Part IB Complex Analysis
Based on lectures by I. Smith
Notes taken by Dexter Chua
Lent 2016
These notes are not endorsed by the lecturers, and I have modified them (often
significantly) after lectures. They are nowhere near accurate representations of what
was actually lectured, and in particular, all errors are almost surely mine.
Analytic functions
Complex differentiation and the Cauchy–Riemann equations. Examples. Conformal
mappings. Informal discussion of branch points, examples of log z and z
c
. [3]
Contour integration and Cauchy’s theorem
Contour integration (for piecewise continuously differentiable curves). Statement and
proof of Cauchy’s theorem for star domains. Cauchy’s integral formula, maximum
modulus theorem, Liouville’s theorem, fundamental theorem of algebra. Morera’s
theorem. [5]
Expansions and singularities
Uniform convergence of analytic functions; local uniform convergence. Differentiability
of a power series. Taylor and Laurent expansions. Principle of isolated zeros. Residue
at an isolated singularity. Classification of isolated singularities. [4]
The residue theorem
Winding numbers. Residue theorem. Jordan’s lemma. Evaluation of definite integrals
by contour integration. Rouch´e’s theorem, principle of the argument. Open mapping
theorem. [4]
Contents
0 Introduction
1 Complex differentiation
1.1 Differentiation
1.2 Conformal mappings
1.3 Power series
1.4 Logarithm and branch cuts
2 Contour integration
2.1 Basic properties of complex integration
2.2 Cauchy’s theorem
2.3 The Cauchy integral formula
2.4 Taylor’s theorem
2.5 Zeroes
2.6 Singularities
2.7 Laurent series
3 Residue calculus
3.1 Winding numbers
3.2 Homotopy of closed curves
3.3 Cauchy’s residue theorem
3.4 Overview
3.5 Applications of the residue theorem
3.6 Rouces theorem
0 Introduction
Complex analysis is the study of complex differentiable functions. While this
sounds like it should be a rather straightforward generalization of real analysis,
it turns out complex differentiable functions behave rather differently. Requir-
ing that a function is complex differentiable is a very strong condition, and
consequently these functions have very nice properties.
One of the most distinguishing results from complex analysis is Liouville’s
theorem, which says that every bounded complex differentiable function
f
:
C C
must be constant. This is very false for real functions (e.g.
sin x
). This
gives a strikingly simple proof of the fundamental theorem of algebra if the
polynomial
p
has no roots, then
1
p
is well-defined on all of
C
, and it is easy to
show this must be bounded. So p is constant.
Many things we hoped were true in real analysis are indeed true in complex
analysis. For example, if a complex function is once differentiable, then it is
infinitely differentiable. In particular, every complex differentiable function has
a Taylor series and is indeed equal to its Taylor series (in reality, to prove these,
we show that every complex differentiable function is equal to its Taylor series,
and then notice that power series are always infinitely differentiable).
Another result we will prove is that the uniform limit of complex differentiable
functions is again complex differentiable. Contrast this with the huge list of
weird conditions we needed for real analysis!
Not only is differentiation nice. It turns out integration is also easier in
complex analysis. In fact, we will exploit this fact to perform real integrals by
pretending they are complex integrals. However, this will not be our main focus
here those belong to the IB Complex Methods course instead.
1 Complex differentiation
1.1 Differentiation
We start with some definitions. As mentioned in the introduction, Liouville’s
theorem says functions defined on the whole of
C
are often not that interesting.
Hence, we would like to work with some subsets of
C
instead. As in real analysis,
for differentiability to be well-defined, we would want a function to be defined
on an open set, so that we can see how
f
:
U C
varies as we approach a point
z
0
U from all different directions.
Definition
(Open subset)
.
A subset
U C
is open if for any
x U
, there is
some ε > 0 such that the open ball B
ε
(x) = B(x; ε) U .
The notation used for the open ball varies form time to time, even within
the same sentence. For example, instead of putting
ε
as the subscript, we could
put
x
as the subscript and
ε
inside the brackets. Hopefully, it will be clear from
context.
This is good, but we also want to rule out some silly cases, such as functions
defined on subsets that look like this:
This would violate results such as functions with zero derivative must be constant.
Hence, we would require our subset to be connected. This means for any two
points in the set, we can find a path joining them. A path can be formally
defined as a function γ : [0, 1] C, with start point γ(0) and end point γ(1).
Definition
(Path-connected subset)
.
A subset
U C
is path-connected if for
any
x, y U
, there is some
γ
: [0
,
1]
U
continuous such that
γ
(0) =
x
and
γ(1) = y.
Together, these define what it means to be a domain. These are (usually)
things that will be the domains of our functions.
Definition
(Domain)
.
A domain is a non-empty open path-connected subset
of C.
With this, we can define what it means to be differentiable at a point. This
is, in fact, exactly the same definition as that for real functions.
Definition
(Differentiable function)
.
Let
U C
be a domain and
f
:
U C
be a function. We say f is differentiable at w U if
f
0
(w) = lim
zw
f(z) f (w)
z w
exists.
Here we implicitly require that the limit does not depend on which direction
we approach
w
from. This requirement is also present for real differentiability, but
there are just two directions we can approach
w
from the positive direction
and the negative direction. For complex analysis, there are infinitely many
directions to choose from, and it turns out this is a very strong condition to
impose.
Complex differentiability at a point
w
is not too interesting. Instead, what we
want is a slightly stronger condition that the function is complex differentiable
in a neighbourhood of w.
Definition
(Analytic/holomorphic function)
.
A function
f
is analytic or holo-
morphic at
w U
if
f
is differentiable on an open neighbourhood
B
(
w, ε
) of
w
(for some ε).
Definition
(Entire function)
.
If
f
:
C C
is defined on all of
C
and is
holomorphic on C, then f is said to be entire.
It is not universally agreed what the words analytic and holomorphic should
mean. Some people take one of these word to mean instead that the function
has a (and is given by the) Taylor series, and then take many pages to prove
that these two notions are indeed the same. But since they are the same, we
shall just opt for the simpler definition.
The goal of the course is to develop the rich theory of these complex dif-
ferentiable functions and see how we can integrate them along continuously
differentiable (C
1
) paths in the complex plane.
Before we try to achieve our lofty goals, we first want to figure out when
a function is differentiable. Sure we can do this by checking the definition
directly, but this quickly becomes cumbersome for more complicated functions.
Instead, we would want to see if we can relate complex differentiability to real
differentiability, since we know how to differentiate real functions.
Given
f
:
U C
, we can write it as
f
=
u
+
iv
, where
u, v
:
U R
are
real-valued functions. We can further view
u
and
v
as real-valued functions of
two real variables, instead of one complex variable.
Then from IB Analysis II, we know this function
u
:
U R
is differentiable
(as a real function) at a point (
c, d
)
U
, with derivative
Du|
(c,d)
= (
λ, µ
), if and
only if
u(x, y) u(c, d) (λ(x c) + µ(y d))
k(x, y) (c, d)k
0 as (x, y) (c, d).
This allows us to come up with a nice criterion for when a complex function is
differentiable.
Proposition.
Let
f
be defined on an open set
U C
. Let
w
=
c
+
id U
and
write
f
=
u
+
iv
. Then
f
is complex differentiable at
w
if and only if
u
and
v
,
viewed as a real function of two real variables, are differentiable at (c, d), and
u
x
= v
y
,
u
y
= v
x
.
These equations are the Cauchy–Riemann equations. In this case, we have
f
0
(w) = u
x
(c, d) + iv
x
(c, d) = v
y
(c, d) iu
y
(c, d).
Proof. By definition, f is differentiable at w with f
0
(w) = p + iq if and only if
lim
zw
f(z) f(w) (p + iq)(z w)
z w
= 0. ()
If z = x + iy, then
(p + iq)(z w) = p(x c) q(y d) + i(q(x c) + p(y d)).
So, breaking into real and imaginary parts, we know () holds if and only if
lim
(x,y)(c,d)
u(x, y) u(c, d) (p(x c) q(y d))
p
(x c)
2
+ (y d)
2
= 0
and
lim
(x,y)(c,d)
v(x, y) v(c, d) (q(x c) + p(y d))
p
(x c)
2
+ (y d)
2
= 0.
Comparing this to the definition of the differentiability of a real-valued function,
we see this holds exactly if u and v are differentiable at (c, d) with
Du|
(c,d)
= (p, q), Dv|
(c,d)
= (q, p).
A standard warning is given that
f
:
U C
can be written as
f
=
u
+
iv
,
where
u
x
=
v
y
and
u
y
=
v
x
at (
c, d
)
U
, we cannot conclude that
f
is complex
differentiable at (
c, d
). These conditions only say the partial derivatives exist,
but this does not imply imply that
u
and
v
are differentiable, as required by the
proposition. However, if the partial derivatives exist and are continuous, then
by IB Analysis II we know they are differentiable.
Example.
(i)
The usual rules of differentiation (sum rule, product, rule, chain rule,
derivative of inverse) all hold for complex differentiable functions, with the
same proof as the real case.
(ii)
A polynomial
p
:
C C
is entire. This can be checked directly from
definition, or using the product rule.
(iii)
A rational function
p(z)
q(z)
:
U C
, where
U C \ {z
:
q
(
z
) = 0
}
, is
holomorphic on any such U. Here p, q are polynomials.
(iv) f
(
z
) =
|z|
is not complex differentiable at any point of
C
. Indeed, we can
write this as f = u + iv, where
u(x, y) =
p
x
2
+ y
2
, v(x, y) = 0.
If (x, y) 6= (0, 0), then
u
x
=
x
p
x
2
+ y
2
, u
y
=
y
p
x
2
+ y
2
.
If we are not at the origin, then clearly we cannot have both vanish, but
the partials of
v
both vanish. Hence the Cauchy–Riemann equations do
not hold and it is not differentiable outside of the origin.
At the origin, we can compute directly that
f(h) f(0)
h
=
|h|
h
.
This is, say, +1 for
h R
+
and
1 for
h R
. So the limit as
h
0 does
not exist.
1.2 Conformal mappings
The course schedules has a weird part where we are supposed to talk about
conformal mappings for a lecture, but not use it anywhere else. We have to put
them somewhere, and we might as well do it now. However, this section will be
slightly disconnected from the rest of the lectures.
Definition
(Conformal function)
.
Let
f
:
U C
be a function holomorphic at
w U . If f
0
(w) 6= 0, we say f is conformal at w.
What exactly does
f
0
(
w
)
6
= 0 tell us? In the real case, if we know a function
f
: (
a, b
)
R
is continuous differentiable, and
f
0
(
c
)
6
= 0, then
f
is locally
increasing or decreasing at
c
, and hence has a local inverse. This is also true in
the case of complex functions.
We write
f
=
u
+
iv
, then viewed as a map
R
2
R
2
, the Jacobian matrix is
given by
Df =
u
x
u
y
v
x
v
y
.
Then
det(Df) = u
x
v
y
u
y
v
x
= u
2
x
+ u
2
y
.
Using the formula for the complex derivative in terms of the partials, this shows
that if
f
0
(
w
)
6
= 0, then
det
(
Df|
w
)
6
= 0. Hence, by the inverse function theorem
(viewing
f
as a function
R
2
R
2
),
f
is locally invertible at
w
(technically, we
need
f
to be continuously differentiable, instead of just differentiable, but we
will later show that
f
in fact must be infinitely differentiable). Moreover, by
the same proof as in real analysis, the local inverse to a holomorphic function is
holomorphic (and conformal).
But being conformal is more than just being locally invertible. An important
property of conformal mappings is that they preserve angles. To give a precise
statement of this, we need to specify how “angles” work.
The idea is to look at tangent vectors of paths. Let
γ
1
, γ
2
: [
1
,
1]
U
be
continuously differentiable paths that intersect when
t
= 0 at
w
=
γ
1
(0) =
γ
2
(0).
Moreover, assume γ
0
i
(0) 6= 0.
Then we can compare the angles between the paths by looking at the difference
in arguments of the tangents at w. In particular, we define
angle(γ
1
, γ
2
) = arg(γ
0
1
(0)) arg(γ
0
2
(0)).
Let
f
:
U C
and
w U
. Suppose
f
is conformal at
w
. Then
f
maps our two
paths to
f γ
i
: [
1
,
1]
C
. These two paths now intersect at
f
(
w
). Then the
angle between them is
angle(f γ
1
, f γ
2
) = arg((f γ
1
)
0
(0)) arg((f γ
2
)
0
(0))
= arg
(f γ
1
)
0
(0)
(f γ
2
)
0
(0)
= arg
γ
0
1
(0)
γ
0
2
(0)
= angle(γ
1
, γ
2
),
using the chain rule and the fact that f
0
(w) 6= 0. So angles are preserved.
What else can we do with conformal maps? It turns out we can use it to
solve Laplace’s equation.
We will later prove that if
f
:
U C
is holomorphic on an open set
U
, then
f
0
: U C is also holomorphic. Hence f is infinitely differentiable.
In particular, if we write
f
=
u
+
iv
, then using the formula for
f
0
in terms
of the partials, we know
u
and
v
are also infinitely differentiable. Differentiating
the Cauchy–Riemann equations, we get
u
xx
= v
yx
= u
yy
.
In other words,
u
xx
+ u
yy
= 0,
We get similar results for
v
instead. Hence
Re
(
f
) and
Im
(
f
) satisfy the Laplace
equation and are hence harmonic (by definition).
Definition
(Conformal equivalence)
.
If
U
and
V
are open subsets of
C
and
f : U V is a conformal bijection, then it is a conformal equivalence.
Note that in general, a bijective continuous map need not have a continuous
inverse. However, if we are given further that it is conformal, then the inverse
mapping theorem tells us there is a local conformal inverse, and if the function
is bijective, these patch together to give a global conformal inverse.
The idea is that if we are required to solve the 2D Laplace’s equation on
a funny domain
U
subject to some boundary conditions, we can try to find a
conformal equivalence
f
between
U
and some other nice domain
V
. We can
then solve Laplace’s equation on
V
subject to the boundary conditions carried
forward by
f
, which is hopefully easier. Afterwards, we pack this solution into
the real part of a conformal function
g
, and then
g f
:
U C
is a holomorphic
function on
U
whose real part satisfies the boundary conditions we want. So we
have found a solution to Laplace’s equation.
