts, which cancel in the limit as theyapproach each other, in pairs; and the large outer curve (which is the same as γin the limit).

γ

z1

z2

z3

41


Note that γ encircles no singularities. So∮γf(z) dz = 0 by Cauchy’s theorem.

So in the limit when the cross cuts cancel, we have∮γ

f(z) dz +

n∑k=1

∮γk

f(z) dz =

∮γ

f(z) dz = 0.

But from what we did in the previous section, we know∮γk

f(z) dz = −2πi resz=zk

f(z),

since γk encircles only one singularity, and we get a negative sign since γk is aclockwise contour. Then the result follows.

This is the key practical result of this course. We are going to be using thisvery extensively to compute integrals.

4.2 Applications of the residue theorem

We now use the residue theorem to evaluate lots of real integrals.

Example. We shall evaluate

I =

∫ ∞0

dx

1 + x2,

which we can already do so by trigonometric substitution. While it is sillyto integrate this with the residue theorem here, since integrating directly ismuch easier, this technique is a much more general method, and can be usedto integrate many other things. On the other hand, our standard tricks easilybecome useless when we change the integrand a bit, and we need to find acompletely different method.

Consider ∮γ

dz

1 + z2,

where γ is the contour shown: from −R to R along the real axis (γ0), thenreturning to −R via a semicircle of radius R in the upper half plane (γR). Thisis known as “closing in the upper-half plane”.

γ0

γR

−R R

×i

Now we have1

1 + z2=

1

(z + i)(z − i).

42


So the only singularity enclosed by γ is a simple pole at z = i, where the residueis

limz→i

1

z + i=

1

2i.

Hence ∫γ0

dz

1 + z2+

∫γR

dz

1 + z2=

∫γ

dz

1 + z2= 2πi · 1

2i= π.

Let’s now look at the terms individually. We know∫γ0

dz

1 + z2=

∫ R

−R

dx

1 + x2→ 2I

as R→∞. Also, ∫γR

dz

1 + z2→ 0

as R→∞ (see below). So we obtain in the limit

2I + 0 = π.

SoI =

π

2.

Finally, we need to show that the integral about γR vanishes as R→∞. This isusually a bit tricky. We can use a formal or informal argument. We first do itformally: by the triangle inequality, we know

|1 + z2| ≥ |1− |z|2|.

On γR, we know |z| = R. So for large R, we get

|1 + z2| ≥ 1−R2 = R2 − 1.

Hence1

|1 + z2|≤ 1

R2 − 1.

Thus we can bound the integral by∣∣∣∣∫γR

dz

1 + z2

∣∣∣∣ ≤ πR · 1

R2 − 1→ 0

as R→∞.We can also do this informally, by writing∣∣∣∣∫

γR

dz

1 + z2

∣∣∣∣ ≤ πR supz∈γR

∣∣∣∣ 1

1 + z2

∣∣∣∣ = πR ·O(R−2) = O(R−1)→ 0.

This example is not in itself impressive, but the method adapts easily to moredifficult integrals. For example, the same argument would allow us to integrate

11+x8 with ease.

Note that we could have “closed in the lower half-plane” instead.

43


−R R

×−i

Most of the argument would be unchanged; the residue would now be

resz=−i

1

1 + z2= − 1

2i,

but the contour is now traversed clockwise. So we collect another minus sign,and obtain the same result.

Let’s do more examples.

Example. To find the integral

I =

∫ ∞0

dx

(x2 + a2)2,

where a > 0 is a real constant, consider the contour integral∫γ

dz

(z2 + a2)2,

where γ is exactly as above. The only singularity within γ is a pole of order 2 atz = ia, at which the residue is

limz→ia

d

dz

1

(z + ia)2= limz→ia

−2

(z + ia)3

=−2

−8ia3

= −1

4ia−3.

We also have to do the integral around γR. This still vanishes as R→∞, since∣∣∣∣∫γR

dz

(z2 + a2)2

∣∣∣∣ ≤ πR ·O(R−4) = O(R−3)→ 0.

Therefore

2I = 2πi

(−1

4ia−3

).

SoI =

π

4a3.

Example. Consider

I =

∫ ∞0

dx

1 + x4.

44


We use the same contour again. There are simple poles of 11+z4 at

eπi/4, e3πi/4, e−πi/4, e−3πi/4,

but only the first two are enclosed by the contour.

