Part IB Complex Methods
Based on lectures by R. E. Hunt
Notes taken by Dexter Chua
Lent 2016
These notes are not endorsed by the lecturers, and I have modified them (often
significantly) after lectures. They are nowhere near accurate representations of what
was actually lectured, and in particular, all errors are almost surely mine.
Analytic functions
Definition of an analytic function. Cauchy-Riemann equations. Analytic functions as
conformal mappings; examples. Application to the solutions of Laplace’s equation in
various domains. Discussion of log z and z
a
. [5]
Contour integration and Cauchy’s Theorem
[Proofs of theorems in this section will not be examined in this course.]
Contours, contour integrals. Cauchy’s theorem and Cauchy’s integral formula. Taylor
and Laurent series. Zeros, poles and essential singularities. [3]
Residue calculus
Residue theorem, calculus of residues. Jordan’s lemma. Evaluation of definite integrals
by contour integration. [4]
Fourier and Laplace transforms
Laplace transform: definition and basic properties; inversion theorem (proof not
required); convolution theorem. Examples of inversion of Fourier and Laplace transforms
by contour integration. Applications to differential equations. [4]
Contents
0 Introduction
1 Analytic functions
1.1 The complex plane and the Riemann sphere
1.2 Complex differentiation
1.3 Harmonic functions
1.4 Multi-valued functions
1.5 obius map
1.6 Conformal maps
1.7 Solving Laplace’s equation using conformal maps
2 Contour integration and Cauchy’s theorem
2.1 Contour and integrals
2.2 Cauchy’s theorem
2.3 Contour deformation
2.4 Cauchy’s integral formula
3 Laurent series and singularities
3.1 Taylor and Laurent series
3.2 Zeros
3.3 Classification of singularities
3.4 Residues
4 The calculus of residues
4.1 The residue theorem
4.2 Applications of the residue theorem
4.3
Further applications of the residue theorem using rectangular
contours
4.4 Jordan’s lemma
5 Transform theory
5.1 Fourier transforms
5.2 Laplace transform
5.3 Elementary properties of the Laplace transform
5.4 The inverse Laplace transform
5.5 Solution of differential equations using the Laplace transform
5.6 The convolution theorem for Laplace transforms
0 Introduction
In Part IA, we learnt quite a lot about differentiating and integrating real
functions. Differentiation was fine, but integration was tedious. Integrals were
very difficult to evaluate.
In this course, we will study differentiating and integrating complex functions.
Here differentiation is nice, and integration is easy. We will show that complex
differentiable functions satisfy many things we hoped were true a complex
differentiable function is automatically infinitely differentiable. Moreover, an
everywhere differentiable function must be constant if it is bounded.
On the integration side, we will show that integrals of complex functions can
be performed by computing things known as residues, which are much easier
to compute. We are not actually interested in performing complex integrals.
Instead, we will take some difficult real integrals, and pretend they are complex
ones.
This is a methods course. By this, we mean we will not focus too much on
proofs. We will at best just skim over the proofs. Instead, we focus on doing
things. We will not waste time proving things people have proved 300 years ago.
If you like proofs, you can go to the IB Complex Analysis course, or look them
up in relevant books.
1 Analytic functions
1.1 The complex plane and the Riemann sphere
We begin with a review of complex numbers. Any complex number
z C
can
be written in the form
x
+
iy
, where
x
=
Re z
,
y
=
Im z
are real numbers. We
can also write it as re
, where
Definition
(Modulus and argument)
.
The modulus and argument of a complex
number z = x + iy are given by
r = |z| =
p
x
2
+ y
2
, θ = arg z,
where x = r cos θ, y = r sin θ.
The argument is defined only up to multiples of 2
π
. So we define the following:
Definition
(Principal value of argument)
.
The principal value of the argument
is the value of θ in the range (π, π].
We might be tempted to write down the formula
θ = tan
1
y
x
,
but this does not always give the right answer it is correct only if
x >
0. If
x 0, then it might be out by ±π (e.g. consider z = 1 + i and z = 1 i).
Definition
(Open set)
.
An open set
D
is one which does not include its boundary.
More technically,
D C
is open if for all
z
0
D
, there is some
δ >
0 such that
the disc |z z
0
| < δ is contained in D.
Definition
(Neighbourhood)
.
A neighbourhood of a point
z C
is an open set
containing z.
The extended complex plane
Often, the complex plane
C
itself is not enough. We want to consider the point
as well. This forms the extended complex plane.
Definition
(The extended complex plane)
.
The extended complex plane is
C
=
C {∞}
. We can reach the “point at infinity” by going off in any direction
in the plane, and all are equivalent. In particular, there is no concept of
−∞
.
All infinities are the same. Operations with are done in the obvious way.
Sometimes, we do write down things like
−∞
. This does not refer to a
different point. Instead, this indicates a limiting process. We mean we are
approaching this infinity from the direction of the negative real axis. However,
we still end up in the same place.
Conceptually, we can visualize this using the Riemann sphere, which is a
sphere resting on the complex plane with its “South Pole” S at z = 0.
S
N
P
z
For any point
z C
, drawing a line through the “North Pole”
N
of the sphere
to
z
, and noting where this intersects the sphere. This specifies an equivalent
point
P
on the sphere. Then
is equivalent to the North Pole of the sphere
itself. So the extended complex plane is mapped bijectively to the sphere.