You will have a chance to try that on the first example sheet. Instead, we
will focus on finding conformal equivalences between different regions, since
examiners like these questions.
Example.
Any obius map
A
(
z
) =
az+b
cz+d
(with
adbc 6
= 0) defines a conformal
equivalence
C {∞} C {∞}
in the obvious sense.
A
0
(
z
)
6
= 0 follows from
the chain rule and the invertibility of A(z).
In particular, the obius group of the disk
ob(D) = {f obius group : f(D) = D}
=
λ
z a
¯az 1
ob : |a| < 1, |λ| = 1
is a group of conformal equivalences of the disk. You will prove that the obius
group of the disk is indeed of this form in the first example sheet, and that these
are all conformal equivalences of the disk on example sheet 2.
Example.
The map
z 7→ z
n
for
n
2 is holomorphic everywhere and conformal
except at z = 0. This gives a conformal equivalence
n
z C
: 0 < arg(z) <
π
n
o
H,
where we adopt the following notation:
Notation. We write C
= C \ {0} and
H = {z C : Im(z) > 0}
is the upper half plane.
Example.
Note that
z H
if and only if
z
is closer to
i
than to
i
. In other
words,
|z i| < |z + i|,
or
z i
z + i
< 1.
So
z 7→
zi
z+i
defines a conformal equivalence
H D
, the unit disk. We know
this is conformal since it is a special case of the obius map.
Example. Consider the map
z 7→ w =
1
2
z +
1
z
,
assuming z C
. This can also be written as
w + 1
w 1
= 1 +
2
w 1
= 1 +
4
z +
1
z
2
= 1 +
4z
z
2
2z + 1
=
z + 1
z 1
2
.
So this is just squaring in some funny coordinates given by
z+1
z1
. This map is
holomorphic (except at z = 0). Also, we have
f
0
(z) = 1
z
2
+ 1
2z
2
.
So f is conformal except at ±1.
Recall that the first thing we learnt about obius maps is that they take
lines and circles to lines and circles. This does something different. We write
z = re
. Then if we write z 7→ w = u + iv, we have
u =
1
2
r +
1
r
cos θ
v =
1
2
r
1
r
sin θ
Fixing the radius and argument fixed respectively, we see that a circle of radius
ρ is mapped to the ellipse
u
2
1
4
ρ +
1
ρ
2
+
v
2
1
4
ρ
1
ρ
2
= 1,
while the half-line arg(z) = µ is mapped to the hyperbola
u
2
cos
2
µ
v
2
sin
2
µ
= 1.
We can do something more interesting. Consider a off-centered circle, chosen to
pass through the point 1 and i. Then the image looks like this:
1
i
f
f(1)
f(i)
Note that we have a singularity at
f
(
1) =
1. This is exactly the point where
f is not conformal, and is no longer required to preserve angles.
This is a crude model of an aerofoil, and the transformation is known as the
Joukowsky transform.
In applied mathematics, this is used to model fluid flow over a wing in terms
of the analytically simpler flow across a circular section.
We interlude with a little trick. Often, there is no simple way to describe
regions in space. However, if the region is bounded by circular arcs, there is a
trick that can be useful.
Suppose we have a circular arc between α and β.
α
z
β
φ
θ
µ
Along this arc,
µ
=
θ φ
=
arg
(
z α
)
arg
(
z β
) is constant, by elementary
geometry. Thus, for each fixed µ, the equation
arg(z α) arg(z β) = µ
determines an arc through the points α, β.
To obtain a region bounded by two arcs, we find the two
µ
and
µ
+
that
describe the boundary arcs. Then a point lie between the two arcs if and only if
its µ is in between µ
and µ
+
, i.e. the region is
z : arg
z α
z β
[µ
, µ
+
]
.
This says the point has to lie in some arc between those given by µ
and µ
+
.
For example, the following region:
1 1
i
can be given by
U =
z : arg
z 1
z + 1
h
π
2
, π
i
.
Thus for instance the map
z 7→
z 1
z + 1
2
is a conformal equivalence from
U
to
H
. This is since if
z U
, then
z1
z+1
has
argument in
π
2
, π
, and can have arbitrary magnitude since
z
can be made
as close to
1 as you wish. Squaring doubles the angle and gives the lower
half-plane, and multiplying by 1 gives the upper half plane.
z 7→
z1
z+1
z 7→ z
2
z 7→ z
In fact, there is a really powerful theorem telling us most things are conformally
equivalent to the unit disk.
Theorem
(Riemann mapping theorem)
.
Let
U C
be the bounded domain
enclosed by a simple closed curve, or more generally any simply connected domain
not equal to all of
C
. Then
U
is conformally equivalent to
D
=
{z
:
|z| <
1
} C
.
This in particular tells us any two simply connected domains are conformally
equivalent.
The terms simple closed curve and simply connected are defined as follows:
Definition
(Simple closed curve)
.
A simple closed curve is the image of an
injective map S
1
C.
It should be clear (though not trivial to prove) that a simple closed curve
separates C into a bounded part and an unbounded part.
The more general statement requires the following definition:
Definition
(Simply connected)
.
A domain
U C
is simply connected if every
continuous map from the circle
f
:
S
1
U
can be extended to a continuous
map from the disk
F
:
D
2
U
such that
F |
D
2
=
f
. Alternatively, any loop
can be continuously shrunk to a point.
Example.
The unit disk is simply-connected, but the region defined by 1
<
|z| <
2 is not, since the circle
|z|
= 1
.
5 cannot be extended to a map from a disk.
We will not prove this statement, but it is nice to know that this is true.
If we believe that the unit disk is relatively simple, then since all simply
connected regions are conformally equivalent to the disk, all simply connected
domains are boring. This suggests we will later encounter domains with holes to
make the course interesting. This is in fact true, and we will study these holes
in depth later.
Example. The exponential function
e
z
= 1 + z +
z
2
2!
+
z
3
3!
+ ···
defines a function
C C
. In fact it is a conformal mapping. This sends
the region
{z
:
Re
(
z
)
[
a, b
]
}
to the annulus
{e
a
|w| e
b
}
. One is simply
connected, but the other is not this is not a problem since
e
z
is not bijective
on the strip.
a
b
e
z
1.3 Power series
Some of our favorite functions are all power series, including polynomials (which
are degenerate power series in some sense), exponential functions and trigono-
metric functions. We will later show that all holomorphic functions are (locally)
given by power series, but without knowing this fact, we shall now study some
of the basic properties of power series.
It turns out power series are nice. The key property is that power series
are infinitely differentiable (as long as it converges), and the derivative is given
pointwise.
We begin by recalling some facts about convergence from IB Analysis II.
Definition
(Uniform convergence)
.
A sequence (
f
n
) of functions converge
uniformly to
f
if for all
ε >
0, there is some
N
such that
n > N
implies
|f
n
(z) f(z)| < ε for all z.
Proposition. The uniform limit of continuous functions is continuous.
Proposition
(Weierstrass M-test)
.
For a sequence of functions
f
n
, if we can
find (
M
n
)
R
>0
such that
|f
n
(
x
)
| < M
n
for all
x
in the domain, then
P
M
n
converges implies
P
f
n
(x) converges uniformly on the domain.
Proposition.
Given any constants
{c
n
}
n0
C
, there is a unique
R
[0
,
]
such that the series
z 7→
P
n=0
c
n
(
z a
)
n
converges absolutely if
|z a| < R
and diverges if
|z a| > R
. Moreover, if 0
< r < R
, then the series converges
uniformly on {z : |z a| < r}. This R is known as the radius of convergence.
So while we don’t necessarily get uniform convergence on the whole domain,
we get uniform convergence on all compact subsets of the domain.
We are now going to look at power series. They will serve as examples, and
as we will see later, universal examples, of holomorphic functions. The most
important result we need is the following result about their differentiability.
Theorem. Let
f(z) =
X
n=0
c
n
(z a)
n
be a power series with radius of convergence R > 0. Then
(i) f is holomorphic on B(a; R) = {z : |z a| < R}.
(ii) f
0
(z) =
P
nc
n
(z 1)
n1
, which also has radius of convergence R.
(iii) Therefore f is infinitely complex differentiable on B(a; R). Furthermore,
c
n
=
f
(n)
(a)
n!
.
Proof.
Without loss of generality, take
a
= 0. The third part obviously follows
from the previous two, and we will prove the first two parts simultaneously. We
would like to first prove that the derivative series has radius of convergence
R
,
so that we can freely happily manipulate it.
Certainly, we have
|nc
n
| |c
n
|
. So by comparison to the series for
f
, we can
see that the radius of convergence of
P
nc
n
z
n1
is at most
R
. But if
|z| < ρ < R
,
then we can see
|nc
n
z
n1
|
|c
n
ρ
n1
|
= n
z
ρ
n1
0
as
n
. So by comparison to
P
c
n
ρ
n1
, which converges, we see that the
radius of convergence of
P
nc
n
z
n1
is at least
ρ
. So the radius of convergence
must be exactly R.
Now we want to show
f
really is differentiable with that derivative. Pick
z, w such that |z|, |w| ρ for some ρ < R as before.
Define a new function
ϕ(z, w) =
X
n=1
c
n
n1
X
j=0
z
j
w
n1j
.
Noting
c
n
n1
X
j=0
z
j
w
n1j
n|c
n
|ρ
n
,
we know the series defining
ϕ
converges uniformly on
{|z| ρ, |w| < ρ}
, and
hence to a continuous limit.
If z 6= w, then using the formula for the (finite) geometric series, we know
ϕ(z, w) =
X
n=1
c
n
z
n
w
n
z w
=
f(z) f(w)
z w
.
On the other hand, if z = w, then
ϕ(z, z) =
X
n=1
c
n
nz
n1
.
Since ϕ is continuous, we know
lim
wz
f(z) f(w)
z w
X
n=1
c
n
nz
n1
.
So f
0
(z) = ϕ(z, z) as claimed. Then (iii) follows from (i) and (ii) directly.
Corollary. Given a power series
f(z) =
X
n0
c
n
(z a)
n
with radius of convergence
R >
0, and given 0
< ε < R
, if
f
vanishes on
B
(
a, ε
),
then f vanishes identically.
Proof.
If
f
vanishes on
B
(
a, ε
), then all its derivatives vanish, and hence the
coefficients all vanish. So it is identically zero.
This is obviously true, but will come up useful some time later.
It is might be useful to have an explicit expression for
R
. For example, by
IA Analysis I, we know
R = sup{r 0 : |c
n
|r
n
0 as n ∞}
=
1
lim sup
n
p
|c
n
|
.
But we probably won’t need these.
1.4 Logarithm and branch cuts
Recall that the exponential function
e
z
= exp(z) = 1 + z +
z
2
2!
+
z
3
3!
+ ···
has a radius of convergence of
. So it is an entire function. We have the usual
standard properties, such as e
z+w
= e
z
e
w
, and also
e
x+iy
= e
x
e
iy
= e
x
(cos y + i sin y).
So given
w C
=
C \ {
0
}
, there are solutions to
e
z
=
w
. In fact, this
has infinitely many solutions, differing by adding integer multiples of 2
πi
. In
particular, e
z
= 1 if and only if z is an integer multiple of 2πi.
This means
e
x
does not have a well-defined inverse. However, we do want to
talk about the logarithm. The solution would be to just fix a particular range
of
θ
allowed. For example, we can define the logarithm as the function sending
re
to
log r
+
, where we now force
π < θ < π
. This is all well, except it
is now not defined at
1 (we can define it as, say,
, but we would then lose
continuity).
There is no reason why we should pick
π < θ < π
. We could as well require
300π < θ < 302π. In general, we make the following definition:
Definition
(Branch of logarithm)
.
Let
U C
be an open subset. A branch of
the logarithm on
U
is a continuous function
λ
:
U C
for which
e
λ(z)
=
z
for
all z U.
This is a partially defined inverse to the exponential function, only defined
on some domain
U
. These need not exist for all
U
. For example, there is no
branch of the logarithm defined on the whole C
, as we will later prove.
Example. Let U = C \ R
0
, a “slit plane”.
Then for each
z U
, we write
z
=
r
, with
π < θ < π
. Then
λ
(
z
) =
log
(
r
)+
is a branch of the logarithm. This is the principal branch.
On
U
, there is a continuous function
arg
:
U
(
π, π
), which is why we can
construct a branch. This is not true on, say, the unit circle.
Fortunately, as long as we have picked a branch, most things we want to be
true about log is indeed true.
Proposition.
On
{z C
:
z 6∈ R
0
}
, the principal branch
log
:
U C
is
holomorphic function. Moreover,
d
dz
log z =
1
z
.
If |z| < 1, then
log(1 + z) =
X
n1
(1)
n1
z
n
n
= z
z
2
2
+
z
3
3
··· .
Proof.
The logarithm is holomorphic since it is a local inverse to a holomorphic
function. Since e
log z
= z, the chain rule tells us
d
dz
(log z) =
1
z
.
To show that
log
(1 +
z
) is indeed given by the said power series, note that
the power series does have a radius of convergence 1 by, say, the ratio test. So
by the previous result, it has derivative
1 z + z
2
+ ··· =
1
1 + z
.
Therefore,
log
(1 +
z
) and the claimed power series have equal derivative, and
hence coincide up to a constant. Since they agree at
z
= 0, they must in fact be
equal.
Having defined the logarithm, we can now define general power functions.
Let
α C
and
log
:
U C
be a branch of the logarithm. Then we can define
z
α
= e
α log z
on U . This is again only defined when log is.
In a more general setting, we can view
log
as an instance of a multi-valued
function on
C
. At each point, the function
log
can take many possible values,
and every time we use
log
, we have to pick one of those values (in a continuous
way).
In general, we say that a point
p C
is a branch point of a multivalued
function if the function cannot be given a continuous single-valued definition in
a (punctured) neighbourhood
B
(
p, ε
)
\ {p}
of
p
for any
ε >
0. For example, 0 is
a branch point of log.
Example. Consider the function
f(z) =
p
z(z 1).
This has two branch points,
z
= 0 and
z
= 1, since we cannot define a square
root consistently near 0, as it is defined via the logarithm.
Note we can define a continuous branch of f on either
10
or we can just kill a finite slit by
10
Why is the second case possible? Note that
f(z) = e
1
2
(log(z)+log(z1))
.