γ0

γR

−R R

××

The residues at these two poles are − 14eπi/4 and + 1

4e−πi/4 respectively. Hence

2I = 2πi

(−1

4eπi/4 +

1

4e−πi/4

).

Working out some algebra, we find

I =π

2√

2.

Example. There is another way of doing the previous integral. Instead, we usethis quarter-circle as shown:

γ0

γ1

γ2

iR

R

×

In words, γ consists of the real axis from 0 to R (γ0); the arc circle from R toiR (γ1); and the imaginary axis from iR to 0 (γ2).

Now ∫γ0

dz

1 + z4→ I as R→∞,

and along γ2 we substitute z = iy to obtain∫γ0

dz

1 + z4=

∫ 0

R

idy

1 + y4→ −iI as R→∞.

Finally, the integral over γ1 vanishes as before. We enclose only one pole, whichmakes the calculation a bit easier than what we did last time. In the limit, weget

I − iI = 2πi

(−1

4eπi/4

),

45


and we again obtain

I =π

2√

2.

Example. We now look at trigonometric integrals of the form :∫ 2π

0

f(sin θ, cos θ) dθ.

We substitute

z = eiθ, cos θ =1

2(z + z−1), sin θ =

1

2i(z − z−1).

We then end up with a closed contour integral.For example, consider the integral

I =

∫ 2π

0

dθ

a+ cos θ,

where a > 1. We substitute z = eiθ, so that dz = iz dθ and cos θ = 12 (z + z−1).

As θ increases from 1 to 2π, z moves round the circle γ of radius 1 in the complexplane. Hence

I =

∮γ

(iz)−1 dz

a+ 12 (z + z−1)

= −2i

∮γ

dz

z2 + 2az + 1.

γ

×z−

×z+

We now solve the quadratic to obtain the poles, which happen to be

z± = −a±√a2 − 1.

With some careful thinking, we realize z+ is inside the circle, while z− is outside.To find the residue, we notice the integrand is equal to

1

(z − z+)(z − z−).

So the residue at z = z+ is

1

z+ − z−=

1

2√a2 − 1

.

46


Hence

I = −2i

(2πi

2√a2 − 1

)=

2π√a2 − 1

.

Example. Integrating a function with a branch cut is a bit more tricky, andrequires a “keyhole contour”. Suppose we want to integrate

I =

∫ ∞0

xα

1 +√

2x+ x2dx,

with −1 < α < 1 so that the integral converges. We need a branch cut for zα.We take our branch cut to be along the positive real axis, and define

zα = rαeiαθ,

where z = reiθ and 0 ≤ θ < 2π. We use the following keyhole contour:

CR

Cε

×e5πi/4

×e3πi/4

This consists of a large circle CR of radius R, a small circle Cε of radius ε, andthe two lines just above and below the branch cut. The idea is to simultaneouslytake the limit ε→ 0 and R→∞.

We have four integrals to work out. The first is∫γR

zα

1 +√

2z + z2dz = O(Rα−2) · 2πR = O(Rα−1)→ 0

as R → ∞. To obtain the contribution from γε, we substitute z = εeiθ, andobtain ∫ 0

2π

εαeiαθ

1 +√

2εeiθ + ε2e2iθiεeiθ dθ = O(εα+1)→ 0.

Finally, we look at the integrals above and below the branch cut. The contributionfrom just above the branch cut is∫ R

ε

xα

1 +√

2x+ x2dx→ I.

Similarly, the integral below is∫ ε

R

xαe2απi

1 +√

2x+ x2→ −e2απiI.

47


So we get ∮γ

zα

1 +√

2z + z2dz → (1− e2απi)I.

All that remains is to compute the residues. We write the integrand as

zα

(z − e3eπi/4)(z − e5πi/4).

So the poles are at z0 = e3πi/4 and z1 = e5πi/4. The residues are

e3απi/4

√2i

,e5απi/4

−√

2i

respectively. Hence we know

(1− e2απi)I = 2πi

(e3απi/4

√2i

+e5απi/4

−√

2i

).

In other words, we get

eαπi(e−απi − eαπi)I =√

2πeαπi(e−απi/4 − eαπi/4).

Thus we have

I =√

2πsin(απ/4)

sin(απ).