This is a useful way to visualize things, but is not as useful when we actually
want to do computations. To investigate properties of
, we use the substitution
ζ
=
1
z
. A function
f
(
z
) is said to have a particular property at
if
f
(
1
ζ
) has
that same property at
ζ
= 0. This vague notion will be made precise when we
have specific examples to play with.
1.2 Complex differentiation
Recall the definition of differentiation for a real function f(x):
f
0
(x) = lim
δx0
f(x + δx) f(x)
δx
.
It is implicit that the limit must be the same whichever direction we approach
from. For example, consider
|x|
at
x
= 0. If we approach from the right, i.e.
δx
0
+
, then the limit is +1, whereas from the left, i.e.
δx
0
, the limit is
1. Because these limits are different, we say that
|x|
is not differentiable at
the origin.
This is obvious and we already know that, but for complex differentiation,
this issue is much more important, since there are many more directions. We
now extend the definition of differentiation to complex number:
Definition
(Complex differentiable function)
.
A complex differentiable function
f : C C is differentiable at z if
f
0
(z) = lim
δz0
f(z + δz) f (z)
δz
exists (and is therefore independent of the direction of approach but now
there are infinitely many possible directions).
This is the same definition as that for a real function. Often, we are not
interested in functions that are differentiable at a point, since this might allow
some rather exotic functions we do not want to consider. Instead, we want the
function to be differentiable near the point.
Definition
(Analytic function)
.
We say
f
is analytic at a point
z
if there
exists a neighbourhood of
z
throughout which
f
0
exists. The terms regular and
holomorphic are also used.
Definition
(Entire function)
.
A complex function is entire if it is analytic
throughout C.
The property of analyticity is in fact a surprisingly strong one! For example,
two consequences are:
(i)
If a function is analytic, then it is differentiable infinitely many times. This
is very very false for real functions. There are real functions differentiable
N
times, but no more (e.g. by taking a non-differentiable function and
integrating it N times).
(ii) A bounded entire function must be a constant.
There are many more interesting properties, but these are sufficient to show us
that complex differentiation is very different from real differentiation.
The Cauchy-Riemann equations
We already know well how to differentiate real functions. Can we use this to
determine whether certain complex functions are differentiable? For example is
the function
f
(
x
+
iy
) =
cos x
+
i sin y
differentiable? In general, given a complex
function
f(z) = u(x, y) + iv(x, y),
where
z
=
x
+
iy
are
u, v
are real functions, is there an easy criterion to determine
whether f is differentiable?
We suppose that
f
is differentiable at
z
. We may take
δz
in any direction
we like. First, we take it to be real, with δz = δx. Then
f
0
(z) = lim
δx0
f(z + δx) f (z)
δx
= lim
δx0
u(x + δx, y) + iv(x + δx, y) (u(x, y) + iv(x, y))
δx
=
u
x
+ i
v
x
.
What this says is something entirely obvious since we are allowed to take the
limit in any direction, we can take it in the
x
direction, and we get the corre-
sponding partial derivative. This is a completely uninteresting point. Instead,
let’s do the really fascinating thing of taking the limit in the y direction!
Let δz = iδy. Then we can compute
f
0
(z) = lim
δy0
f(z + y) f(z)
y
= lim
δy0
u(x, y + δy) + iv(x, y + δy) (u(x, y) + iv(x, y))
y
=
v
y
i
u
y
.
By the definition of differentiability, the two results for
f
0
(
z
) must agree! So we
must have
u
x
+ i
v
x
=
v
y
i
u
y
.
Taking the real and imaginary components, we get
Proposition
(Cauchy-Riemann equations)
.
If
f
=
u
+
iv
is differentiable, then
u
x
=
v
y
,
u
y
=
v
x
.
Is the converse true? If these equations hold, does it follow that
f
is differ-
entiable? This is not always true. This holds only if
u
and
v
themselves are
differentiable, which is a stronger condition that the partial derivatives exist, as
you may have learnt from IB Analysis II. In particular, this holds if the partial
derivatives u
x
, u
y
, v
x
, v
y
are continuous (which implies differentiability). So
Proposition.
Given a complex function
f
=
u
+
iv
, if
u
and
v
are real differen-
tiable at a point z and
u
x
=
v
y
,
u
y
=
v
x
,
then f is differentiable at z.
We will not prove this proofs are for IB Complex Analysis.
Examples of analytic functions
Example.
(i) f
(
z
) =
z
is entire, i.e. differentiable everywhere. Here
u
=
x, v
=
y
. Then
the Cauchy-Riemann equations are satisfied everywhere, since
u
x
=
v
y
= 1,
u
y
=
v
x
= 0,
and these are clearly continuous. Alternatively, we can prove this directly
from the definition.
(ii) f(z) = e
z
= e
x
(cos y + i sin y) is entire since
u
x
= e
x
cos y =
v
y
,
u
y
= e
x
sin y =
v
x
.
The derivative is
f
0
(z) =
u
x
+ i
v
x
= e
x
cos y + ie
x
sin y = e
z
,
as expected.
(iii) f
(
z
) =
z
n
for
n N
is entire. This is less straightforward to check. Writing
z = r(cos θ + i sin θ), we obtain
u = r
n
cos , v = r
n
sin .
We can check the Cauchy-Riemann equation using the chain rule, writing
r
=
p
x
2
= y
2
and
tan θ
=
y
x
. This takes quite a while, and it’s not worth
your time. But if you really do so, you will find the derivative to be
nz
n1
.