If we move around a path encircling the finite slit, the argument of each of
log
(
z
)
and
log
(
z
1) will jump by 2
πi
, and the total change in the exponent is 2
πi
. So
the expression for f(z) becomes uniquely defined.
While these two ways of cutting slits look rather different, if we consider this
to be on the Riemann sphere, then these two cuts now look similar. It’s just
that one passes through the point , and the other doesn’t.
The introduction of these slits is practical and helpful for many of our
problems. However, theoretically, this is not the best way to think about multi-
valued functions. A better treatment will be provided in the IID Riemann
Surfaces course.
2 Contour integration
In the remaining of the course, we will spend all our time studying integration of
complex functions. At first, you might think this is just an obvious generalization
of integration of real functions. This is not true. Starting from Cauchy’s theorem,
beautiful and amazing properties of complex integration comes one after another.
Using these, we can prove many interesting properties of holomorphic functions
as well as do lots of integrals we previously were not able to do.
2.1 Basic properties of complex integration
We start by considering functions
f
: [
a, b
]
C
. We say such a function is
Riemann integrable if
Re
(
f
) and
Im
(
f
) are individually, and the integral is
defined to be
Z
b
a
f(t) dt =
Z
b
a
Re(f(t)) dt + i
Z
b
a
Im(f(t)) dt.
While Riemann integrability is a technical condition to check, we know that all
continuous functions are integrable, and this will apply in most cases we care
about in this course. After all, this is not a course on exotic functions.
We start from some less interesting facts, and slowly develop and prove some
really amazing results.
Lemma. Suppose f : [a, b] C is continuous (and hence integrable). Then
Z
b
a
f(t) dt
(b a) sup
t
|f(t)|
with equality if and only if f is constant.
Proof. We let
θ = arg
Z
b
a
f(t) dt
!
,
and
M = sup
t
|f(t)|.
Then we have
Z
b
a
f(t) dt
=
Z
b
a
e
f(t) dt
=
Z
b
a
Re(e
f(t)) dt
(b a)M,
with equality if and only if
|f
(
t
)
|
=
M
and
arg f
(
t
) =
θ
for all
t
, i.e.
f
is
constant.
Integrating functions of the form
f
: [
a, b
]
C
is easy. What we really care
about is integrating a genuine complex function
f
:
U C C
. However,
we cannot just “integrate” such a function. There is no given one-dimensional
domain we can integrate along. Instead, we have to make some up ourselves.
We have to define some paths in the complex plane, and integrate along them.
Definition
(Path)
.
A path in
C
is a continuous function
γ
: [
a, b
]
C
, where
a, b R.
For general paths, we just require continuity, and do not impose any conditions
about, say, differentiability.
Unfortunately, the world is full of weird paths. There are even paths that fill
up the whole of the unit square. So we might want to look at some nicer paths.
Definition
(Simple path)
.
A path
γ
: [
a, b
]
C
is simple if
γ
(
t
1
) =
γ
(
t
2
) only
if t
1
= t
2
or {t
1
, t
2
} = {a, b}.
In other words, it either does not intersect itself, or only intersects itself at
the end points.
Definition (Closed path). A path γ : [a, b] C is closed if γ(a) = γ(b).
Definition
(Contour)
.
A contour is a simple closed path which is piecewise
C
1
,
i.e. piecewise continuously differentiable.
For example, it can look something like this:
Most of the time, we are just interested in integration along contours. However,
it is also important to understand integration along just simple
C
1
smooth paths,
since we might want to break our contour up into different segments. Later, we
will move on to consider more general closed piecewise
C
1
paths, where we can
loop around a point many many times.
We can now define what it means to integrate along a smooth path.
Definition
(Complex integration)
.
If
γ
: [
a, b
]
U C
is
C
1
-smooth and
f : U C is continuous, then we define the integral of f along γ as
Z
γ
f(z) dz =
Z
b
a
f(γ(t))γ
0
(t) dt.
By summing over subdomains, the definition extends to piecewise
C
1
-smooth
paths, and in particular contours.
We have the following elementary properties:
(i)
The definition is insensitive to reparametrization. Let
φ
: [
a
0
, b
0
]
[
a, b
]
be
C
1
such that
φ
(
a
0
) =
a, φ
(
b
0
) =
b
. If
γ
is a
C
1
path and
δ
=
γ φ
, then
Z
γ
f(z) dz =
Z
δ
f(z) dz.
This is just the regular change of variables formula, since
Z
b
0
a
0
f(γ(φ(t)))γ
0
(φ(t))φ
0
(t) dt =
Z
b
a
f(γ(u))γ
0
(u) du
if we let u = φ(t).
(ii) If a < u < b, then
Z
γ
f(z) dz =
Z
γ|
[a,u]
f(z) dz +
Z
γ|
[u,b]
f(z) dz.
These together tells us the integral depends only on the path itself, not how we
look at the path or how we cut up the path into pieces.
We also have the following easy properties:
(iii) If γ is γ with reversed orientation, then
Z
γ
f(z) dz =
Z
γ
f(z) dz.
(iv) If we set for γ : [a, b] C the length
length(γ) =
Z
b
a
|γ
0
(t)| dt,
then
Z
γ
f(z) dz
length(γ) sup
t
|f(γ(t))|.
Example.
Take
U
=
C
, and let
f
(
z
) =
z
n
for
n Z
. We pick
φ
: [0
,
2
π
]
U
that sends θ 7→ e
. Then
Z
φ
f(z) dz =
(
2πi n = 1
0 otherwise
To show this, we have
Z
φ
f(z) dz =
Z
2π
0
e
inθ
ie
dθ
= i
Z
2π
0
e
i(n+1)θ
dθ.
If
n
=
1, then the integrand is constantly 1, and hence gives 2
πi
. Otherwise, the
integrand is a non-trivial exponential which is made of trigonometric functions,
and when integrated over 2π gives zero.
Example. Take γ to be the contour
γ
1
γ
2
RR
iR
We parametrize the path in segments by
γ
1
: [R, R] C γ
2
: [0, 1] C
t 7→ t t 7→ Re
t
Consider the function f(z) = z
2
. Then the integral is
Z
γ
f(z) dz =
Z
R
R
t
2
dt +
Z
1
0
R
2
e
2πit
Re
t
dt
=
2
3
R
3
+ R
3
Z
1
0
e
3πit
dt
=
2
3
R
3
+ R
3
e
3πit
3πi
1
0
= 0
We worked this out explicitly, but we have just wasted our time, since this is
just an instance of the fundamental theorem of calculus!
Definition
(Antiderivative)
.
Let
U C
and
f
:
U C
be continuous. An
antiderivative of
f
is a holomorphic function
F
:
U C
such that
F
0
(
z
) =
f
(
z
).
Then the fundamental theorem of calculus tells us:
Theorem
(Fundamental theorem of calculus)
.
Let
f
:
U C
be continuous
with antiderivative F . If γ : [a, b] U is piecewise C
1
-smooth, then
Z
γ
f(z) dz = F (γ(b)) F (γ(a)).
In particular, the integral depends only on the end points, and not the path
itself. Moreover, if γ is closed, then the integral vanishes.
Proof. We have
Z
γ
f(z) dz =
Z
b
a
f(γ(t))γ
0
(t) dt =
Z
b
a
(F γ)
0
(t) dt.
Then the result follows from the usual fundamental theorem of calculus, applied
to the real and imaginary parts separately.
Example.
This allows us to understand the first example we had. We had the
function f (z) = z
n
integrated along the path φ(t) = e
it
(for 0 t 2π).
If n 6= 1, then
f =
d
dt
z
n+1
n + 1
.
So
f
has a well-defined antiderivative, and the integral vanishes. On the other
hand, if n = 1, then
f(z) =
d
dz
(log z),
where
log
can only be defined on a slit plane. It is not defined on the whole unit
circle. So we cannot apply the fundamental theorem of calculus.
Reversing the argument around, since
R
φ
z
1
d
z
does not vanish, this implies
there is not a continuous branch of log on any set U containing the unit circle.
2.2 Cauchy’s theorem
A question we might ask ourselves is when the anti-derivative exists. A necessary
condition, as we have seen, is that the integral around any closed curve has to
vanish. This is also sufficient.
Proposition.
Let
U C
be a domain (i.e. path-connected non-empty open
set), and f : U C be continuous. Moreover, suppose
Z
γ
f(z) dz = 0
for any closed piecewise C
1
-smooth path γ in U. Then f has an antiderivative.
This is more-or-less the same proof we gave in IA Vector Calculus that a real
function is a gradient if and only if the integral about any closed path vanishes.
Proof.
Pick our favorite
a
0
U
. For
w U
, we choose a path
γ
w
: [0
,
1]
U
such that γ
w
(0) = a
0
and γ
w
(1) = w.
We first go through some topological nonsense to show we can pick
γ
w
such
that this is piecewise
C
1
. We already know a continuous path
γ
: [0
,
1]
U
from
a
0
to
w
exists, by definition of path connectedness. Since
U
is open, for
all
x
in the image of
γ
, there is some
ε
(
x
)
>
0 such that
B
(
x, ε
(
x
))
U
. Since
the image of
γ
is compact, it is covered by finitely many such balls. Then it is
trivial to pick a piecewise straight path living inside the union of these balls,
which is clearly piecewise smooth.
γ
a
0
w
γ
w
We thus define
F (w) =
Z
γ
w
f(z) dz.
Note that this
F
(
w
) is independent of the choice of
γ
w
, by our hypothesis on
f
given another choice
˜γ
w
, we can form the new path
γ
w
(
˜γ
w
), namely the
path obtained by concatenating γ
w
with ˜γ
w
.
a
0
w
˜γ
w
γ
w
This is a closed piecewise C
1
-smooth curve. So
Z
γ
w
(˜γ
w
)
f(z) dz = 0.
The left hand side is
Z
γ
w
f(z) dz +
Z
˜γ
w
f(z) dz =
Z
γ
w
f(z) dz
Z
˜γ
w
f(z) dz.
So the two integrals agree.
Now we need to check that
F
is complex differentiable. Since
U
is open, we
can pick
θ >
0 such that
B
(
w
;
ε
)
U
. Let
δ
h
be the radial path in
B
(
w, ε
) from
W to w + h, with |h| < ε.
a
γ
w
w
δ
h
w + h
Now note that γ
w
δ
h
is a path from a
0
to w + h. So
F (w + h) =
Z
γ
w
δ
h
f(z) dz
= F (w) +
Z
δ
h
f(z) dz
= F (w) + hf(w) +
Z
δ
h
(f(z) f(w)) dz.
Thus, we know
F (w + h) F (w)
h
f(w)
1
|h|
Z
δ
h
f(z) f(w) dz
1
|h|
length(δ
h
) sup
δ
h
|f(z) f(w)|
= sup
δ
h
|f(z) f(w)|.
Since
f
is continuous, as
h
0, we know
f
(
z
)
f
(
w
)
0. So
F
is differentiable
with derivative f.
To construct the anti-derivative, we assumed
R
γ
f
(
z
) d
z
= 0. But we didn’t
really need that much. To simplify matters, we can just consider curves consisting
of straight line segments. To do so, we need to make sure we really can draw
line segments between two points.
You might think aha! We should work with convex spaces. No. We do not
need such a strong condition. Instead, all we need is that we have a distinguished
point a
0
such that there is a line segment from a
0
to any other point.
Definition
(Star-shaped domain)
.
A star-shaped domain or star domain is a
domain
U
such that there is some
a
0
U
such that the line segment [
a
0
, w
]
U
for all w U .
a
0
w
This is weaker than requiring
U
to be convex, which says any line segment
between any two points in U, lies in U.
In general, we have the implications
U is a disc U is convex U is star-shaped U is path-connected,
and none of the implications reverse.
In the proof, we also needed to construct a small straight line segment
δ
h
.
However, this is a non-issue. By the openness of
U
, we can pick an open ball
B(w, ε) U , and we can certainly construct the straight line in this ball.
Finally, we get to the integration part. Suppose we picked all our
γ
w
to be the
fixed straight line segment from
a
0
. Then for antiderivative to be differentiable,
we needed
Z
γ
w
δ
h
f(z) dz =
Z
γ
w+h
f(z) dz.
In other words, we needed to the integral along the path
γ
w
δ
h
(
γ
w+h
) to
vanish. This is a rather simple kind of paths. It is just (the boundary of) a
triangle, consisting of three line segments.
Definition
(Triangle)
.
A triangle in a domain
U
is what it ought to be the
Euclidean convex hull of 3 points in
U
, lying wholly in
U
. We write its boundary
as T , which we view as an oriented piecewise C
1
path, i.e. a contour.
good
bad
very bad
Our earlier result on constructing antiderivative then shows:
Proposition. If U is a star domain, and f : U C is continuous, and if
Z
T
f(z) dz = 0
for all triangles T U, then f has an antiderivative on U.
Proof. As before, taking γ
w
= [a
0
, w] U if U is star-shaped about a
0
.
This is in some sense a weaker proposition while our hypothesis only
requires the integral to vanish over triangles, and not arbitrary closed loops, we
are restricted to star domains only.
But well, this is technically a weakening, but how is it useful? Surely if we
can somehow prove that the integral of a particular function vanishes over all
triangles, then we can easily modify the proof such that it works for all possible
curves.
Turns out, it so happens that for triangles, we can fiddle around with some
geometry to prove the following result:
Theorem
(Cauchy’s theorem for a triangle)
.
Let
U
be a domain, and let
f : U C be holomorphic. If T U is a triangle, then
R
T
f(z) dz = 0.
So for holomorphic functions, the hypothesis of the previous theorem auto-
matically holds.
We immediately get the following corollary, which is what we will end up
using most of the time.
Corollary
(Convex Cauchy)
.
If
U
is a convex or star-shaped domain, and
f
:
U C
is holomorphic, then for any closed piecewise
C
1
paths
γ U
, we
must have
Z
γ
f(z) dz = 0.
Proof of corollary.
If
f
is holomorphic, then Cauchy’s theorem says the integral
over any triangle vanishes. If
U
is star shaped, our proposition says
f
has an
antiderivative. Then the fundamental theorem of calculus tells us the integral
around any closed path vanishes.
Hence, all we need to do is to prove that fact about triangles.
Proof of Cauchy’s theorem for a triangle. Fix a triangle T. Let
η =
Z
T
f(z) dz
, ` = length(T ).