Note that we we labeled the poles as e3πi/4 and e5πi/4. The second point is thesame point as e−3πi/4, but it would be wrong to label it like that. We havedecided at the beginning to pick the branch such that 0 ≤ θ < 2π, and −3πi/4is not in that range. If we wrote it as e−3πi/4 instead, we might have got theresidue as e−3απi/4/(−

√2i), which is not the same as e5απi/4/(−

√2i).

Note that the need of a branch cut does not mean we must use a keyholecontour. Sometimes, we can get away with the branch cut and contour chosenas follows:

−R ε ε R

×i

48


4.3 Further applications of the residue theorem using rect-angular contours

Example. We want to calculate

I =

∫ ∞−∞

eαx

coshxdx,

where α ∈ (−1, 1). We notice this has singularities at x =(n+ 1

2

)πi for all

n. So if we used our good, old semi-circular contours, then we would run intoinfinitely many singularities as R→∞.

Instead, we abuse the periodic nature of cosh, and consider the followingrectangular contour:

−R γ0 R

γ+R

γ1

γ−R ×πi2

πi

We see that ∫γ0

eαz

cosh zdz → I as R→∞.

Also, ∫γ1

eαz

cosh zdz =

∫ −RR

eα(x+πi)

cosh(x+ πi)dx

= eαπi∫ −RR

eαx

− coshxdx

→ eαπiI.

On the γ+R , we have

cosh z = cosh(R+ iy) = coshR cos y + i sinhR sin y.

So

| cosh z| =√

cosh2R cos2 y + sinh2R sin2 y.

We now use the formula cosh2R = 1 + sinh2R, and magically obtain

| cosh z| =√

cos2 y + sinh2R ≥ sinhR.

Hence ∣∣∣∣ eαz

cosh z

∣∣∣∣ ≤ |eαReαiy|sinhR=

eαR

sinhR= O(e(α−1)R)→ 0.

49


Hence as R→∞, ∫γ+

eαz

cosh zdz → 0.

Similarly, the integral along γ− vanishes.Finally, we need to find the residue. The only singularity inside the contour

is at πi2 , where the residue is

eαπi/2

sinh πi2

= −ieαπi/2.

Hence we getI(1 + eαπi) = 2πi(−ieαπi/2) = 2πieαπi/2.

So we findI =

π

cos(απ/2).

These rectangular contours tend to be useful for trigonometric and hyperbolicfunctions.

Example. Consider the integral∫γ

cotπz

z2dz,

where γ is the square contour shown with corners at (N + 12 )(±1± i), where N

is a large integer, avoiding the singularities

× × × × × × × × × × ×

(N + 12 )i

−(N + 12 )i

N + 12−(N + 1

2 )

There are simple poles at z = n ∈ Z \ {0}, with residues 1n2π , and a triple pole

at z = 0 with residue − 13π (from the Taylor series for z2 tanπz). It turns out

the integrals along the sides all vanish as N →∞ (see later). So we know

2πi

(2

N∑n=1

1

n2π − π

3

)→ 0

as N →∞. In other words,N∑n=1

1

n2=π2

6.

50


This is probably the most complicated and most inefficient way of computingthis series. However, notice we can easily modify this to fund the sum of 1

n3 , or1n4 , or any complicated sum we can think of.

Hence all that remains is to show that the integrals along the sides vanish.On the right-hand side, we can write z = N + 1

2 + iy. Then

| cotπz| =∣∣∣∣cot

((N +

1

2

)+ iπy

)∣∣∣∣ = | − tan iπy| = | tanhπy| ≤ 1.

So cotπz is bounded on the vertical side. Since we are integrating cotπzz2 , the

integral vanishes as N →∞.Along the top, we get z = x+

(N + 1

2

)i. This gives

| cotπz| =

√cosh2

(N + 1

2

)π − sin2 πx√

sinh2(N + 1

2

)π + sin2 πx

≤ coth

(N +

1

2

)π ≤ coth

π

2.

So again cotπz is bounded on the top side. So again, the integral vanishes asN →∞.

4.4 Jordan’s lemma

So far, we have bounded the semi-circular integral in a rather crude way, with∣∣∣∣∫γR

f(z)eiλz dz

∣∣∣∣ ≤ R sup|z|=R

|f(z)|.

This works for most cases. However, sometimes we need a more subtle argument.

Lemma (Jordan’s lemma). Suppose that f is an analytic function, except for afinite number of singularities, and that f(z)→ 0 as |z| → ∞.