(iv)
Any rational function, i.e.
f
(
z
) =
P (z)
Q(z)
where
P, Q
are polynomials, is
analytic except at points where
Q
(
z
) = 0 (where it is not even defined).
For instance,
f(z) =
z
z
2
+ 1
is analytic except at ±i.
(v)
Many standard functions can be extended naturally to complex functions
and obey the usual rules for their derivatives. For example,
sin z
=
e
iz
e
iz
2i
is differentiable with derivative
cos z
=
e
iz
+e
iz
2
. We
can also write
sin z = sin(x + iy)
= sin x cos iy + cos x sin iy
= sin x cosh y + i cos x sinh y,
which is sometimes convenient.
Similarly
cos z, sinh z, cosh z
etc. differentiate to what we expect them
to differentiate to.
log z = log |z| + i arg z has derivative
1
z
.
The product rule, quotient rule and chain rule hold in exactly the
same way, which allows us to prove (iii) and (iv) easily.
Examples of non-analytic functions
Example.
(i) Let f(z) = Re z. This has u = x, v = 0. But
u
x
= 1 6= 0 =
v
y
.
So Re z is nowhere analytic.
(ii)
Consider
f
(
z
) =
|z|
. This has
u
=
p
x
2
+ y
2
, v
= 0. This is thus nowhere
analytic.
(iii)
The complex conjugate
f
(
z
) =
¯z
=
z
=
x iy
has
u
=
x, v
=
y
. So the
Cauchy-Riemann equations don’t hold. Hence this is nowhere analytic.
We could have deduced (ii) from this if
|z|
were analytic, then so would
|z|
2
, and hence ¯z =
|z|
2
z
also has to be analytic, which is not true.
(iv)
We have to be a bit more careful with
f
(
z
) =
|z|
2
=
x
2
+
y
2
. The
Cauchy-Riemann equations are satisfied only at the origin. So
f
is only
differentiable at
z
= 0. However, it is not analytic since there is no
neighbourhood of 0 throughout which f is differentiable.
1.3 Harmonic functions
This is the last easy section of the course.
Definition
(Harmonic conjugates)
.
Two functions
u, v
satisfying the Cauchy-
Riemann equations are called harmonic conjugates.
If we know one, then we can find the other up to a constant. For example, if
u(x, y) = x
2
y
2
, then v must satisfy
v
y
=
u
x
= 2x.
So we must have
v
= 2
xy
+
g
(
x
) for some function
g
(
x
). The other Cauchy-
Riemann equation gives
2y =
u
y
=
v
x
= 2y g
0
(x).
This tells us
g
0
(
x
) = 0. So
g
must be a genuine constant, say
α
. The corresponding
analytic function whose real part is u is therefore
f(z) = x
2
y
2
+ 2ixy + = (x + iy)
2
+ = z
2
+ iα.
Note that in an exam, if we were asked to find the analytic function
f
with real
part
u
(where
u
is given), then we must express it in terms of
z
, and not
x
and
y, or else it is not clear this is indeed analytic.
On the other hand, if we are given that
f
(
z
) =
u
+
iv
is analytic, then we
can compute
2
u
x
2
=
x
u
x
=
x
v
y
=
y
v
x
=
y
u
y
=
2
u
y
2
.
So u satisfies Laplace’s equation in two dimensions, i.e.
2
u =
2
u
x
2
+
2
u
y
2
= 0.
Similarly, so does v.
Definition
(Harmonic function)
.
A function satisfying Laplace’s equation equa-
tion in an open set is said to be harmonic.
Thus we have shown the following:
Proposition.
The real and imaginary parts of any analytic function are har-
monic.
1.4 Multi-valued functions
For
z
=
r
, we define
log z
=
log r
+
. There are infinitely many values of
log z, for every choice of θ. For example,
log i =
πi
2
or
5πi
2
or
3πi
2
or ··· .
This is fine, right? Functions can be multi-valued. Nothing’s wrong.
Well, when we write down an expression, it’d better be well-defined. So we
really should find some way to deal with this.
This section is really more subtle than it sounds like. It turns out it is non-
trivial to deal with these multi-valued functions. We can’t just, say, randomly
require
θ
to be in, say,
(0, 2π]
, or else we will have some continuity problems, as
we will later see.
Branch points
Consider the three curves shown in the diagram.
C
3
C
1
C
2
In
C
1
, we could always choose
θ
to be always in the range
0,
π
2
, and then
log z
would be continuous and single-valued going round C
1
.
On
C
2
, we could choose
θ
π
2
,
3π
2
and
log z
would again be continuous
and single-valued.
However, this doesn’t work for
C
3
. Since this encircles the origin, there is no
such choice. Whatever we do,
log z
cannot be made continuous and single-valued
around
C
3
. It must either “jump” somewhere, or the value has to increase by
2πi every time we go round the circle, i.e. the function is multi-valued.
We now define what a branch point is. In this case, it is the origin, since
that is where all our problems occur.
Definition
(Branch point)
.
A branch point of a function is a point which is
impossible to encircle with a curve on which the function is both continuous and
single-valued. The function is said to have a branch point singularity there.
Example.
(i) log(z a) has a branch point at z = a.
(ii) log
z1
z+1
= log(z 1) log(z + 1) has two branch points at ±1.