The idea is to show to bound
η
by
ε
, for every
ε >
0, and hence we must have
η = 0. To do so, we subdivide our triangles.
Before we start, it helps to motivate the idea of subdividing a bit. By
subdividing the triangle further and further, we are focusing on a smaller and
smaller region of the complex plane. This allows us to study how the integral
behaves locally. This is helpful since we are given that
f
is holomorphic, and
holomorphicity is a local property.
We start with T = T
0
:
We then add more lines to get
T
0
a
, T
0
b
, T
0
c
, T
0
d
(it doesn’t really matter which is
which).
We orient the middle triangle by the anti-clockwise direction. Then we have
Z
T
0
f(z) dz =
X
a,b,c,d
Z
T
0
·
f(z) dz,
since each internal edge occurs twice, with opposite orientation.
For this to be possible, if
η
=
R
T
0
f(z) dz
, then there must be some
subscript in {a, b, c, d} such that
Z
T
0
·
f(z) dz
η
4
.
We call this T
0
·
= T
1
. Then we notice T
1
has length
length(T
1
) =
`
2
.
Iterating this, we obtain triangles
T
0
T
1
T
2
···
such that
Z
T
i
f(z) dz
η
4
i
, length(T
i
) =
`
2
i
.
Now we are given a nested sequence of closed sets. By IB Metric and Topological
Spaces (or IB Analysis II), there is some z
0
T
i0
T
i
.
Now fix an
ε >
0. Since
f
is holomorphic at
z
0
, we can find a
δ >
0 such that
|f(w) f(z
0
) (w z
0
)f
0
(z
0
)| ε|w z
0
|
whenever
|w z
0
| < δ
. Since the diameters of the triangles are shrinking each
time, we can pick an
n
such that
T
n
B
(
z
0
, ε
). We’re almost there. We just
need to do one last thing that is slightly sneaky. Note that
Z
T
n
1 dz = 0 =
Z
T
n
z dz,
since these functions certainly do have anti-derivatives on
T
n
. Therefore, noting
that f (z
0
) and f
0
(z
0
) are just constants, we have
Z
T
n
f(z) dz
=
Z
T
n
(f(z) f(z
0
) (z z
0
)f
0
(z
0
)) dz
Z
T
n
|f(z) f(z
0
) (z z
0
)f
0
(z
0
)| dz
length(T
n
)ε sup
zT
n
|z z
0
|
ε length(T
n
)
2
,
where the last line comes from the fact that
z
0
T
n
, and the distance between
any two points in the triangle cannot be greater than the perimeter of the
triangle. Substituting our formulas for these in, we have
η
4
n
1
4
n
`
2
ε.
So
η `
2
ε.
Since ` is fixed and ε was arbitrary, it follows that we must have η = 0.
Is this the best we can do? Can we formulate this for an arbitrary domain,
and not just star-shaped ones? It is obviously not true if the domain is not
simply connected, e.g. for
f
(
z
) =
1
z
defined on
C \ {
0
}
. However, it turns out
Cauchy’s theorem holds as long as the domain is simply connected, as we will
show in a later part of the course. However, this is not surprising given the
Riemann mapping theorem, since any simply connected domain is conformally
equivalent to the unit disk, which is star-shaped (and in fact convex).
We can generalize our result when
f
:
U C
is continuous on the whole
of
U
, and holomorphic except on finitely many points. In this case, the same
conclusion holds
R
γ
f(z) dz = 0 for all piecewise smooth closed γ.
Why is this? In the proof, it was sufficient to focus on showing
R
T
f
(
z
) d
z
= 0
for a triangle
T U
. Consider the simple case where we only have a single point
of non-holomorphicity a T . The idea is again to subdivide.
a
We call the center triangle
T
0
. Along all other triangles in our subdivision, we
get
R
f(z) dz = 0, as these triangles lie in a region where f is holomorphic. So
Z
T
f(z) dz =
Z
T
0
f(z) dz.
Note now that we can make T
0
as small as we like. But
Z
T
0
f(z) dz
length(T
0
) sup
zT
0
|f(z)|.
Since
f
is continuous, it is bounded. As we take smaller and smaller subdivisions,
length(T
0
) 0. So we must have
R
T
f(z) dz = 0.
From here, it’s straightforward to conclude the general case with many points
of non-holomorphicity we can divide the triangle in a way such that each
small triangle contains one bad point.
2.3 The Cauchy integral formula
Our next amazing result will be Cauchy’s integral formula. This formula allows
us to find the value of
f
inside a ball
B
(
z
0
, r
) just given the values of
f
on the
boundary B(z
0
, r).
Theorem
(Cauchy integral formula)
.
Let
U
be a domain, and
f
:
U C
be
holomorphic. Suppose there is some
B(z
0
; r) U
for some
z
0
and
r >
0. Then
for all z B(z
0
; r), we have
f(z) =
1
2πi
Z
B(z
0
;r)
f(w)
w z
dw.
Recall that we previously computed
R
B(0,1)
1
z
d
z
= 2
πi
. This is indeed a
special case of the Cauchy integral formula. We will provide two proofs. The
first proof relies on the above generalization of Cauchy’s theorem.
Proof.
Since
U
is open, there is some
δ >
0 such that
B(z
0
; r + δ) U
. We
define g : B(z
0
; r + δ) C by
g(w) =
(
f(w)f (z)
wz
w 6= z
f
0
(z) w = z
,
where we have fixed
z B
(
z
0
;
r
) as in the statement of the theorem. Now
note that
g
is holomorphic as a function of
w B
(
z
0
, r
+
δ
), except perhaps at
w
=
z
. But since
f
is holomorphic, by definition
g
is continuous everywhere on
B(z
0
, r + δ). So the previous result says
Z
B(z
0
;r)
g(w) dw = 0.
This is exactly saying that
Z
B(z
0
;r)
f(w)
w z
dw =
Z
B(z
0
;r)
f(z)
w z
dw.
We now rewrite
1
w z
=
1
w z
0
·
1
1
zz
0
wz
0
=
X
n=0
(z z
0
)
n
(w z
0
)
n+1
.
Note that this sum converges uniformly on B(z
0
; r) since
z z
0
w z
0
< 1
for w on this circle.
By uniform convergence, we can exchange summation and integration. So
Z
B(z
0
;r)
f(w)
w z
dw =
X
n=0
Z
B(z
0
,r)
f(z)
(z z
0
)
n
(w z
0
)
n+1
dw.
We note that
f
(
z
)(
z z
0
)
n
is just a constant, and that we have previously proven
Z
B(z
0
;r)
(w z
0
)
k
dw =
(
2πi k = 1
0 k 6= 1
.
So the right hand side is just 2πif(z). So done.
Corollary
(Local maximum principle)
.
Let
f
:
B
(
z, r
)
C
be holomorphic.
Suppose
|f
(
w
)
| |f
(
z
)
|
for all
w B
(
z
;
r
). Then
f
is constant. In other words,
a non-constant function cannot achieve an interior local maximum.
Proof. Let 0 < ρ < r. Applying the Cauchy integral formula, we get
|f(z)| =
1
2πi
Z
B(z;ρ)
f(w)
w z
dw
Setting w = z + ρe
2π
, we get
=
Z
1
0
f(z + ρe
2π
) dθ
sup
|zw|=ρ
|f(w)|
f(z).
So we must have equality throughout. When we proved the supremum bound
for the integral, we showed equality can happen only if the integrand is constant.
So
|f
(
w
)
|
is constant on the circle
|z w|
=
ρ
, and is equal to
f
(
z
). Since this
is true for all
ρ
(0
, r
), it follows that
|f|
is constant on
B
(
z
;
r
). Then the
Cauchy–Riemann equations then entail that
f
must be constant, as you have
shown in example sheet 1.
Going back to the Cauchy integral formula, recall that we had
B(z
0
; r) U
,
f : U C holomorphic, and we want to show
f(z) =
1
2πi
Z
B(z
0
;r)
f(w)
w z
dw.
When we proved it last time, we remember we know how to integrate things of
the form
1
(wz
0
)
n
, and manipulated the formula such that we get the integral is
made of things like this.
The second strategy is to change the contour of integration instead of changing
the integrand. If we can change it so that the integral is performed over a circle
around z instead of z
0
, then we know what to do.
z
0
z
Proof.
(of Cauchy integral formula again) Given
ε >
0, we pick
δ >
0 such that
B(z, δ) B
(
z
0
, r
), and such that whenever
|w z| < δ
, then
|f
(
w
)
f
(
z
)
| < ε
.
This is possible since
f
is uniformly continuous on the neighbourhood of
z
. We
now cut our region apart:
z
0
z
z
0
z
We know
f(w)
wz
is holomorphic on sufficiently small open neighbourhoods of
the half-contours indicated. The area enclosed by the contours might not be
star-shaped, but we can definitely divide it once more so that it is. Hence the
integral of
f(w)
wz
around the half-contour vanishes by Cauchy’s theorem. Adding
these together, we get
Z
B(z
0
,r)
f(w)
w z
dw =
Z
B(z,δ)
f(w)
w z
dw,
where the balls are both oriented anticlockwise. Now we have
f(z)
1
2πi
Z
B(z
0
,r)
f(w)
w z
dw
=
f(z)
1
2πi
Z
B(z,δ)
f(w)
w z
dw
.
Now we once again use the fact that
Z
B(z,δ)
1
w z
dz = 2πi
to show this is equal to
1
2πi
Z
B(z,δ)
f(z) f(w)
w z
dw
1
2π
· 2πδ ·
1
δ
· ε = ε.
Taking ε 0, we see that the Cauchy integral formula holds.
Note that the subdivision we did above was something we can do in general.
Definition
(Elementary deformation)
.
Given a pair of
C
1
-smooth (or piecewise
smooth) closed paths
φ, ψ
: [0
,
1]
U
, we say
ψ
is an elementary deformation of
φ
if there exists convex open sets
C
1
, ··· , C
n
U
and a division of the interval
0 =
x
0
< x
1
< ··· < x
n
= 1 such that on [
x
i1
, x
i
], both
φ
(
t
) and
ψ
(
t
) belong
to C
i
.
φ
ψ
φ(x
i1
)
φ(x
i
)
φ(x
i1
)
φ(x
i
)
Then there are straight lines
γ
i
:
φ
(
x
i
)
ψ
(
x
i
) lying inside
C
i
. If
f
is holomor-
phic on U, considering the shaded square, we find
Z
φ
f(z) dz =
Z
ψ
f(z) dz
when φ and ψ are convex deformations.
We now explore some classical consequences of the Cauchy Integral formula.
The next is Liouville’s theorem, as promised.
Theorem
(Liouville’s theorem)
.
Let
f
:
C C
be an entire function (i.e.
holomorphic everywhere). If f is bounded, then f is constant.
This, for example, means there is no interesting holomorphic period functions
like sin and cos that are bounded everywhere.
Proof.
Suppose
|f
(
z
)
| M
for all
z C
. We fix
z
1
, z
2
C
, and estimate
|f(z
1
) f(z
2
)| with the integral formula.
Let R > max{2|z
1
|, 2|z
2
|}. By the integral formula, we know
|f(z
1
) f(z
2
)| =
1
2πi
Z
B(0,R)
f(w)
w z
1
f(w)
w z
2
dw
=
1
2πi
Z
B(0,R)
f(w)(z
1
z
2
)
(w z
1
)(w z
2
)
dw
1
2π
· 2πR ·
M|z
1
z
2
|
(R/2)
2
=
4M|z
1
z
2
|
R
.
Note that we get the bound on the denominator since
|w|
=
R
implies
|wz
i
| >
R
2
by our choice of
R
. Letting
R
, we know we must have
f
(
z
1
) =
f
(
z
2
). So
f
is constant.
Corollary
(Fundamental theorem of algebra)
.
A non-constant complex polyno-
mial has a root in C.
Proof. Let
P (z) = a
n
z
n
+ a
n1
z
n1
+ ··· + a
0
,
where
a
n
6
= 0 and
n >
0. So
P
is non-constant. Thus, as
|z|
,
|P
(
z
)
|
.
In particular, there is some R such that for |z| > R, we have |P (z)| 1.
Now suppose for contradiction that
P
does not have a root in
C
. Then
consider
f(z) =
1
P (z)
,
which is then an entire function, since it is a rational function. On
B(0, R)
, we
know
f
is certainly continuous, and hence bounded. Outside this ball, we get
|f
(
z
)
|
1. So
f
(
z
) is constant, by Liouville’s theorem. But
P
is non-constant.
This is absurd. Hence the result follows.
There are many many ways we can prove the fundamental theorem of algebra.
However, none of them belong wholely to algebra. They all involve some analysis
or topology, as you might encounter in the IID Algebraic Topology and IID
Riemann Surface courses.
This is not surprising since the construction of
R
, and hence
C
, is intrinsically
analytic we get from
N
to
Z
by requiring it to have additive inverses;
Z
to
Q
by requiring multiplicative inverses;
R
to
C
by requiring the root to
x
2
+ 1 = 0.
These are all algebraic. However, to get from
Q
to
R
, we are requiring something
about convergence in
Q
. This is not algebraic. It requires a particular of metric
on Q. If we pick a different metric, then you get a different completion, as you
may have seen in IB Metric and Topological Spaces. Hence the construction of
R is actually analytic, and not purely algebraic.
2.4 Taylor’s theorem
When we first met Taylor series, we were happy, since we can express anything
as a power series. However, we soon realized this is just a fantasy the Taylor
series of a real function need not be equal to the function itself. For example,
the function
f
(
x
) =
e
x
2
has vanishing Taylor series at 0, but does not vanish
in any neighbourhood of 0. What we do have is Taylor’s theorem, which gives
you an expression for what the remainder is if we truncate our series, but is
otherwise completely useless.
In the world of complex analysis, we are happy once again. Every holomorphic
function can be given by its Taylor series.
Theorem
(Taylor’s theorem)
.
Let
f
:
B
(
a, r
)
C
be holomorphic. Then
f
has
a convergent power series representation
f(z) =
X
n=0
c
n
(z a)
n
on B(a, r). Moreover,
c
n
=
f
(n)
(a)
n!
=
1
2πi
Z
B(a,ρ)
f(z)
(z a)
n+1
dz
for any 0 < ρ < r.
Note that the very statement of the theorem already implies any holomorphic
function has to be infinitely differentiable. This is a good world.