γR

γ′R

−R R

×

×

×

Then for any real constant λ > 0, we have∫γR

f(z)eiλz dz → 0

51


as R→∞, where γR is the semicircle of radius R in the upper half-plane.For λ < 0, the same conclusion holds for the semicircular γ′R in the lower

half-plane.

Such integrals arise frequently in Fourier transforms, as we shall see in thenext chapter.

How can we prove this result?

The result is easy to show if in fact f(z) = o(|z|−1) as |z| → ∞, i.e. f(z)|z| → 0

as |z| → ∞, since |eiλz| = e−λ Im z ≤ 1 on γR. So∣∣∣∣∫γR

f(z)eiλz dz

∣∣∣∣ ≤ πR · o(R−1) = o(1)→ 0.

But for functions decaying less rapidly than o(|z|−1) (e.g. if f(z) = 1z ), we need

to prove Jordan’s lemma, which extends the result to any function f(z) thattends to zero at infinity.

Proof. The proof relies on the fact that for θ ∈[0, π2

], we have

sin θ ≥ 2θ

π.

So we get ∣∣∣∣∫γR

f(z)eiλz dz

∣∣∣∣ =

∣∣∣∣∫ π

0

f(Reiθ)eiλReiθ

iReiθ dθ

∣∣∣∣≤ R

∫ π

0

|f(Reiθ)|∣∣∣eiλReiθ ∣∣∣ dθ

≤ 2R supz∈γR

|f(z)|∫ π/2

0

e−λR sin θ dθ

≤ 2R supz∈γR

|f(z)|∫ π/2

0

e−2λRθ/π dθ

=π

λ(1− e−λR) sup

z∈γR|f(z)|

→ 0,

as required. Same for γ′R when λ < 0.

Note that for most cases, we don’t actually need it, but can just bound it byR sup|z|=R f(z).

Example. Suppose we want to compute

I =

∫ ∞0

cosαx

1 + x2dx,

where α is a positive real constant. We consider the following contour:

52


γ0

γR

−R R

×i

and we compute

Re

∫γ

eiαz

1 + z2dz.

Along γ0, we obtain 2I as R → ∞. On γR, we do not actually need Jordan’slemma to show that we obtain zero in the limit, but we can still use it to saveink. So

I =1

2Re

(2πi res

z=i

eiαz

1 + z2

)=

1

2Re

(2πi

e−α

2i

)=

1

2πe−α.

Note that taking the real part does not do anything here — the result of the

integral is completely real. This is since the imaginary part of∫γeiαz

1+z2 dz is

integrating sinαz1+z2 , which is odd and vanishes.

Note that if we had attempted to integrate∫γ

cosαz1+z2 dz directly, we would

have found that∫γR6→ 0. In fact, cosαz is unbounded at ∞.

Example. We want to find

I =

∫ ∞−∞

sinx

xdx.

This time, we do require Jordan’s lemma. Here we have an extra complication— while sin x

x is well-behaved at the origin, to perform the contour integral, we

need to integrate eiz

z instead, which is singular at the origin.Instead, we split the integral in half, and write∫ ∞

−∞= limε→0R→∞

(∫ −ε−R

sinx

xdx+

∫ R

ε

sinx

xdx

)

= Im limε→0R→∞

(∫ −ε−R

eiz

xdx+

∫ R

ε

eiz

xdx

)

We now let C be the contour from −R to −ε, then round a semi-circle Cε to ε,then to R, then returning via a semi-circle CR of radius R. Then C misses allour singularities.

53


−R −εCε

ε R

CR

×

Since C misses all singularities, we must have∫ −ε−R

eiz

zdz +

∫ R

ε

eiz

zdz = −

∫Cε

eiz

zdz −

∫CR

eiz

zdz.

By Jordan’s lemma, the integral around CR vanishes as R → ∞. On Cε, wesubstitute z = εeiθ and eiz = 1 +O(ε) to obtain∫

Cε

eiz

zdz =

∫ 0

π

1 +O(ε)

εeiθiεeiθ dθ = −iπ +O(ε).

Hence, in the limit ε→ 0 and R→∞, we get∫ ∞−∞

sinx

xdx = Im(iπ) = π.

Similarly, we can compute ∫ ∞−∞

sin2 x

x2dx = π.

Alternatively, we notice that sin zz has a removable singularity at the origin.