(iii) z
α
=
r
α
e
iαθ
has a branch point at the origin as well for
α 6∈ Z
consider
a circle of radius of
r
0
centered at 0, and wlog that we start at
θ
= 0 and
go once round anticlockwise. Just as before,
θ
must vary continuous to
ensure continuity of
e
iαθ
. So as we get back almost to where we started,
θ
will approach 2
π
, and there will be a jump in
θ
from 2
π
back to 0. So
there will be a jump in
z
α
from
r
α
0
e
2π
to
r
α
0
. So
z
α
is not continuous if
e
2π
6= 1, i.e. α is not an integer.
(iv) log z
also has a branch point at
. Recall that to investigate the properties
of a function
f
(
z
) at infinity, we investigate the property of
f
1
z
at zero.
If
ζ
=
1
z
, then
log z
=
log ζ
, which has a branch point at
ζ
= 0. Similarly,
z
α
has a branch point at for α 6∈ Z.
(v)
The function
log
z1
z+1
does not have a branch point at infinity, since if
ζ =
1
z
, then
log
z 1
z + 1
= log
1 ζ
1 + ζ
.
For
ζ
close to zero,
1ζ
1+ζ
remains close to 1, and therefore well away from
the branch point of
log
at the origin. So we can encircle
ζ
= 0 without
log
1ζ
1+ζ
being discontinuous.
So we’ve identified the points where the functions have problems. How do
we deal with these problems?
Branch cuts
If we wish to make
log z
continuous and single valued, therefore, we must stop
any curve from encircling the origin. We do this by introducing a branch cut
from −∞ on the real axis to the origin. No curve is allowed to cross this cut.
z
θ
Once we’ve decided where our branch cut is, we can use it to fix on values of
θ
lying in the range (
π, π
], and we have defined a branch of
log z
. This branch
is single-valued and continuous on any curve
C
that does not cross the cut.
This branch is in fact analytic everywhere, with
d
dz
log z
=
1
z
, except on the
non-positive real axis, where it is not even continuous.
Note that a branch cut is the squiggly line, while a branch is a particular
choice of the value of log z.
The cut described above is the canonical (i.e. standard) branch cut for
log z
.
The resulting value of log z is called the principal value of the logarithm.
What are the values of
log z
just above and just below the branch cut?
Consider a point on the negative real axis,
z
=
x <
0. Just above the cut, at
z
=
x
+
i
0
+
, we have
θ
=
π
. So
log z
=
log |x|
+
. Just below it, at
z
=
x
+
i
0
,
we have log z = log |x| . Hence we have a discontinuity of 2πi.
We have picked an arbitrary branch cut and branch. We can pick other
branch cuts or branches. Even with the same branch cut, we can still have a
different branch we can instead require
θ
to fall in (
π,
3
π
]. Of course, we can
also pick other branch cuts, e.g. the non-negative imaginary axis. Any cut that
stops curves wrapping around the branch point will do.
Here we can choose θ
3π
2
,
π
2
. We can also pick a branch cut like this:
The exact choice of
θ
is more difficult to write down, but this is an equally valid
cut, since it stops curves from encircling the origin.
Exactly the same considerations (and possible branch cuts) apply for
z
α
(for
α 6∈ Z).
In practice, whenever a problem requires the use of a branch, it is important
to specify it clearly. This can be done in two ways:
(i) Define the function and parameter range explicitly, e.g.
log z = log |z| + i arg z, arg z (π, π].
(ii)
Specify the location of the branch cut and give the value of the required
branch at a single point not on the cut. The values everywhere else are
then defined uniquely by continuity. For example, we have
log z
with a
branch cut along
R
0
and
log
1 = 0. Of course, we could have defined
log 1 = 2πi as well, and this would correspond to picking arg z (π, 3π].
Either way can be used, but it must be done properly.
Riemann surfaces*
Instead of this brutal way of introducing a cut and forbidding crossing, Riemann
imagined different branches as separate copies of
C
, all stacked on top of each
other but each one joined to the next at the branch cut. This structure is a
Riemann surface.
C
C
C
C
C
The idea is that traditionally, we are not allowed to cross branch cuts. Here,
when we cross a branch cut, we will move to a different copy of
C
, and this
corresponds to a different branch of our function.
We will not say any more about this there is a whole Part II course
devoted to these, uncreatively named IID Riemann Surfaces.
Multiple branch cuts
When there is more than one branch point, we may need more than one branch
cut. For
f(z) = (z(z 1))
1
3
,
there are two branch points, at 0 and 1. So we need two branch cuts. A possibility
is shown below. Then no curve can wrap around either 0 or 1.
10
z
r
r
1
θ
θ
1
For any
z
, we write
z
=
re
and
z
1 =
r
1
e
1
with
θ
(
π, π
] and
θ
1
[0
,
2
π
),
and define
f(z) =
3
rr
1
e
i(θ+θ
1
)/3
.
This is continuous so long as we don’t cross either branch cut. This is all and
simple.
However, sometimes, we need fewer branch cuts than we might think. Con-
sider instead the function
f(z) = log
z 1
z + 1
.
Writing z + 1 = re
and z 1 = r
1
e
1
, we can write this as
f(z) = log(z 1) log(z + 1)
= log(r
1
/r) + i(θ
1
θ).