Proof. We’ll use Cauchy’s integral formula. If |w a| < ρ < r, then
f(w) =
1
2πi
Z
B(a,ρ)
f(z)
z w
dz.
Now (cf. the first proof of the Cauchy integral formula), we note that
1
z w
=
1
(z a)
1
wa
za
=
n
X
n=0
(w a)
n
(z a)
n+1
.
This series is uniformly convergent everywhere on the
ρ
disk, including its
boundary. By uniform convergence, we can exchange integration and summation
to get
f(w) =
X
n=0
1
2πi
Z
B(a,ρ)
f(z)
(z a)
n+1
dz
!
(w a)
n
=
X
n=0
c
n
(w a)
n
.
Since
c
n
does not depend on
w
, this is a genuine power series representation,
and this is valid on any disk B(a, ρ) B(a, r).
Then the formula for
c
n
in terms of the derivative comes for free since that’s
the formula for the derivative of a power series.
This tells us every holomorphic function behaves like a power series. In
particular, we do not get weird things like
e
x
2
on
R
that have a trivial Taylor
series expansion, but is itself non-trivial. Similarly, we know that there are no
“bump functions” on
C
that are non-zero only on a compact set (since power
series don’t behave like that). Of course, we already knew that from Liouville’s
theorem.
Corollary.
If
f
:
B
(
a, r
)
C
is holomorphic on a disc, then
f
is infinitely
differentiable on the disc.
Proof.
Complex power series are infinitely differentiable (and
f
had better be
infinitely differentiable for us to write down the formula for
c
n
in terms of
f
(n)
).
This justifies our claim from the very beginning that
Re
(
f
) and
Im
(
f
) are
harmonic functions if f is holomorphic.
Corollary.
If
f
:
U C
is a complex-valued function, then
f
=
u
+
iv
is
holomorphic at
p U
if and only if
u, v
satisfy the Cauchy–Riemann equations,
and that u
x
, u
y
, v
x
, v
y
are continuous in a neighbourhood of p.
Proof.
If
u
x
, u
y
, v
x
, v
y
exist and are continuous in an open neighbourhood of
p
,
then
u
and
v
are differentiable as functions
R
2
R
2
at
p
, and then we proved
that the Cauchy–Riemann equations imply differentiability at each point in the
neighbourhood of p. So f is differentiable at a neighbourhood of p.
On the other hand, if
f
is holomorphic, then it is infinitely differentiable. In
particular,
f
0
(
z
) is also holomorphic. So
u
x
, u
y
, v
x
, v
y
are differentiable, hence
continuous.
We also get the following (partial) converse to Cauchy’s theorem.
Corollary
(Morera’s theorem)
.
Let
U C
be a domain. Let
f
:
U C
be
continuous such that
Z
γ
f(z) dz = 0
for all piecewise-C
1
closed curves γ U. Then f is holomorphic on U.
Proof.
We have previously shown that the condition implies that
f
has an
antiderivative
F
:
U C
, i.e.
F
is a holomorphic function such that
F
0
=
f
.
But F is infinitely differentiable. So f must be holomorphic.
Recall that Cauchy’s theorem required
U
to be sufficiently nice, e.g. being
star-shaped or just simply-connected. However, Morera’s theorem does not. It
just requires that
U
is a domain. This is since holomorphicity is a local property,
while vanishing on closed curves is a global result. Cauchy’s theorem gets us
from a local property to a global property, and hence we need to assume more
about what the “globe” looks like. On the other hand, passing from a global
property to a local one does not. Hence we have this asymmetry.
Corollary.
Let
U C
be a domain,
f
n
;
U C
be a holomorphic function. If
f
n
f uniformly, then f is in fact holomorphic, and
f
0
(z) = lim
n
f
0
n
(z).
Proof. Given a piecewise C
1
path γ, uniformity of convergence says
Z
γ
f
n
(z) dz
Z
γ
f(z) dz
uniformly. Since
f
being holomorphic is a local condition, so we fix
p U
and
work in some small, convex disc
B
(
p, ε
)
U
. Then for any curve
γ
inside this
disk, we have
Z
γ
f
n
(z) dz = 0.
Hence we also have
R
γ
f
(
z
) d
z
= 0. Since this is true for all curves, we conclude
f
is holomorphic inside
B
(
p, ε
) by Morera’s theorem. Since
p
was arbitrary, we
know f is holomorphic.
We know the derivative of the limit is the limit of the derivative since we can
express f
0
(a) in terms of the integral of
f(z)
(za)
2
, as in Taylor’s theorem.
There is a lot of passing between knowledge of integrals and knowledge of
holomorphicity all the time, as we can see in these few results. These few sections
are in some sense the heart of the course, where we start from Cauchy’s theorem
and Cauchy’s integral formula, and derive all the other amazing consequences.
2.5 Zeroes
Recall that for a polynomial
p
(
z
), we can talk about the order of its zero at
z
=
a
by looking at the largest power of (
z a
) dividing
p
. A priori, it is not
clear how we can do this for general functions. However, given that everything
is a Taylor series, we know how to do this for holomorphic functions.
Definition
(Order of zero)
.
Let
f
:
B
(
a, r
)
C
be holomorphic. Then we
know we can write
f(z) =
X
n=0
c
n
(z a)
n
as a convergent power series. Then either all
c
n
= 0, in which case
f
= 0 on
B
(
a, r
), or there is a least
N
such that
c
N
6
= 0 (
N
is just the smallest
n
such
that f
(n)
(a) 6= 0).
If N > 0, then we say f has a zero of order N.
If f has a zero of order N at a, then we can write
f(z) = (z a)
N
g(z)
on B(a, r), where g(a) = c
N
6= 0.
Often, it is not the actual order that is too important. Instead, it is the
ability to factor f in this way. One of the applications is the following:
Lemma
(Principle of isolated zeroes)
.
Let
f
:
B
(
a, r
)
C
be holomorphic and
not identically zero. Then there exists some 0
< ρ < r
such that
f
(
z
)
6
= 0 in the
punctured neighbourhood B(a, ρ) \ {a}.
Proof. If f (a) 6= 0, then the result is obvious by continuity of f.
The other option is not too different. If
f
has a zero of order
N
at
a
, then
we can write
f
(
z
) = (
z a
)
N
g
(
z
) with
g
(
a
)
6
= 0. By continuity of
g
,
g
does
not vanish on some small neighbourhood of
a
, say
B
(
a, ρ
). Then
f
(
z
) does not
vanish on B(a, ρ) \ {a}.
A consequence is that given two holomorphic functions on the same domain,
if they agree on sufficiently many points, then they must in fact be equal.
Corollary
(Identity theorem)
.
Let
U C
be a domain, and
f, g
:
U C
be
holomorphic. Let
S
=
{z U
:
f
(
z
) =
g
(
z
)
}
. Suppose
S
contains a non-isolated
point, i.e. there exists some
w S
such that for all
ε >
0,
S B
(
w, ε
)
6
=
{w}
.
Then f = g on U .
Proof.
Consider the function
h
(
z
) =
f
(
z
)
g
(
z
). Then the hypothesis says
h
(
z
)
has a non-isolated zero at
w
, i.e. there is no non-punctured neighbourhood of
w
on which
h
is non-zero. By the previous lemma, this means there is some
ρ >
0
such that h = 0 on B(w, ρ) U.
Now we do some topological trickery. We let
U
0
= {a U : h = 0 on some neighbourhood B(a, ρ) of a in U},
U
1
= {a U : there exists n 0 such that h
(n)
6= 0}.
Clearly,
U
0
U
1
=
, and the existence of Taylor expansions shows
U
0
U
1
=
U
.
Moreover,
U
0
is open by definition, and
U
1
is open since
h
(n)
(
z
) is continuous
near any given
a U
1
. Since
U
is (path) connected, such a decomposition can
happen if one of
U
0
and
U
1
is empty. But
w U
0
. So in fact
U
0
=
U
, i.e.
h
vanishes on the whole of U. So f = g.
In particular, if two holomorphic functions agree on some small open subset
of the domain, then they must in fact be identical. This is a very strong result,
and is very false for real functions. Hence, to specify, say, an entire function, all
we need to do is to specify it on an arbitrarily small domain we like.
Definition
(Analytic continuiation)
.
Let
U
0
U C
be domains, and
f
:
U
0
C
be holomorphic. An analytic continuation of
f
is a holomorphic function
h : U C such that h|
U
0
= f, i.e. h(z) = f(z) for all z U
0
.
By the identity theorem, we know the analytic continuation is unique if it
exists.
Thus, given any holomorphic function
f
:
U C
, it is natural to ask how
far we can extend the domain, i.e. what is the largest
U
0
U
such that there is
an analytic continuation of f to U
0
.
There is no general method that does this for us. However, one useful trick
is to try to write our function
f
in a different way so that it is clear how we can
extend it to elsewhere.
Example. Consider the function
f(z) =
X
n0
z
n
= 1 + z + z
2
+ ···
defined on B(0, 1).
By itself, this series diverges for
z
outside
B
(0
,
1). However, we know well
that this function is just
f(z) =
1
1 z
.
This alternative representation makes sense on the whole of
C
except at
z
= 1.
So we see that
f
has an analytic continuation to
C \ {
1
}
. There is clearly no
extension to the whole of C, since it blows up near z = 1.
Example. Alternatively, consider
f(z) =
X
n0
z
2
n
.
Then this again converges on
B
(0
,
1). You will show in example sheet 2 that
there is no analytic continuation of f to any larger domain.
Example. The Riemann zeta function
ζ(z) =
X
n=1
n
z
defines a holomorphic function on
{z
:
Re
(
z
)
>
1
} C
. Indeed, we have
|n
z
|
=
|n
Re(z)
|
, and we know
P
n
t
converges for
t R
>1
, and in fact does so
uniformly on any compact domain. So the corollary of Morera’s theorem tells us
that ζ(z) is holomorphic on Re(z) > 1.
We know this cannot converge as
z
1, since we approach the harmonic
series which diverges. However, it turns out
ζ
(
z
) has an analytic continuation to
C \ {1}. We will not prove this.
At least formally, using the fundamental theorem of arithmetic, we can
expand n as a product of its prime factors, and write
ζ(z) =
Y
primes p
(1 + p
z
+ p
2z
+ ···) =
Y
primes p
1
1 p
z
.
If there were finitely many primes, then this would be a well-defined function
on all of
C
, since this is a finite product. Hence, the fact that this blows up at
z = 1 implies that there are infinitely many primes.
2.6 Singularities
The next thing to study is singularities of holomorphic functions. These are
places where the function is not defined. There are many ways a function can
be ill-defined. For example, if we write
f(z) =
1 z
1 z
,
then on the face of it, this function is not defined at
z
= 1. However, elsewhere,
f
is just the constant function 1, and we might as well define
f
(1) = 1. Then we
get a holomorphic function. These are rather silly singularities, and are singular
solely because we were not bothered to define f there.
Some singularities are more interesting, in that they are genuinely singular.
For example, the function
f(z) =
1
1 z
is actually singular at
z
= 1, since
f
is unbounded near the point. It turns out
these are the only possibilities.
Proposition
(Removal of singularities)
.
Let
U
be a domain and
z
0
U
. If
f
:
U \{z
0
} C
is holomorphic, and
f
is bounded near
z
0
, then there exists an
a such that f(z) a as z z
0
.
Furthermore, if we define
g(z) =
(
f(z) z U \{z
0
}
a z = z
0
,
then g is holomorphic on U.
Proof. Define a new function h : U C by
h(z) =
(
(z z
0
)
2
f(z) z 6= z
0
0 z = z
0
.
Then since
f
is holomorphic away from
z
0
, we know
h
is also holomorphic away
from z
0
.
Also, we know
f
is bounded near
z
0
. So suppose
|f
(
z
)
| < M
in some
neighbourhood of z
0
. Then we have
h(z) h(z
0
)
z z
0
|z z
0
|M.
So in fact
h
is also differentiable at
z
0
, and
h
(
z
0
) =
h
0
(
z
0
) = 0. So near
z
0
,
h
has a Taylor series
h(z) =
X
n0
a
n
(z z
0
)
n
.
Since we are told that a
0
= a
1
= 0, we can define a g(z) by
g(z) =
X
n0
a
n+2
(z z
0
)
n
,
defined on some ball
B
(
z
0
, ρ
), where the Taylor series for
h
is defined. By
construction, on the punctured ball
B
(
z
0
, ρ
)
\{z
0
}
, we get
g
(
z
) =
f
(
z
). Moreover,
g(z) a
2
as z z
0
. So f(z) a
2
as z z
0
.
Since g is a power series, it is holomorphic. So the result follows.
This tells us the only way for a function to fail to be holomorphic at an
isolated point is that it blows up near the point. This won’t happen because
f
fails to be continuous in some weird ways.
However, we are not yet done with our classification. There are many ways
in which things can blow up. We can further classify these into two cases the
case where
|f
(
z
)
|
as
z z
0
, and the case where
|f
(
z
)
|
does not converge as
z z
0
. It happens that the first case is almost just as boring as the removable
ones.
Proposition.
Let
U
be a domain,
z
0
U
and
f
:
U \{z
0
} C
be holomorphic.
Suppose
|f
(
z
)
|
as
z z
0
. Then there is a unique
k Z
1
and a unique
holomorphic function g : U C such that g(z
0
) 6= 0, and
f(z) =
g(z)
(z z
0
)
k
.
Proof.
We shall construct
g
near
z
0
in some small neighbourhood, and then
apply analytic continuation to the whole of
U
. The idea is that since
f
(
z
) blows
up nicely as z z
0
, we know
1
f(z)
behaves sensibly near z
0
.
We pick some
δ >
0 such that
|f
(
z
)
|
1 for all
z B
(
z
0
;
δ
)
\ {z
0
}
. In
particular, f (z) is non-zero on B(z
0
; δ) \ {z
0
}. So we can define
h(z) =
(
1
f(z)
z B(z
0
; δ) \ {z
0
}
0 z = z
0
.
Since
|
1
f(z)
|
1 on
B
(
z
0
;
δ
)
\{z
0
}
, by the removal of singularities,
h
is holomorphic
on
B
(
z
0
, δ
). Since
h
vanishes at the
z
0
, it has a unique definite order at
z
0
, i.e.
there is a unique integer
k
1 such that
h
has a zero of order
k
at
z
0
. In other
words,
h(z) = (z z
0
)
k
`(z),
for some holomorphic ` : B(z
0
; δ) C and `(z
0
) 6= 0.