Removing the singularity, the integrand is completely analytic. Therefore theoriginal integral is equivalent to the integral along this path:

We can then write sin z = eiz−e−iz2i , and then apply our standard techniques and

Jordan’s lemma.

54

5 Transform theory IB Complex Methods

5 Transform theory

We are now going to consider two different types of “transforms”. The firstis the Fourier transform, which we already met in IB methods. The second isthe Laplace transform. While the formula of a Laplace transform is completelyreal, the inverse Laplace transform involves a more complicated contour integral,which is why we did not do it in IB Methods. In either case, the new tool ofcontour integrals allows us to compute more transforms.

Apart from that, the two transforms are pretty similar, and most propertiesof the Fourier transform also carry over to the Laplace transform.

5.1 Fourier transforms

Definition (Fourier transform). The Fourier transform of a function f(x) thatdecays sufficiently as |x| → ∞ is defined as

f(k) =

∫ ∞−∞

f(x)e−ikx dx,

and the inverse transform is

f(x) =1

2π

∫ ∞−∞

f(k)eikx dk.

It is common for the terms e−ikx and eikx to be swapped around in thesedefinitions. It might even be swapped around by the same author in the samepaper — for some reason, if we have a function in two variables, then it istraditional to transform one variable with e−ikx and the other with eikx, just toconfuse people. More rarely, factors of 2π or

√2π are rearranged.

Traditionally, if f is a function of position x, then the transform variable iscalled k; while if f is a function of time t, then it is called ω. You don’t have tostick to this notation, if you like being confusing.

In fact, a more precise version of the inverse transform is

1

2(f(x+) + f(x−)) =

1

2πPV

∫ ∞−∞

f(k)eikx dk.

The left-hand side indicates that at a discontinuity, the inverse Fourier transformgives the average value. The right-hand side shows that only the Cauchy principalvalue of the integral (denoted PV

∫, P∫

or −∫

) is required, i.e. the limit

limR→∞

∫ R

−Rf(k)eikx dk,

rather than

limR→∞S→−∞

∫ R

S

f(k)eikx dk.

Several functions have PV integrals, but not normal ones. For example,

PV

∫ ∞−∞

x

1 + x2dx = 0,

55


since it is an odd function, but ∫ ∞−∞

x

1 + x2dx

diverges at both −∞ and ∞. So the normal proper integral does not exist.So for the inverse Fourier transform, we only have to care about the Cauchy

principal value. This is convenient because that’s how we are used to computecontour integrals all the time!

Notation. The Fourier transform can also be denoted by f = F(f) or f(k) =F(f)(k). In a slight abuse of notation, we often write f(k) = F(f(x)), but thisis not correct notation, since F takes in a function at a parameter, not a functionevaluated at a particular point.

Note that in the Tripos exam, you are expected to know about all theproperties of the Fourier transform you have learned from IB Methods.

We now calculate some Fourier transforms using the calculus of residues.

Example. Consider f(x) = e−x2/2. Then

f(k) =

∫ ∞−∞

e−x2/2e−ikx dx

=

∫ ∞−∞

e−(x+ik)2/2e−k2/2 dx

= e−k2/2

∫ ∞+ik

−∞+ik

e−z2/2 dz

We create a rectangular contour that looks like this:

−R γ1 R

γ+R

γ0

γ−R

ik

The integral we want is the integral along γ0 as shown, in the limit as R→∞. Wecan show that

∫γ+R→ 0 and

∫γ−R→ 0. Then we notice there are no singularities

inside the contour. So ∫γ0

e−z2/2 dz = −

∫γ1

e−z2/2 dz,

in the limit. Since γ1 is traversed in the reverse direction, we have

f(k) = e−k2/2

∫ ∞−∞

e−z2/2 dz =

√2πe−k

2/2,

using a standard result from real analysis.

56


When inverting Fourier transforms, we generally use a semicircular contour(in the upper half-plane if x > 0, lower if x < 0), and apply Jordan’s lemma.

Example. Consider the real function

f(x) =

{0 x < 0

e−ax x > 0,

where a > 0 is a real constant. The Fourier transform of f is

f(k) =

∫ ∞−∞

f(x)e−ikx dx

=

∫ ∞0

e−ax−ikx dx

= − 1

a+ ik[e−ax−ikx]∞0

=1

a+ ik.

We shall compute the inverse Fourier transform by evaluating

1

2π

∫ ∞−∞

f(k)eikx dk.