This has branch points at
±
1. We can, of course, pick our branch cut as above.
However, notice that these two cuts also make it impossible for
z
to “wind
around
(e.g. moving around a circle of arbitrarily large radius). Yet
is not
a branch point, and we don’t have to make this unnecessary restriction. Instead,
we can use the following branch cut:
11
z
r
r
1
θ
θ
1
Drawing this branch cut is not hard. However, picking the values of
θ, θ
1
is
more tricky. What we really want to pick is
θ, θ
1
[0
,
2
π
). This might not look
intuitive at first, but we will shortly see why this is the right choice.
Suppose that we are unlawful and cross the branch cut. Then the value of
θ
passes through the branch cut, while the value of
θ
1
varies smoothly. So the
value of
f
(
z
) jumps. This is expected since we have a branch cut there. If we
pass through the negative real axis on the left of the branch cut, then nothing
happens, since θ = θ
1
= π are not at a point of discontinuity.
The interesting part is when we pass through the positive real axis on the
right of branch cut. When we do this, both
θ
and
θ
1
jump by 2
π
. However, this
does not induce a discontinuity in
f
(
z
), since
f
(
z
) depends on the difference
θ
1
θ, which has not experienced a jump.
1.5 obius map
We are now going to consider a special class of maps, namely the obius maps, as
defined in IA Groups. While these maps have many many different applications,
the most important thing we are going to use it for is to define some nice
conformal mappings in the next section.
We know from general theory that the obius map
z 7→ w =
az + b
cz + d
with
ad bc 6
= 0 is analytic except at
z
=
d
c
. It is useful to consider it as a
map from C
C
= C {∞}, with
d
c
7→ , 7→
a
c
.
It is then a bijective map between C
and itself, with the inverse being
w 7→
dw + b
cw a
,
another obius map. These are all analytic everywhere when considered as a
map C
C
.
Definition (Circline). A circline is either a circle or a line.
The key property of obius maps is the following:
Proposition. obius maps take circlines to circlines.
Note that if we start with a circle, we might get a circle or a line; if we start
with a line, we might get a circle or a line.
Proof. Any circline can be expressed as a circle of Apollonius,
|z z
1
| = λ|z z
2
|,
where z
1
, z
2
C and λ R
+
.
This was proved in the first example sheet of IA Vectors and Matrices. The
case
λ
= 1 corresponds to a line, while
λ 6
= 1 corresponds to a circle. Substituting
z in terms of w, we get
dw + b
cw a
z
1
= λ
dw + b
cw a
z
2
.
Rearranging this gives
|(cz
1
+ d)w (az
1
+ b)| = λ|(cz
2
+ d)w (az
2
+ b)|. ()
A bit more rearranging gives
w
az
1
+ b
cz
1
+ d
= λ
cz
2
+ d
cz
1
+ d
w
az
2
+ b
cz
2
+ d
.
This is another circle of Apollonius.
Note that the proof fails if either
cz
1
+
d
= 0 or
cz
2
+
d
= 0, but then (
)
trivially represents a circle.
Geometrically, it is clear that choosing three distinct points in
C
uniquely
specifies a circline (if one of the points is
, then we have specified the straight
line through the other two points).
Also,
Proposition.
Given six points
α, β, γ, α
0
, β
0
, γ
0
C
, we can find a obius
map which sends α 7→ α
0
, β 7→ β
0
and γ γ
0
.
Proof. Define the obius map
f
1
(z) =
β γ
β α
z α
z γ
.
By direct inspection, this sends α 0, β 1 and γ . Again, we let
f
2
(z) =
β
0
γ
0
β
0
α
0
z α
0
z γ
0
.
This clearly sends
α
0
0
, β
0
1 and
γ
0
. Then
f
1
2
f
1
is the required
mapping. It is a obius map since obius maps form a group.
Therefore, we can therefore find a obius map taking any given circline to
any other, which is convenient.
1.6 Conformal maps
Sometimes, we might be asked to solve a problem on some complicated subspace
U C
. For example, we might need to solve Laplace’s equation subject to
some boundary conditions. In such cases, it is often convenient to transform
our space
U
into some nicer space
V
, such as the open disk. To do so, we will
need a complex function
f
that sends
U
to
V
. For this function to preserve our
properties such that the solution on
V
can be transferred back to a solution on
U
, we would of course want
f
to be differentiable. Moreover, we would like it to
have non-vanishing derivative, so that it is at least locally invertible.
Definition
(Conformal map)
.
A conformal map
f
:
U V
, where
U, V
are
open subsets of C, is one which is analytic with non-zero derivative.
In reality, we would often want the map to be a bijection. We sometimes call
these conformal equivalences.
Unfortunately, after many hundred years, we still haven’t managed to agree
on what being conformal means. An alternative definition is that a conformal
map is one that preserves the angle (in both magnitude and orientation) between
intersecting curves.
We shall show that our definition implies this is true; the converse is also
true, but the proof is omitted. So the two definitions are equivalent.
Proposition.
A conformal map preserves the angles between intersecting curves.
Proof.
Suppose
z
1
(
t
) is a curve in
C
, parameterised by
t R
, which passes
through a point
z
0
when
t
=
t
1
. Suppose that its tangent there,
z
0
1
(
t
1
), has a
well-defined direction, i.e. is non-zero, and the curve makes an angle
φ
=
arg z
0
1
(
t
1
)
to the x-axis at z
0
.