Now by continuity of
`
, there is some 0
< ε < δ
such that
`
(
z
)
6
= 0 for all
z B(z
0
, ε). Now define g : B(z
0
; ε) C by
g(z) =
1
`(z)
.
Then g is holomorphic on this disc.
By construction, at least away from z
0
, we have
g(z) =
1
`(z)
=
1
h(z)
· (z z
0
)
k
= (z z
0
)
k
f(z).
g
was initially defined on
B
(
z
0
;
ε
)
C
, but now this expression certainly makes
sense on all of
U
. So
g
admits an analytic continuation from
B
(
z
0
;
ε
) to
U
. So
done.
We can start giving these singularities different names. We start by formally
defining what it means to be a singularity.
Definition
(Isolated singularity)
.
Given a domain
U
and
z
0
U
, and
f
:
U \ {z
0
} C holomorphic, we say z
0
is an isolated singularity of f .
Definition
(Removable singularity)
.
A singularity
z
0
of
f
is a removable singu-
larity if f is bounded near z
0
.
Definition
(Pole)
.
A singularity
z
0
is a pole of order
k
of
f
if
|f
(
z
)
|
as
z z
0
and one can write
f(z) =
g(z)
(z z
0
)
k
with g : U C, g(z
0
) 6= 0.
Definition
(Isolated essential singularity)
.
An isolated singularity is an isolated
essential singularity if it is neither removable nor a pole.
It is easy to give examples of removable singularities and poles. So let’s look
at some essential singularities.
Example. z 7→ e
1/z
has an isolated essential singularity at z = 0.
Note that if
B
(
z
0
, ε
)
\{z
0
} C
has a pole of order
h
at
z
0
, then
f
naturally
defines a map
ˆ
f : B(z
0
; ε) CP
1
= C {∞}, the Riemann sphere, by
f(z) =
(
z = z
0
f(z) z 6= z
0
.
This is then a “continuous” function. So a singularity is just a point that gets
mapped to the point .
As was emphasized in IA Groups, the point at infinity is not a special point
in the Riemann sphere. Similarly, poles are also not really singularities from the
viewpoint of the Riemann sphere. It’s just that we are looking at it in a wrong
way. Indeed, if we change coordinates on the Riemann sphere so that we label
each point
w CP
1
by
w
0
=
1
w
instead, then
f
just maps
z
0
to 0 under the new
coordinate system. In particular, at the point
z
0
, we find that
f
is holomorphic
and has an innocent zero of order k.
Since poles are not bad, we might as well allow them.
Definition
(Meromorphic function)
.
If
U
is a domain and
S U
is a finite or
discrete set, a function
f
:
U \ S C
which is holomorphic and has (at worst)
poles on S is said to be meromorphic on U.
The requirement that
S
is discrete is so that each pole in
S
is actually an
isolated singularity.
Example.
A rational function
P (z)
Q(z)
, where
P, Q
are polynomials, is holomorphic
on
C \ {z
:
Q
(
z
) = 0
}
, and meromorphic on
C
. More is true it is in fact
holomorphic as a function CP
1
CP
1
.
These ideas are developed more in depth in the IID Riemann Surfaces course.
As an aside, if we want to get an interesting holomorphic function with
domain
CP
1
, its image must contain the point
, or else its image will be
a compact subset of
C
(since
CP
1
is compact), thus bounded, and therefore
constant by Liouville’s theorem.
At this point, we really should give essential singularities their fair share of
attention. Not only are they bad. They are bad spectacularly.
Theorem
(Casorati-Weierstrass theorem)
.
Let
U
be a domain,
z
0
U
, and
suppose
f
:
U \ {z
0
} C
has an essential singularity at
z
0
. Then for all
w C
,
there is a sequence z
n
z
0
such that f (z
n
) w.
In other words, on any punctured neighbourhood
B
(
z
0
;
ε
)
\ {z
0
}
, the image
of f is dense in C.
This is not actually too hard to proof.
Proof. See example sheet 2.
If you think that was bad, actually essential singularities are worse than that.
The theorem only tells us the image is dense, but not that we will hit every
point. It is in fact not true that every point will get hit. For example
e
1
z
can
never be zero. However, this is the worst we can get
Theorem
(Picard’s theorem)
.
If
f
has an isolated essential singularity at
z
0
, then
there is some
b C
such that on each punctured neighbourhood
B
(
z
0
;
ε
)
\ {z
0
}
,
the image of f contains C \ {b}.
The proof is beyond this course.
2.7 Laurent series
If f is holomorphic at z
0
, then we have a local power series expansion
f(z) =
X
n=0
c
n
(z z
0
)
n
near
z
0
. If
f
is singular at
z
0
(and the singularity is not removable), then there
is no hope we can get a Taylor series, since the existence of a Taylor series would
imply f is holomorphic at z = z
0
.
However, it turns out we can get a series expansion if we allow ourselves to
have negative powers of z.
Theorem (Laurent series). Let 0 r < R < , and let
A = {z C : r < |z a| < R}
denote an annulus on C.
Suppose
f
:
A C
is holomorphic. Then
f
has a (unique) convergent series
expansion
f(z) =
X
n=−∞
c
n
(z a)
n
,
where
c
n
=
1
2πi
Z
B(a,ρ)
f(z)
(z a)
n+1
dz
for
r < ρ < R
. Moreover, the series converges uniformly on compact subsets of
the annulus.
The Laurent series provides another way of classifying singularities. In the
case where r = 0, we just have
f(z) =
X
−∞
c
n
(z a)
n
on B(a, R) \ {a}, then we have the following possible scenarios:
(i) c
n
= 0 for all
n <
0. Then
f
is bounded near
a
, and hence this is a
removable singularity.
(ii)
Only finitely many negative coefficients are non-zero, i.e. there is a
k
1
such that
c
n
= 0 for all
n < k
and
c
k
6
= 0. Then
f
has a pole of order
k at a.
(iii)
There are infinitely many non-zero negative coefficients. Then we have an
isolated essential singularity.
So our classification of singularities fit nicely with the Laurent series expansion.
We can interpret the Laurent series as follows we can write
f(z) = f
in
(z) + f
out
(z),
where
f
in
consists of the terms with positive power and
f
out
consists of those with
negative power. Then
f
in
is the part that is holomorphic on the disk
|z a| < R
,
while
f
out
(
z
) is the part that is holomorphic on
|z a| > r
. These two combine
to give an expression holomorphic on
r < |z a| < R
. This is just a nice way
of thinking about it, and we will not use this anywhere. So we will not give a
detailed proof of this interpretation.
Proof.
The proof looks very much like the blend of the two proofs we’ve given for
the Cauchy integral formula. In one of them, we took a power series expansion
of the integrand, and in the second, we changed our contour by cutting it up.
This is like a mix of the two.
Let w A. We let r < ρ
0
< |w a| < ρ
00
< R.
a
˜γ
˜
˜γ
w
ρ
0
ρ
00
We let ˜γ be the contour containing w, and
˜
˜γ be the other contour.
Now we apply the Cauchy integral formula to say
f(w) =
1
2πi
Z
˜γ
f(z)
z w
dz
and
0 =
1
2πi
Z
˜
˜γ
f(z)
z w
dz.
So we get
f(w) =
1
2πi
Z
B(a,ρ
00
)
f(z)
z w
dz
1
2πi
Z
B(a,ρ
0
)
f(z)
z w
dz.
As in the first proof of the Cauchy integral formula, we make the following
expansions: for the first integral, we have w a < z a. So
1
z w
=
1
z a
1
1
wa
za
!
=
X
n=0
(w a)
n
(z a)
n+1
,
which is uniformly convergent on z B(a, ρ
00
).
For the second integral, we have w a > z a. So
1
z w
=
1
w a
1
1
za
wa
!
=
X
m=1
(z a)
m1
(w a)
m
,
which is uniformly convergent for z B(a, ρ
0
).
By uniform convergence, we can swap summation and integration. So we get
f(w) =
X
n=0
1
2πi
Z
B(a,ρ
00
)
f(z)
(z a)
n+1
dz
!
(w a)
n
+
X
m=1
1
2πi
Z
B(a,ρ
0
)
f(z)
(z a)
m+1
dz
!
(w a)
m
.
Now we substitute n = m in the second sum, and get
f(w) =
X
n=−∞
˜c
n
(w a)
n
,
for the integrals
˜c
n
. However, some of the coefficients are integrals around the
ρ
00
circle, while the others are around the
ρ
0
circle. This is not a problem. For
any
r < ρ < R
, these circles are convex deformations of
|z a|
=
ρ
inside the
annulus A. So
Z
B(a,ρ)
f(z)
(z a)
n+1
dz
is independent of ρ as long as ρ (r, R). So we get the result stated.
Definition
(Principal part)
.
If
f
:
B
(
a, r
)
\ {a} C
is holomorphic and if
f
has Laurent series
f(z) =
X
n=−∞
c
n
(z a)
n
,
then the principal part of f at a is
f
principal
=
1
X
n=−∞
c
n
(z a)
n
.
So
f f
principal
is holomorphic near
a
, and
f
principal
carries the information
of what kind of singularity f has at a.
When we talked about Taylor series, if
f
:
B
(
a, r
)
C
is holomorphic
with Taylor series
f
(
z
) =
P
n=0
c
n
(
z a
)
n
, then we had two possible ways of
expressing the coefficients of c
n
. We had
c
n
=
1
2πi
Z
B(a,ρ)
f(z)
(z a)
n+1
dz =
f
(n)
(a)
n!
.
In particular, the second expansion makes it obvious the Taylor series is uniquely
determined by f .
For the Laurent series, we cannot expect to have a simple expression of the
coefficients in terms of the derivatives of the function, for the very reason that
f
is not even defined, let alone differentiable, at
a
. So is the Laurent series unique?
Lemma. Let f : A C be holomorphic, A = {r < |z a| < R}, with
f(z) =
X
n=−∞
c
n
(z a)
n
Then the coefficients c
n
are uniquely determined by f .
Proof. Suppose also that
f(z) =
X
n=−∞
b
n
(z a)
n
.
Using our formula for c
k
, we know
2πic
k
=
Z
B(a,ρ)
f(z)
(z a)
k+1
dz
=
Z
B(a,ρ)
X
n
b
n
(z a)
nk1
!
dz
=
X
n
b
n
Z
B(a,ρ)
(z a)
nk1
dz
= 2πib
k
.
So c
k
= b
k
.
While we do have uniqueness, we still don’t know how to find a Laurent
series. For a Taylor series, we can just keep differentiating and then get the
coefficients. For Laurent series, the above integral is often almost impossible to
evaluate. So the technique to compute a Laurent series is blind guesswork.
Example. We know
sin z = z
z
3
3!
+
z
5
5!
···
defines a holomorphic function, with a radius of convergence of
. Now consider
cosec z =
1
sin z
,
which is holomorphic except for
z
=
kπ
, with
k Z
. So
cosec z
has a Laurent
series near z = 0. Using
sin z = z
1
z
2
6
+ O(z
4
)
,
we get
cosec z =
1
z
1 +
z
2
6
+ O(z
4
)
.
From this, we can read off that the Laurent series has
c
n
= 0 for all
n
2,
c
1
= 1,
c
1
=
1
5
. If we want, we can go further, but we already see that
cosec
has a simple pole at z = 0.
By periodicity, cosec has a simple pole at all other singularities.
Example. Consider instead
sin
1
z
=
1
z
1
3!z
3
+
1
5!z
5
··· .
We see this is holomorphic on
C
, with
c
n
6
= 0 for infinitely many
n <
0. So this
has an isolated essential singularity.
Example.
Consider
cosec
1
z
. This has singularities at
z
=
1
for
k N
=
{
1
,
2
,
3
, ···}
. So it is not holomorphic at any punctured neighbourhood
B
(0
, r
)
\
{
0
}
of zero. So this has a non-isolated singularity at zero, and there is no Laurent
series in a neighbourhood of zero.
We’ve already done most of the theory. In the remaining of the course, we
will use these techniques to do stuff. We will spend most of our time trying to
evaluate integrals, but before that, we will have a quick look on how we can use
Laurent series to evalue some series.
Example (Series summation). We claim that
f(z) =
X
n=−∞
1
(z n)
2
is holomorphic on C \ Z, and moreover if we let
f(z) =
π
2
sin
2
(πz)
,
We will reserve the name
f
for the original series, and refer to the function
z 7→
π
2
sin
2
(πz)
as g instead, until we have proven that they are the same.
Our strategy is as follows we first show that
f
(
z
) converges and is holo-
morphic, which is not hard, given the Weierstrass
M
-test and Morera’s theorem.
To show that indeed we have
f
(
z
) =
g
(
z
), we first show that they have equal
principal part, so that
f
(
z
)
g
(
z
) is entire. We then show it is zero by proving
f g
is bounded, hence constant, and that
f
(
z
)
g
(
z
)
0 as
z
(in some
appropriate direction).
For any fixed
w C \ Z
, we can compare it with
P
1
n
2
and apply the
Weierstrass
M
-test. We pick
r >
0 such that
|w n| >
2
r
for all
n Z
. Then
for all z B(w; r), we have
|z n| max{r, n |w| r}.
Hence
1
|z n|
2
min
1
r
2
,
1
(n |w| r)
2
= M
n
.
By comparison to
P
1
n
2
, we know
P
n
M
n
converges. So by the Weierstrass
M-test, we know our series converges uniformly on B(w, r).
By our results around Morera’s theorem, we see that
f
is a uniform limit of
holomorphic functions
P
N
n=N
1
(zn)
2
, and hence holomorphic.
Since
w
was arbitrary, we know
f
is holomorphic on
C \ Z
. Note that we
do not say the sum converges uniformly on
C \ Z
. It’s just that for any point
w C \ Z
, there is a small neighbourhood of
w
on which the sum is uniformly
convergent, and this is sufficient to apply the result of Morera’s.
For the second part, note that
f
is periodic, since
f
(
z
+ 1) =
f
(
z
). Also, at
0,
f
has a double pole, since
f
(
z
) =
1
z
2
+ holomorphic stuff near
z
= 0. So
f
has
a double pole at each
k Z
. Note that
1
sin
2
(πz)
also has a double pole at each
k Z.