In the complex plane, we let γ0 be the contour from −R to R in the real axis;γR be the semicircle of radius R in the upper half-plane, γ′R the semicircle inthe lower half-plane. We let γ = γ0 + γR and γ′ = γ0 + γ′R.

γ0

γR

γ′R

−R R

×i

We see f(k) has only one pole, at k = ia, which is a simple pole. So we get∮γ

f(k)eikx dk = 2πi resk=ia

eikx

i(k − ia)= 2πe−ax,

while ∮γ′f(k)eikx dk = 0.

57


Now if x > 0, applying Jordan’s lemma (with λ = x) to CR shows that∫CR

f(k)eikx dk → 0 as R→∞. Hence we get

1

2π

∫ ∞−∞

=1

2πlimR→∞

∫γ0

f(k)eikx dk

=1

2πlimR→∞

(∫γ

f(k)eikx dk −∫γR

f(k)eikx dk

)= e−ax.

For x < 0, we have to close in the lower half plane. Since there are no singularities,we get

1

2π

∫ ∞−∞

f(k)eikx dk = 0.

Combining these results, we obtain

1

2π

∫ ∞−∞

f(k)eikx dk =

{0 x < 0

e−ax x > 0,

We’ve already done Fourier transforms in IB Methods, so we will no spendany more time on it. We move on to the new exciting topic of Laplace transforms.

5.2 Laplace transform

The Fourier transform is a powerful tool for solving differential equations andinvestigating physical systems, but it has two key restrictions:

(i) Many functions of interest grow exponentially (e.g. ex), and so do not haveFourier transforms;

(ii) There is no way of incorporating initial or boundary conditions in thetransform variable. When used to solve an ODE, the Fourier transformmerely gives a particular integral: there are no arbitrary constants producedby the method.

So for solving differential equations, the Fourier transform is pretty limited.Right into our rescue is the Laplace transform, which gets around these tworestrictions. However, we have to pay the price with a different restriction — itis only defined for functions f(t) which vanishes for t < 0 (by convention).

From now on, we shall make this assumption, so that if we refer to thefunction f(t) = et for instance, we really mean f(t) = etH(t), where H(t) is theHeaviside step function,

H(t) =

{1 t > 0

0 t < 0.

Definition (Laplace transform). The Laplace transform of a function f(t) suchthat f(t) = 0 for t < 0 is defined by

f(p) =

∫ ∞0

f(t)e−pt dt.

This exists for functions that grow no more than exponentially fast.

58


There is no standard notation for the Laplace transform.

Notation. We sometimes write

f = L(f),

orf(p) = L(f(t)).

The variable p is also not standard. Sometimes, s is used instead.Many functions (e.g. t and et) which do not have Fourier transforms do have

Laplace transforms.Note that f(p) = f(−ip), where f is the Fourier transform, provided that

both transforms exist.

Example. Let’s do some exciting examples.

(i)

L(1) =

∫ ∞0

e−pt dt =1

p.

(ii) Integrating by parts, we find

L(t) =1

p2.

(iii)

L(eλt) =

∫ ∞0

e(λ−p)t dt =1

p− λ.

(iv) We have

L(sin t) = L(

1

2i

(eit − e−it

))=

1

2i

(1

p− i− 1

p+ i

)=

1

p2 + 1.

Note that the integral only converges if Re p is sufficiently large. For example,in (iii), we require Re p > Reλ. However, once we have calculated f in thisdomain, we can consider it to exist everywhere in the complete p-plane, exceptat singularities (such as at p = λ in this example). This process of extending acomplex function initially defined in some part of the plane to a larger part isknown as analytic continuation.

So far, we haven’t done anything interesting with Laplace transform, andthis is going to continue in the next section!

59


5.3 Elementary properties of the Laplace transform

We will come up with seven elementary properties of the Laplace transform. Thefirst 4 properties are easily proved by direct substitution

Proposition.

(i) Linearity:L(αf + βg) = αL(f) + βL(g).

(ii) Translation:

L(f(t− t0)H(t− t0)) = e−pt0 f(p).

(iii) Scaling:

L(f(λt)) =1

λf( pλ

),

where we require λ > 0 so that f(λt) vanishes for t < 0.

(iv) Shifting:

L(ep0tf(t)) = f(p− p0).

(v) Transform of a derivative:

L(f ′(t)) = pf(p)− f(0).