Consider the image of the curve,
Z
1
(
t
) =
f
(
z
1
(
t
)). Its tangent direction at
t = t
1
is
Z
0
1
(t
1
) = z
0
1
(t
1
)f
0
(z
1
(t
1
)) = z
0
1
(t
0
)f
0
(z
0
),
and therefore makes an angle with the x-axis of
arg(Z
0
1
(t
1
)) = arg(z
0
1
(t
1
)f
0
(z
0
)) = φ + arg f
0
(z
0
),
noting that arg f
0
(z
0
) exists since f is conformal, and hence f
0
(z
0
) 6= 0.
In other words, the tangent direction has been rotated by
arg f
0
(
z
0
), and this
is independent of the curve we started with.
Now if
z
2
(
t
) is another curve passing through
z
0
. Then its tangent direction
will also be rotated by arg f
0
(z
0
). The result then follows.
Often, the easiest way to find the image set of a conformal map acting on a
set
U
is first to find the image of its boundary,
U
, which will form the boundary
V
of
V
; but, since this does not reveal which side of
V V
lies on, take a point
of your choice within U, whose image will lie within V .
Example.
(i)
The map
z 7→ az
+
b
, where
a, b C
and
a 6
= 0, is a conformal map. It
rotates by
arg a
, enlarges by
|a|
, and translates by
b
. This is conformal
everywhere.
(ii) The map f(z) = z
2
is a conformal map from
U =
n
z : 0 < |z| < 1, 0 < arg z <
π
2
o
to
V = {w : 0 < |w| < 1, 0 < arg w < π}.
1
U
f
1
V
Note that the right angles between the boundary curves at
z
= 1 and
i
are
preserved, because
f
is conformal there; but the right angle at
z
= 0 is not
preserved because
f
is not conformal there (
f
0
(0) = 0). Fortunately, this
does not matter, because U is an open set and does not contain 0.
(iii) How could we conformally map the left-hand half-plane
U = {z : Re z < 0}
to a wedge
V =
n
w :
π
4
< arg w
π
4
o
.
U
V
We need to halve the angle. We saw that
z 7→ z
2
doubles then angle, so we
might try
z
1
2
, for which we need to choose a branch. The branch cut must
not lie in
U
, since
z
1
2
is not analytic on the branch cut. In particular, the
principal branch does not work.
So we choose a cut along the negative imaginary axis, and the function is
defined by
re
7→
re
iθ/2
, where
θ
π
2
,
3π
2
. This produces the wedge
{z
0
:
π
4
< arg z
0
<
3π
4
}
. This isn’t exactly the wedge we want. So we need
to rotate it through
π
2
. So the final map is
f(z) = iz
1
2
.
(iv) e
z
takes rectangles conformally to sectors of annuli:
U
iy
1
iy
2
x
1
x
2
e
x
1
e
x
2
y
1
y
2
V
With an appropriate choice of branch, log z does the reverse.
(v)
obius maps (which are conformal equivalence except at the point that is
sent to
) are very useful in taking circles, or parts of them to straight
lines, or vice versa.
Consider
f
(
z
) =
z1
z+1
acting on the unit disk
U
=
{z
:
|z| <
1
}
. The
boundary of
U
is a circle. The three points
1
, i
and +1 lie on this circle,
and are mapped to , i and 0 respectively.
Since obius maps take circlines to circlines, the image of
U
is the
imaginary axis. Since
f
(0) =
1, we see that the image of
U
is the
left-hand half plane.
U
V
We can derive this alternatively by noting
w =
z 1
z + 1
z =
w + 1
w 1
.
So
|z| < 1 |w + 1| < |w 1|,
i.e.
w
is closer to
1 than it is to +1, which describes precisely the left-hand
half plane.
In fact, this particular map
f
(
z
) =
z1
z+1
can be deployed more generally on
quadrants, because it permutes 8 divisions on the complex plane as follows:
23
67
14
58
The map sends 1
7→
2
7→
3
7→
4
7→
1 and 5
7→
6
7→
7
7→
8
7→
5. In
particular, this agrees with what we had above it sends the complete
circle to the left hand half plane.
(vi)
Consider the map
f
(
z
) =
1
z
. This is just another obius map! Hence
everything we know about obius maps apply to this. In particular, it is
useful for acting on vertical and horizontal lines. Details are left for the
first example sheet.
In practice, complicated conformal maps are usually built up from individual
building blocks, each a simple conformal map. The required map is the com-
position of these. For this to work, we have to note that the composition of
conformal maps is conformal, by the chain rule.
Example.
Suppose we want to map the upper half-disc
|z| <
1,
Im z >
0 to the
full disc
|z| <
1. We might want to just do
z 7→ z
2
. However, this does not work,
since the image does not include the non-negative real axis, say
z
=
1
2
. Instead,
we need to do something more complicated. We will do this in several steps:
(i) We apply f
1
(z) =
z1
z+1
to take the half-disc to the second quadrant.
(ii)
We now recall that
f
1
also takes the right-hand half plane to the disc. So we
square and rotate to get the right-hand half plane. We apply
f
2
(
z
) =
iz
2
.
(iii) We apply f
3
(z) = f
1
(z) again to obtain the disc.
Then the desired conformal map is
f
3
f
2
f
1
, you can, theoretically, expand
this out and get an explicit expression, but that would be a waste of time.
z 7→
z1
z+1
z 7→ iz
2
z 7→
z1
z+1
1.7 Solving Laplace’s equation using conformal maps
As we have mentioned, conformal maps are useful for transferring problems from
a complicated domain to a simple domain. For example, we can use it to solve
Laplace’s equation, since solutions to Laplace’s equations are given by real and
imaginary parts of holomorphic functions.