Now, consider the principal parts of our functions at
k Z
,
f
(
z
) has
principal part
1
(zk)
2
. Looking at our previous Laurent series for cosec(z), if
g(z) =
π
sin πz
2
,
then
lim
z0
z
2
g
(
z
) = 1. So
g
(
z
) must have the same principal part at 0 and
hence at k for all k Z.
Thus
h
(
z
) =
f
(
z
)
g
(
z
) is holomorphic on
C \Z
. However, since its principal
part vanishes at the integers, it has at worst a removable singularity. Removing
the singularity, we know h(z) is entire.
Since we want to prove f(z) = g(z), we need to show h(z) = 0.
We first show it is boundedness. We know
f
and
g
are both periodic with
period 1. So it suffices to focus attention on the strip
1
2
x = Re(z)
1
2
.
To show this is bounded on the rectangle, it suffices to show that
h
(
x
+
iy
)
0
as
y ±∞
, by continuity. To do so, we show that
f
and
g
both vanish as
y .
So we set z = x + iy, with |x|
1
2
. Then we have
|g(z)|
4π
2
|e
πy
e
πy
|
0
as y . Exactly analogously,
|f(z)|
X
nZ
1
|x + iy n|
2
1
y
2
+ 2
X
n=1
1
(n
1
2
)
2
+ y
2
0
as
y
. So
h
is bounded on the strip, and tends to 0 as
y
, and is hence
constant by Liouville’s theorem. But if
h
0 as
y
, then the constant
better be zero. So we get
h(z) = 0.
3 Residue calculus
3.1 Winding numbers
Recall that the type of the singularity of a point depends on the coefficients in
the Laurent series, and these coefficients play an important role in determining
the behaviour of the functions. Among all the infinitely many coefficients, it
turns out the coefficient of
z
1
is the most important one, as we will soon see.
We call this the residue of f .
Definition
(Residue)
.
Let
f
:
B
(
a, r
)
\ {a} C
be holomorphic, with Laurent
series
f(z) =
X
n=−∞
c
n
(z a)
n
.
Then the residue of f at a is
Res(f, a) = Res
f
(a) = c
1
.
Note that if ρ < r, then by definition of the Laurent coefficients, we know
Z
B(a,ρ)
f(z) dz = 2πic
1
.
So we can alternatively write the residue as
Res
f
(a) =
1
2πi
Z
B(a,ρ)
f(z) dz.
This gives us a formulation of the residue without reference to the Laurent series.
Deforming paths if necessary, it is not too far-fetching to imagine that for
any simple curve γ around the singularity a, we have
Z
γ
f(z) dz = 2πi Res(f, a).
Moreover, if the path actually encircles two singularities
a
and
b
, then deforming
the path, we would expect to have
Z
γ
f(z) dz = 2πi(Res(f, a) + Res(f, b)),
and this generalizes to multiple singularities in the obvious way.
If this were true, then it would be very helpful, since this turns integration
into addition, which is (hopefully) much easier!
Indeed, we will soon prove that this result holds. However, we first get rid of
the technical restriction that we only work with simple (i.e. non-self intersecting)
curves. This is completely not needed. We are actually not really worried in
the curve intersecting itself. The reason why we’ve always talked about simple
closed curves is that we want to avoid the curve going around the same point
many times.
There is a simple workaround to this problem we consider arbitrary curves,
and then count how many times we are looping around the point. If we are
looping around it twice, then we count its contribution twice!
Indeed, suppose we have the following curve around a singularity:
a
We see that the curve loops around
a
twice. Also, by the additivity of the
integral, we can break this curve into two closed contours. So we have
1
2πi
Z
γ
f(z) dz = 2 Res
f
(a).
So what we want to do now is to define properly what it means for a curve to
loop around a point n times. This will be called the winding number.
There are many ways we can define the winding number. The definition we
will pick is based on the following observation suppose, for convenience, that
the point in question is the origin. As we move along a simple closed curve around
0, our argument will change. If we keep track of our argument continuously,
then we will find that when we return to starting point, the argument would
have increased by 2
π
. If we have a curve that winds around the point twice,
then our argument will increase by 4π.
What we do is exactly the above given a path, find a continuous function
that gives the “argument” of the path, and then define the winding number to
be the difference between the argument at the start and end points, divided by
2π.
For this to make sense, there are two important things to prove. First, we
need to show that there is indeed a continuous “argument” function of the curve,
in a sense made precise in the lemma below. Then we need to show the winding
number is well-defined, but that is easier.
Lemma.
Let
γ
: [
a, b
]
C
be a continuous closed curve, and pick a point
w C \ image
(
γ
). Then there are continuous functions
r
: [
a, b
]
R >
0 and
θ : [a, b] R such that
γ(t) = w + r(t)e
(t)
.
Of course, at each point
t
, we can find
r
and
θ
such that the above holds.
The key point of the lemma is that we can do so continuously.
Proof.
Clearly
r
(
t
) =
|γ
(
t
)
w|
exists and is continuous, since it is the composi-
tion of continuous functions. Note that this is never zero since
γ
(
t
) is never
w
.
The actual content is in defining θ.
To define
θ
(
t
), we for simplicity assume
w
= 0. Furthermore, by considering
instead the function
γ(t)
r(t)
, which is continuous and well-defined since
r
is never
zero, we can assume |γ(t)| = 1 for all t.
Recall that the principal branch of
log
, and hence of the argument
Im
(
log
),
takes values in (π, π) and is defined on C \ R
0
.
If
γ
(
t
) always lied in, say, the right-hand half plane, we would have no problem
defining θ consistently, since we can just let
θ(t) = arg(γ(t))
for
arg
the principal branch. There is nothing special about the right-hand half
plane. Similarly, if γ lies in the region as shaded below:
α
i.e. we have
γ(t)
n
z : Re
z
e
> 0
o
for a fixed α, we can define
θ(t) = α + arg
γ(t)
e
.
Since γ : [a, b] C is continuous, it is uniformly continuous, and we can find a
subdivision
a = a
0
< a
1
< ··· < a
m
= b,
such that if
s, t
[
a
i1
, a
i
], then
|γ
(
s
)
γ
(
t
)
| <
2
, and hence
γ
(
s
) and
γ
(
t
)
belong to such a half-plane.
So we define θ
j
: [a
j1
, a
j
] R such that
γ(t) = e
j
(t)
for t [a
j1
, a
j
], and 1 j n 1.
On each region [
a
j1
, a
j
], this gives a continuous argument function. We
cannot immediately extend this to the whole of [
a, b
], since it is entirely possible
that
θ
j
(
a
j
) =
θ
j+1
(
a
j
). However, we do know that
θ
j
(
a
j
) are both values of the
argument of
γ
(
a
j
). So they must differ by an integer multiple of 2
π
, say 2
.
Then we can just replace
θ
j+1
by
θ
j+1
2
, which is an equally valid argument
function, and then the two functions will agree at a
j
.
Hence, for
j >
1, we can successively re-define
θ
j
such that the resulting map
θ is continuous. Then we are done.
We can thus use this to define the winding number.
Definition
(Winding number)
.
Given a continuous path
γ
: [
a, b
]
C
such
that γ(a) = γ(b) and w 6∈ image(γ), the winding number of γ about w is
θ(b) θ(a)
2π
,
where
θ
: [
a, b
]
R
is a continuous function as above. This is denoted by
I
(
γ, w
)
or n
γ
(W ).
I and n stand for index and number respectively.
Note that we always have
I
(
γ, w
)
Z
, since
θ
(
b
) and
θ
(
a
) are arguments
of the same number. More importantly,
I
(
γ, w
) is well-defined suppose
γ
(
t
) =
r
(
t
)
e
1
(t)
=
r
(
t
)
e
2
(t)
for continuous functions
θ
1
, θ
2
: [
a, b
]
R
. Then
θ
1
θ
2
: [
a, b
]
R
is continuous, but takes values in the discrete set 2
πZ
. So it
must in fact be constant, and thus θ
1
(b) θ
1
(a) = θ
2
(b) θ
2
(a).
So far, what we’ve done is something that is true for arbitrary continuous
closed curve. However, if we focus on piecewise
C
1
-smooth closed path, then we
get an alternative expression:
Lemma.
Suppose
γ
: [
a, b
]
C
is a piecewise
C
1
-smooth closed path, and
w 6∈ image(γ). Then
I(γ, w) =
1
2πi
Z
γ
1
z w
dz.
Proof. Let γ(t) w = r(t)e
(t)
, with now r and θ piecewise C
1
-smooth. Then
Z
γ
1
z w
dz =
Z
b
a
γ
0
(t)
γ(t) w
dt
=
Z
b
a
r
0
(t)
r(t)
+
0
(t)
dt
= [ln r(t) + (t)]
b
a
= i(θ(b) θ(a))
= 2πiI(γ, w).
So done.
In some books, this integral expression is taken as the definition of the
winding number. While this is elegant in complex analysis, it is not clear a priori
that this is an integer, and only works for piecewise
C
1
-smooth closed curves,
not arbitrary continuous closed curves.
On the other hand, what is evident from this expression is that
I
(
γ, w
) is
continuous as a function of
w C \ image
(
γ
), since it is even holomorphic as a
function of
w
. Since
I
(
γ
;
w
) is integer valued,
I
(
γ
) must be locally constant on
path components of C \ image(γ).
We can quickly verify that this is a sensible definition, in that the winding
number around a point “outside” the curve is zero. More precisely, since
image
(
γ
)
is compact, all points of sufficiently large modulus in
C
belong to one component
of
C \ image
(
γ
). This is indeed the only path component of
C \ image
(
γ
) that is
unbounded.
To find the winding number about a point in this unbounded component, note
that
I
(
γ
;
w
) is consistent on this component, and so we can consider arbitrarily
larger w. By the integral formula,
|I(γ, w)|
1
2π
length(γ) max
zγ
1
|w z|
0
as
w
. So it does vanish outside the curve. Of course, inside the other path
components, we can still have some interesting values of the winding number.
3.2 Homotopy of closed curves
The last ingredient we need before we can get to the residue theorem is the
idea of homotopy. Recall we had this weird, ugly definition of elementary
deformation of curves given
φ, ψ
: [
a, b
]
U
, which are closed, we say
ψ
is an
elementary deformation or convex deformation of
φ
if there exists a decomposition
a
=
x
0
< x
1
< ··· < x
n
=
b
and convex open sets
C
1
, ··· , C
n
U
such that for
x
i1
t x
i
, we have φ(t) and ψ(t) in C
i
.
It was a rather unnatural definition, since we have to make reference to
this arbitrarily constructed dissection of [
a, b
] and convex sets
C
i
. Moreover,
this definition fails to be transitive (e.g. on
R \ {
0
}
, rotating a circle about the
center by, say,
π
10
is elementary, but rotating by
π
is not). Yet, this definition
was cooked up just so that it immediately follows that elementary deformations
preserve integrals of holomorphic functions around the loop.
The idea now is to define a more general and natural notion of deforming a
curve, known as “homotopy”. We will then show that each homotopy can be
given by a sequence of elementary deformations. So homotopies also preserve
integrals of holomorphic functions.
Definition
(Homotopy of closed curves)
.
Let
U C
be a domain, and let
φ
: [
a, b
]
U
and
ψ
: [
a, b
]
U
be piecewise
C
1
-smooth closed paths. A
homotopy from φ : ψ is a continuous map F : [0, 1] × [a, b] U such that
F (0, t) = φ(t), F (1, t) = ψ(t),
and moreover, for all
s
[0
, t
], the map
t 7→ F
(
s, t
) viewed as a map [
a, b
]
U
is closed and piecewise C
1
-smooth.
We can imagine this as a process of “continuously deforming” the path
φ
to
ψ, with a path F (s, ·) at each point in time s [0, 1].
Proposition.
Let
φ, ψ
: [
a, b
]
U
be homotopic (piecewise
C
1
) closed paths in
a domain
U
. Then there exists some
φ
=
φ
0
, φ
1
, ··· , φ
N
=
ψ
such that each
φ
j
is piecewise
C
1
closed and
φ
i+1
is obtained from
φ
i
by elementary deformation.
Proof.
This is an exercise in uniform continuity. We let
F
: [0
,
1]
×
[
a, b
]
U
be a homotopy from
φ
to
ψ
. Since
image
(
F
) is compact and
U
is open, there
is some
ε >
0 such that
B
(
F
(
s, t
)
, ε
)
U
for all (
s, t
)
[0
,
1]
×
[
a, b
] (for each
s, t
, pick the maximum
ε
s,t
>
0 such that
B
(
F
(
s, t
)
, ε
s,t
)
U
. Then
ε
s,t
varies
continuously with
s, t
, hence attains its minimum on the compact set [0
,
1]
×
[
a, b
].
Then picking ε to be the minimum works).
Since
F
is uniformly continuous, there is some
δ
such that
k
(
s, t
)
(
s
0
, t
0
)
k < δ
implies |F (s, t) F (s
0
, t
0
)| < ε.
Now we pick n N such that
1+(ba)
n
< δ, and let
x
j
= a + (b a)
j
n
φ
i
(t) = F
i
n
, t
C
ij
= B
F
i
n
, x
j
, ε
Then
C
ij
is clearly convex. These definitions are cooked up precisely so that if
s
i1
n
,
i
n
and t [x
j1
, x
j
], then F (s, t) C
ij
. So the result follows.
Corollary.
Let
U
be a domain,
f
:
U C
be holomorphic, and
γ
1
, γ
2
be
homotopic piecewise C
1
-smooth closed curves in U. Then
Z
γ
1
f(z) dz =
Z
γ
2
f(z) dz.
This means the integral around any path depends only on the homotopy
class of the path, and not the actual path itself.
We can now use this to “upgrade” our Cauchy’s theorem to allow arbitrary
simply connected domains. The theorem will become immediate if we adopt the
following alternative definition of a simply connected domain:
Definition
(Simply connected domain)
.
A domain
U
is simply connected if
every C
1
smooth closed path is homotopic to a constant path.
This is in fact equivalent to our earlier definition that every continuous
map
S
1
U
can be extended to a continuous map
D
2
U
. This is almost
immediately obvious, except that our old definition only required the map to be
continuous, while the new definition only works with piecewise
C
1
paths. We
will need something that allows us to approximate any continuous curve with a
piecewise
C
1
-smooth one, but we shall not do that here. Instead, we will just
forget about the old definition and stick to the new one.