Repeating the process,

L(f ′′(t)) = pL(f ′(t))− f ′(0) = p2f(p)− pf(0)− f ′(0),

and so on. This is the key fact for solving ODEs using Laplace transforms.

(vi) Derivative of a transform:

f ′(p) = L(−tf(t)).

Of course, the point of this is not that we know what the derivative of fis. It is we know how to find the Laplace transform of tf(t)! For example,this lets us find the derivative of t2 with ease.

In general,f (n)(p) = L((−t)nf(t)).

(vii) Asymptotic limits

pf(p)→

{f(0) as p→∞f(∞) as p→ 0

,

where the second case requires f to have a limit at ∞.

Proof.

(v) We have∫ ∞0

f ′(t)e−pt dt = [f(t)e−pt]∞0 + p

∫ ∞0

f(t)e−pt dt = pf(p)− f(0).

60


(vi) We have

f(p) =

∫ ∞0

f(t)e−pt dt.

Differentiating with respect to p, we have

f ′(p) = −∫ ∞

0

tf(t)e−pt dt.

(vii) Using (v), we know

pf(p) = f(0) +

∫ ∞0

f ′(t)e−pt dt.

As p→∞, we know e−pt → 0 for all t. So pf(p)→ f(0). This proof looksdodgy, but is actually valid since f ′ grows no more than exponentially fast.

Similarly, as p→ 0, then e−pt → 1. So

pf(p)→ f(0) +

∫ ∞0

f ′(t) dt = f(∞).

Example. We can compute

L(t sin t) = − d

dpL(sin t) = − d

dp

1

p2 + 1=

2p

(p2 + 1)2.

5.4 The inverse Laplace transform

It’s no good trying to find the Laplace transforms of functions if we can’t invertthem. Given f(p), we can calculate f(t) using the Bromwich inversion formula.

Proposition. The inverse Laplace transform is given by

f(t) =1

2πi

∫ c+i∞

c−i∞f(p)ept dp,

where c is a real constant such that the Bromwich inversion contour γ given byRe p = c lies to the right of all the singularities of f(p).

Proof. Since f has a Laplace transform, it grows no more than exponentially.So we can find a c ∈ R such that

g(t) = f(t)e−ct

decays at infinity (and is zero for t < 0, of course). So g has a Fourier transform,and

g(ω) =

∫ ∞−∞

f(t)e−cte−iωt dt = f(c+ iω).

Then we use the Fourier inversion formula to obtain

g(t) =1

2π

∫ ∞−∞

f(c+ iω)eiωt dω.

61


So we make the substitution p = c+ iω, and thus obtain

f(t)e−ct =1

2πi

∫ c+i∞

c−i∞f(p)e(p−c)t dp.

Multiplying both sides by ect, we get the result we were looking for (the re-quirement that c lies to the right of all singularities is to fix the “constant ofintegration” so that f(t) = 0 for all t < 0, as we will soon see).

In most cases, we don’t have to use the full inversion formula. Instead, weuse the following special case:

Proposition. In the case that f(p) has only a finite number of isolated singu-

larities pk for k = 1, · · · , n, and f(p)→ 0 as |p| → ∞, then

f(t) =

n∑k=1

resp=pk

(f(p)ept)

for t > 0, and vanishes for t < 0.

Proof. We first do the case where t < 0, consider the contour γ0 + γ′R as shown,which encloses no singularities.

c− iR

γ0

c+ iR

γ′R

×

××

Now if f(p) = o(|p|−1) as |p| → ∞, then∣∣∣∣∣∫γ′R

f(p)ept dp

∣∣∣∣∣ ≤ πRect supp∈γ′R

|f(p)| → 0 as R→∞.

Here we used the fact |ept| ≤ ect, which arises from Re(pt) ≤ ct, noting thatt < 0.

If f decays less rapidly at infinity, but still tends to zero there, the sameresult holds, but we need to use a slight modification of Jordan’s lemma. So ineither case, the integral ∫

γ′R

f(p)ept dp→ 0 as R→∞.

62


Thus, we know∫γ0→∫γ. Hence, by Cauchy’s theorem, we know f(t) = 0 for

t < 0. This is in agreement with the requirement that functions with Laplacetransform vanish for t < 0.

Here we see why γ must lie to the right of all singularities. If not, then thecontour would encircle some singularities, and then the integral would no longerbe zero.