More concretely, the following algorithm can be used to solve Laplace’s
Equation
2
φ
(
x, y
) = 0 on a tricky domain
U R
2
with given Dirichlet
boundary conditions on
U
. We now pretend
R
2
is actually
C
, and identify
subsets of R
2
with subsets of C in the obvious manner.
(i)
Find a conformal map
f
:
U V
, where
U
is now considered a subset of
C
, and
V
is a “nice” domain of our choice. Our aim is to find a harmonic
function Φ in V that satisfies the same boundary conditions as φ.
(ii)
Map the boundary conditions on
U
directly to the equivalent points on
V .
(iii) Now solve
2
Φ = 0 in V with the new boundary conditions.
(iv) The required harmonic function φ in U is then given by
φ(x, y) = Φ(Re(f(x + iy)), Im f(x + iy)).
To prove this works, we can take the
2
of this expression, write
f
=
u
+
iv
, use
the Cauchy-Riemann equation, and expand the mess.
Alternatively, we perform magic. Note that since Φ is harmonic, it is the
real part of some complex analytic function
F
(
z
) = Φ(
x, y
) +
i
Ψ(
x, y
), where
z
=
x
+
iy
. Now
F
(
f
(
z
)) is analytic, as it is a composition of analytic functions.
So its real part, which is Φ(Re f, Im f), is harmonic.
Let’s do an example. In this case, you might be able to solve this directly
just by looking at it, using what you’ve learnt from IB Methods. However, we
will do it with this complex methods magic.
Example.
We want to find a bounded solution of
2
φ
= 0 on the first quadrant
of R
2
subject to φ(x, 0) = 0 and φ(0, y) = 1 when, x, y > 0.
This is a bit silly, since our
U
is supposed to be a nasty region, but our
U
is
actually quite nice. Nevertheless, we still do this since this is a good example.
We choose f(z) = log z, which maps U to the strip 0 < Im z <
π
2
.
U
1
0
z 7→ log z
V
i
π
2
00
1
Recall that we said
log
maps an annulus to a rectangle. This is indeed the case
here
U
is an annulus with zero inner radius and infinite outer radius;
V
is an
infinitely long rectangle.
Now, we must now solve
2
Φ = 0 in V subject to
Φ(x, 0) = 0, Φ
x,
π
2
= 1
for all
x R
. Note that we have these boundary conditions since
f
(
z
) takes
positive real axis of
V
to the line
Im z
= 0, and the positive imaginary axis to
Im z =
π
2
.
By inspection, the solution is
Φ(x, y) =
2
π
y.
Hence,
Φ(x, y) = Φ(Re log z, Im log z)
=
2
π
Im log z
=
2
π
tan
1
y
x
.
Notice this is just the argument θ.
2 Contour integration and Cauchy’s theorem
In the remaining of the course, we will spend all our time studying integration
of complex functions, and see what we can do with it. At first, you might think
this is just an obvious generalization of integration of real functions. This is not
true. Complex integrals have many many nice properties, and it turns out there
are some really convenient tricks for evaluating complex integrals. In fact, we
will learn how to evaluate certain real integrals by pretending they are complex.
2.1 Contour and integrals
With real functions, we can just integrate a function, say, from 0 to 1, since there
is just one possible way we can get from 0 to 1 along the real line. However, in
the complex plane, there are many paths we can take to get from a point to
another. Integrating along different paths may produce different results. So we
have to carefully specify our path of integration.
Definition (Curve). A curve γ(t) is a (continuous) map γ : [0, 1] C.
Definition (Closed curve). A closed curve is a curve γ such that γ(0) = γ(1).
Definition
(Simple curve)
.
A simple curve is one which does not intersect itself,
except at t = 0, 1 in the case of a closed curve.
Definition (Contour). A contour is a piecewise smooth curve.
Everything we do is going to be about contours. We shall, in an abuse of
notation, often use the symbol
γ
to denote both the map and its image, namely
the actual curve in C traversed in a particular direction.
Notation.
The contour
γ
is the contour
γ
traversed in the opposite direction.
Formally, we say
(γ)(t) = γ(1 t).
Given two contours
γ
1
and
γ
2
with
γ
1
(1) =
γ
2
(0),
γ
1
+
γ
2
denotes the two
contours joined end-to-end. Formally,
(γ
1
+ γ
2
)(t) =
(
γ
1
(2t) t <
1
2
γ
2
(2t 1) t
1
2
.
Definition
(Contour integral)
.
The contour integral
R
γ
f
(
z
) d
z
is defined to be
the usual real integral
Z
γ
f(z) dz =
Z
1
0
f(γ(t))γ
0
(t) dt.
Alternatively, and equivalently, dissect [0
,
1] into 0 =
t
0
< t
1
< ··· < t
n
= 1,
and let z
n
= γ(t
n
) for n = 0, ··· , N. We define
δt
n
= t
n+1
t
n
, δz
n
= z
n+1
z
n
.
Then
Z
γ
f(z) dz = lim
0
N1
X
n=0
f(z
n
)δz
n
,
where
∆ = max
n=0,··· ,N 1
δt
n
,
and as 0, N .
All this says is that the integral is what we expect it to be an infinite sum.
The result of a contour integral between two points in
C
may depend on the
choice of contour.