When t > 0, we close the contour to the left.

c− iR

γ0

c+ iR

γR

×

××

This time, our γ does enclose some singularities. Since there are only finitelymany singularities, we enclose all singularities for sufficiently large R. Onceagain, we get

∫γR→ 0 as R→∞. Thus, by the residue theorem, we know∫

γ

f(p)ept dp = limR→∞

∫γ0

f(p)ept dp = 2πi

n∑k=1

resp=pk

(f(p)ept).

So from the Bromwich inversion formula,

f(t) =

n∑k=1

resp=pk

(f(p)ept),

as required.

Example. We know

f(p) =1

p− 1

has a pole at p = 1. So we must use c > 1. We have f(p)→ 0 as |p| → ∞. SoJordan’s lemma applies as above. Hence f(t) = 0 for t < 0, and for t > 0, wehave

f(t) = resp=1

(ept

p− 1

)= et.

This agrees with what we computed before.

Example. Consider f(p) = p−n. This has a pole of order n at p = 0. So wepick c > 0. Then for t > 0, we have

f(t) = resp=0

(ept

pn

)= limp→0

(1

(n− 1)!

dn−1

dpn−1ept)

=tn−1

(n− 1)!.

63


This again agrees with what we computed before.

Example. In the case where

f(p) =e−p

p,

then we cannot use the standard result about residues, since f(p) does not vanishas |p| → ∞. But we can use the original Bromwich inversion formula to get

f(t) =1

2πi

∫γ

e−p

pept dp

=1

2πi

∫γ

1

pept′

dp,

where t′ = t− 1. Now we can close to the right when t′ < 0, and to the left whent′ > 0, picking up the residue from the pole at p = 0. Then we get

f(t) =

{0 t′ < 0

1 t′ > 0=

{0 t < 1

1 t > 1= H(t− 1).

This again agrees with what we’ve got before.

Example. If f(p) has a branch point (at p = 0, say), then we must use aBromwich keyhole contour as shown.

5.5 Solution of differential equations using the Laplacetransform

The Laplace transform converts ODEs to algebraic equations, and PDEs toODEs. We will illustrate this by an example.

Example. Consider the differential equation

ty − ty + y = 0,

64


with y(0) = 2 and y(0) = −1. Note that

L(ty) = − d

dpL(y)− d

dp(py − y(0)) = −py′ − y.

Similarly, we findL(ty) = −p2y′ − 2py + y(0).

Substituting and rearranging, we obtain

py′ + 2y =2

p,

which is a simpler differential equation. We can solve this using an integratingfactor to obtain

y =2

p+A

p2,

where A is an arbitrary constant. Hence we have

y = 2 +At,

and A = −1 from the initial conditions.

Example. A system of ODEs can always be written in the form

x = Mx, x(0) = x0,

where x ∈ Rn and M is an n × n matrix. Taking the Laplace transform, weobtain

px− x0 = M x.

So we getx = (pI −M)−1x0.

This has singularities when p is equal to an eigenvalue of M .

5.6 The convolution theorem for Laplace transforms

Finally, we recall from IB Methods that the Fourier transforms turn convolutionsinto products, and vice versa. We will now prove an analogous result for Laplacetransforms.

Definition (Convolution). The convolution of two functions f and g is definedas

(f ∗ g)(t) =

∫ ∞−∞

f(t− t′)g(t′) dt′.

When f and g vanish for negative t, this simplifies to

(f ∗ g)(t) =

∫ t

0

f(t− t′)g(t) dt.

Theorem (Convolution theorem). The Laplace transform of a convolution isgiven by

L(f ∗ g)(p) = f(p)g(p).

65


Proof.

L(f ∗ g)(p) =

∫ ∞0

(∫ t

0

f(t− t′)g(t′) dt′)e−pt dt

We change the order of integration in the (t, t′) plane, and adjust the limitsaccordingly (see picture below)

=

∫ ∞0

(∫ ∞t′

f(t− t′)g(t′)e−pt dt

)dt′

We substitute u = t− t′ to get

=

∫ ∞0

(∫ ∞0

f(u)g(t′)e−pue−pt′

du

)dt′

=

∫ ∞0

(∫ ∞0

f(u)e−pu du

)g(t′)e−pt

′dt′

= f(p)g(p).

Note the limits of integration are correct since they both represent the regionbelow

t

t′

66

part ib - complex methods · part ib | complex methods based on lectures by r. e. hunt notes taken...

Documents