Example. Consider
I
1
=
Z
γ
1
dz
z
, I
2
=
Z
γ
2
dz
z
,
where the paths are given by
γ
1
γ
2
θ
11 0
In both cases, we integrate from
z
=
1 to +1 around a unit circle:
γ
1
above,
γ
2
below the real axis. Substitute z = e
, dz = ie
dθ. Then we get
I
1
=
Z
0
π
ie
dθ
e
=
I
2
=
Z
0
π
ie
dθ
e
= .
So they can in fact differ.
Elementary properties of the integral
Contour integrals behave as we would expect them to.
Proposition.
(i) We write γ
1
+ γ
2
for the path obtained by joining γ
1
and γ
2
. We have
Z
γ
1
+γ
2
f(z) dz =
Z
γ
1
f(z) dz +
Z
γ
2
f(z) dz
Compare this with the equivalent result on the real line:
Z
c
a
f(x) dx =
Z
b
a
f(x) dx +
Z
c
b
f(x) dx.
(ii) Recall γ is the path obtained from reversing γ. Then we have
Z
γ
f(z) dz =
Z
γ
f(z) dz.
Compare this with the real result
Z
b
a
f(x) dx =
Z
a
b
f(x) dx.
(iii) If γ is a contour from a to b in C, then
Z
γ
f
0
(z) dz = f(b) f(a).
This looks innocuous. This is just the fundamental theorem of calculus.
However, there is some subtlety. This requires
f
to be differentiable at
every point on
γ
. In particular, it must not cross a branch cut. For example,
our previous example had
log z
as the antiderivative of
1
z
. However, this
does not imply the integrals along different paths are the same, since we
need to pick different branches of
log
for different paths, and things become
messy.
(iv)
Integration by substitution and by parts work exactly as for integrals on
the real line.
(v) If γ has length L and |f (z)| is bounded by M on γ, then
Z
γ
f(z) dz
LM.
This is since
Z
γ
f(z) dz
Z
γ
|f(z)|| dz| M
Z
γ
|dz| = ML.
We will be using this result a lot later on.
We will not prove these. Again, if you like proofs, go to IB Complex Analysis.
Integrals on closed contours
If
γ
is a closed contour, then it doesn’t matter where we start from on
γ
;
H
γ
f
(
z
) d
z
means the same thing in any case, so long as we go all the way round
(
H
denotes an integral around a closed contour).
The usual direction of traversal is anticlockwise (the “positive sense”). If
we traverse
γ
in a negative sense (clockwise), then we get negative the previous
result. More technically, the positive sense is the direction that keeps the interior
of the contour on the left. This “more technical” definition might seem pointless
if you can’t tell what anticlockwise is, then you probably can’t tell which is
the left. However, when we deal with more complicated structures in the future,
it turns out it is easier to define what is “on the left” than “anticlockwise”.
Simply connected domain
Definition
(Simply connected domain)
.
A domain
D
(an open subset of
C
) is
simply connected if it is connected and every closed curve in
D
encloses only
points which are also in D.
In other words, it does not have holes. For example, this is not simply-
connected:
These “holes” need not be big holes like this, but just individual points at which
a function under consider consideration is singular.
2.2 Cauchy’s theorem
We now come to the highlight of the course Cauchy’s theorem. Most of the
things we do will be based upon this single important result.
Theorem
(Cauchy’s theorem)
.
If
f
(
z
) is analytic in a simply-connected domain
D, then for every simple closed contour γ in D, we have
I
γ
f(z) dz = 0.
This is quite a powerful statement, and will allow us to do a lot! On the other
hand, this tells us functions that are analytic everywhere are not too interesting.
Instead, we will later look at functions like
1
z
that have singularities.
Proof.
(non-examinable) The proof of this remarkable theorem is simple (with a
catch), and follows from the Cauchy-Riemann equations and Green’s theorem.
Recall that Green’s theorem says
I
S
(P dx + Q dy) =
ZZ
S
Q
x
P
y
dx dy.
Let u, v be the real and imaginary parts of f. Then
I
γ
f(z) dz =
I
γ
(u + iv)(dx + i dy)
=
I
γ
(u dx v dy) + i
I
γ
(v dx + u dy)
=
ZZ
S
v
x
u
y
dx dy + i
ZZ
S
u
x
v
y
dx dy
But both integrands vanish by the Cauchy-Riemann equations, since
f
is differ-
entiable throughout S. So the result follows.
Actually, this proof requires
u
and
v
to have continuous partial derivatives
in
S
, otherwise Green’s theorem does not apply. We shall see later that in fact
f
is differentiable infinitely many time, so actually
u
and
v
do have continuous
partial derivatives. However, our proof of that will utilize Cauchy’s theorem! So
we are trapped.
Thus a completely different proof (and a very elegant one!) is required if we
do not wish to make assumptions about
u
and
v
. However, we shall not worry
about this in this course since it is easy to verify that the functions we use do
have continuous partial derivatives. And we are not doing Complex Analysis.
2.3 Contour deformation
One useful consequence of Cauchy’s theorem is that we can freely deform contours
along regions where f is defined without changing the value of the integral.
Proposition.
Suppose that
γ
1
and
γ
2
are contours from
a
to
b
, and that
f
is
analytic on the contours and between the contours. Then
Z
γ
1
f(z) dz =
Z
γ
2
f(z) dz.
a
b
γ
2
γ
1
Proof.
Suppose first that
γ
1
and
γ
2
do not cross. Then
γ
1
γ
2
is a simple c