Rewriting history, we get the following corollary:
Corollary
(Cauchy’s theorem for simply connected domains)
.
Let
U
be a simply
connected domain, and let
f
:
U C
be holomorphic. If
γ
is any piecewise
C
1
-smooth closed curve in U, then
Z
γ
f(z) dz = 0.
We will sometimes refer to this theorem as “simply-connected Cauchy”, but
we are not in any way suggesting that Cauchy himself is simply connected.
Proof.
By definition of simply-connected,
γ
is homotopic to the constant path,
and it is easy to see the integral along a constant path is zero.
3.3 Cauchy’s residue theorem
We finally get to Cauchy’s residue theorem. This in some sense a mix of all the
results we’ve previously had. Simply-connected Cauchy tells us the integral of a
holomorphic
f
around a closed curve depends only on its homotopy class, i.e. we
can deform curves by homotopy and this preserves the integral. This means the
value of the integral really only depends on the “holes” enclosed by the curve.
We also had the Cauchy integral formula. This says if
f
:
B
(
a, r
)
C
is
holomorphic, w B(a, ρ) and ρ < r, then
f(w) =
1
2πi
Z
B(a,ρ)
f(z)
z w
dz.
Note that
f
(
w
) also happens to be the residue of the function
f(z)
zw
. So this
really says if
g
has a simple pole at
a
inside the region bounded by a simple
closed curve γ, then
1
2π
Z
γ
g(z) dz = Res(g, a).
The Cauchy’s residue theorem says the result holds for any type of singularities,
and any number of singularities.
Theorem
(Cauchy’s residue theorem)
.
Let
U
be a simply connected domain,
and
{z
1
, ··· , z
k
} U
. Let
f
:
U \ {z
1
, ··· , z
k
} C
be holomorphic. Let
γ
: [
a, b
]
U
be a piecewise
C
1
-smooth closed curve such that
z
i
6
=
image
(
γ
)
for all i. Then
1
2πi
Z
γ
f(z) dz =
k
X
j=1
I(γ, z
i
) Res(f; z
i
).
The Cauchy integral formula and simply-connected Cauchy are special cases
of this.
Proof. At each z
i
, f has a Laurent expansion
f(z) =
X
nZ
c
(i)
n
(z z
i
)
n
,
valid in some neighbourhood of z
i
. Let g
i
(z) be the principal part, namely
g
i
(z) =
1
X
n=−∞
c
(i)
n
(z z
i
)
n
.
From the proof of the Laurent series, we know
g
i
(
z
) gives a holomorphic function
on U \ {z
i
}.
We now consider
f g
1
g
2
···g
k
, which is holomorphic on
U \{z
1
, ··· , z
k
}
,
and has a removable singularity at each z
i
. So
Z
γ
(f g
1
··· g
k
)(z) dz = 0,
by simply-connected Cauchy. Hence we know
Z
γ
f(z) dz =
k
X
j=1
Z
γ
g
j
(z) dz.
For each
j
, we use uniform convergence of the series
P
n≤−1
c
(j)
n
(
z z
j
)
n
on
compact subsets of U \ {z
j
}, and hence on γ, to write
Z
γ
g
j
(z) dz =
X
n≤−1
c
(j)
n
Z
γ
(z z
j
)
n
dz.
However, for
n 6
=
1, the function (
z z
j
)
n
has an antiderivative, and hence
the integral around γ vanishes. So this is equal to
c
(j)
1
Z
γ
1
z z
j
dz.
But
c
(j)
1
is by definition the residue of
f
at
z
j
, and the integral is just the integral
definition of the winding number (up to a factor of 2πi). So we get
Z
γ
f(z) dz = 2πi
k
X
j=1
Res(f; z
j
)I(γ, z
j
).
So done.
3.4 Overview
We’ve done most of the theory we need. In the remaining of the time, we are
going to use these tools to do something useful. In particular, we will use the
residue theorem heavily to compute integrals.
But before that, we shall stop and look at what we have done so far.
Our first real interesting result was Cauchy’s theorem for a triangle, which
had a rather weird hypothesis if
f
:
U C
is holomorphic and
U
is at
triangle, then
Z
f(z) dz = 0.
To prove this, we dissected our triangle into smaller and smaller triangles, and
then the result followed how the numbers and bounds magically fit in together.
To accompany this, we had another theorem that used triangles. Suppose
U
is a star domain and f : U C is continuous. Then if
Z
f(z) dz = 0
for all triangles, then there is a holomorphic
F
with
F
0
(
z
) =
f
(
z
). Here we
defined F by
F (z) =
Z
z
z
0
f(z) dz,
where
z
0
is the “center” of the star, and we integrate along straight lines. The
triangle condition ensures this is well-defined.
These are the parts where we used some geometric insight in the first
case we thought of subdividing, and in the second we decided to integrate along
paths.
These two awkward theorems about triangles fit in perfectly into the convex
Cauchy theorem, via the fundamental theorem of calculus. This tells us that if
f : U C is holomorphic and U is convex, then
Z
γ
f(z) dz = 0
for all closed γ U.
We then noticed this allows us to deform paths nicely and still preserve the
integral. We called these nice deformations elementary deformations, and then
used it to obtain the Cauchy integral formula, namely
f(w) =
1
2πi
Z
B(a,ρ)
f(z)
z w
dz
for f : B(a, r) C, ρ < r and w B(a, ρ).
This formula led us to some classical theorems like the Liouville theorem and
the maximum principle. We also used the power series trick to prove Taylor’s
theorem, saying any holomorphic function is locally equal to some power series,
which we call the Taylor series. In particular, this shows that holomorphic
functions are infinitely differentiable, since all power series are.
We then notice that for
U
a convex domain, if
f
:
U C
is continuous and
Z
γ
f(z) dz = 0
for all curves
γ
, then
f
has an antiderivative. Since
f
is the derivative of its
antiderivative (by definition), it is then (infinitely) differentiable. So a function
is holomorphic on a simply connected domain if and only if the integral along
any closed curve vanishes. Since the latter property is easily shown to be
conserved by uniform limits, we know the uniform limit of holomorphic functions
is holomorphic.
Then we figured out that we can use the same power series expansion trick
to deal with functions with singularities. It’s just that we had to include
negative powers of
z
. Adding in the ideas of winding numbers and homotopies,
we got the residue theorem. We showed that if
U
is simply connected and
f : U \ {z
1
, ··· , z
k
} C is holomorphic, then
1
2πi
Z
γ
f(z) dz =
X
Res(f, z
i
)I(γ, z
i
).
This will further lead us to Rouce’s theorem and the argument principle, to be
done later.
Throughout the course, there weren’t too many ideas used. Everything was
built upon the two “geometric” theorems of Cauchy’s theorem for triangles and
the antiderivative theorem. Afterwards, we repeatedly used the idea of deforming
and cutting paths, as well as the power series expansion of
1
zw
, and that’s it.
3.5 Applications of the residue theorem
This section is more accurately described as “Integrals, integrals, integrals”. Our
main objective is to evaluate real integrals, but to do so, we will pretend they
are complex integrals, and apply the residue theorem.
Before that, we first come up with some tools to compute residues, since we
will have to do that quite a lot.
Lemma.
Let
f
:
U \ {a} C
be holomorphic with a pole at
a
, i.e
f
is
meromorphic on U.
(i) If the pole is simple, then
Res(f, a) = lim
za
(z a)f(z).
(ii) If near a, we can write
f(z) =
g(z)
h(z)
,
where
g
(
a
)
6
= 0 and
h
has a simple zero at
a
, and
g, h
are holomorphic on
B(a, ε) \ {a}, then
Res(f, a) =
g(a)
h
0
(a)
.
(iii) If
f(z) =
g(z)
(z a)
k
near a, with g(a) 6= 0 and g is holomorphic, then
Res(f, a) =
g
(k1)
(a)
(k 1)!
.
Proof.
(i) By definition, if f has a simple pole at a, then
f(z) =
c
1
(z a)
+ c
0
+ c
1
(z a) + ··· ,
and by definition c
1
= Res(f, a). Then the result is obvious.
(ii) This is basically L’Hˆopital’s rule. By the previous part, we have
Res(f; a) = lim
za
(z a)
g(z)
h(z)
= g(a) lim
za
z a
h(z) h(a)
=
g(a)
h
0
(a)
.
(iii)
We know the residue
Res
(
f
;
a
) is the coefficient of (
z a
)
k1
in the Taylor
series of g at a, which is exactly
1
(k1)!
g
(k1)
(a).
Example. We want to compute the integral
Z
0
1
1 + x
4
dx.
We consider the following contour:
R R
×
e
/4
×
e
3/4
××
We notice
1
1+x
4
has poles at
x
4
=
1, as indicated in the diagram. Note that
the two of the poles lie in the unbounded region. So I(γ, ·) = 0 for these.
We can write the integral as
Z
γ
R
1
1 + z
4
dz =
Z
R
R
1
1 + x
4
dx +
Z
π
0
iRe
1 + R
4
e
4
dθ.
The first term is something we care about, while the second is something we
despise. So we might want to get rid of it. We notice the integrand of the second
integral is O(R
3
). Since we are integrating it over something of length R, the
whole thing tends to 0 as R .
We also know the left hand side is just
Z
γ
R
1
1 + z
4
dz = 2πi(Res(f, e
/4
) + Res(f, e
3/4
)).
So we just have to compute the residues. But our function is of the form given
by part (ii) of the lemma above. So we know
Res(f, e
/4
) =
1
4z
3
z=e
iπ/4
=
1
4
e
3πi/4
,
and similarly at
e
i3π/4
. On the other hand, as
R
, the first integral on the
right is
R
−∞
1
1+x
4
dx, which is, by evenness, twice of what we want. So
2
Z
0
1
1 + x
4
dx =
Z
−∞
1
1 + x
4
dx =
2πi
4
(e
/4
+ e
3πi/4
) =
π
2
.
Hence our integral is
Z
0
1
1 + x
4
dx =
π
2
2
.
When computing contour integrals, there are two things we have to decide.
First, we need to pick a nice contour to integrate along. Secondly, as we will see
in the next example, we have to decide what function to integrate.
Example. Suppose we want to integrate
Z
R
cos(x)
1 + x + x
2
dx.
We know
cos
, as a complex function, is everywhere holomorphic, and 1 +
x
+
x
2
have two simple zeroes, namely at the cube roots of unity. We pick the same
contour, and write ω = e
2πi/3
. Then we have
R R
×
ω
Life would be good if
cos
were bounded, for the integrand would then be
O
(
R
2
),
and the circular integral vanishes. Unfortunately, at, say,
iR
,
cos
(
z
) is large. So
instead, we consider
f(z) =
e
iz
1 + z + z
2
.
Now, again by the previous lemma, we get
Res(f; ω) =
e
2ω + 1
.
On the semicircle, we have
Z
π
0
f(Re
)Re
dθ
Z
π
0
Re
sin θR
|R
2
e
2
+ Re
+ 1|
dθ,
which is O(R
1
). So this vanishes as R .
The remaining is not quite the integral we want, but we can just take the
real part. We have
Z
R
cos x
1 + x + x
2
dx = Re
Z
R
f(z) dz
= Re lim
R→∞
Z
γ
R
f(z) dz
= Re(2πi Res(f, ω))
=
2π
3
e
3/2
cos
1
2
.
Another class of integrals that often come up are integrals of trigonometric
functions, where we are integrating along the unit circle.
Example. Consider the integral
Z
π/2
0
1
1 + sin
2
(t)
dt.
We use the expression of sin in terms of the exponential function, namely
sin(t) =
e
it
e
it
2i
.
So if we are on the unit circle, and z = e
it
, then
sin(t) =
z z
1
2
.
Moreover, we can check
dz
dt
= ie
it
.
So
dt =
dz
iz
.
Hence we get
Z
π/2
0
1
1 + sin
2
(t)
dt =
1
4
Z
2π
0
1
1 + sin
2
(t)
dt
=
1
4
Z
|z|=1
1
1 +
(zz
1
)
2
4
dz
iz
=
Z
|z|=1
iz
z
4
6z
2
+ 1
dz.
The base is a quadratic in
z
2
, which we can solve. We find the roots to be 1
±
2
and 1 ±
2.
× × ××
The residues at the point
2
1 and
2
+ 1 give
i
2
16
. So the integral we
want is
Z
π/2
0
1
1 + sin
2
(t)
= 2πi
2i
16
+
2i
16
!
=
π
2
2
.
Most rational functions of trigonometric functions can be integrated around
|z| = 1 in this way, using the fact that
sin(kt) =
e
ikt
e
ikt
2i
=
z
k
z
k
2
, cos(kt) =
e
ikt
+ e
ikt
2
=
z
k
+ z
k
2
.
We now develop a few lemmas that help us evaluate the contributions of certain
parts of contours, in order to simplify our work.
Lemma.
Let
f
:
B
(
a, r
)
\{a} C
be holomorphic, and suppose
f
has a simple
pole at a. We let γ
ε
: [α, β] C be given by
t 7→ a + εe
it
.
γ
ε
a
β
α
ε
Then
lim
ε0
Z
γ
ε
f(z) dz = (β α) · i · Res(f, a).
Proof. We can write
f(z) =
c
z a
+ g(z)
near
a
, where
c
=
Res
(
f
;
a
), and
g
:
B
(
a, δ
)
C
is holomorphic near
a
. We
take ε < δ. Then
Z
γ
ε
g(z) dz
(β α) · ε sup
zγ
ε
|g(z)|.
But
g
is bounded on
B
(
α, δ
). So this vanishes as
ε
0. So the remaining
integral is
lim
ε0
Z
γ
ε
c
z a
dz = c lim
ε0
Z
γ
ε
1
z a
dz
= c lim
ε0
Z
β
α
1
εe
it
· iεe
it
dt
= i(β α)c,
as required.
A lemma of a similar flavor allows us to consider integrals on expanding
semicircles.
Lemma
(Jordan’s lemma)
.
Let
f
be holomorphic on a neighbourhood of infinity
in
C
, i.e. on
{|z| > r}
for some
r >
0. Assume that
zf
(
z
) is bounded in this
region. Then for α > 0, we have
Z
γ
R
f(z)e
iαz
dz 0
as
R
, where
γ
R
(
t
) =
Re
it
for
t
[0
, π
] is the semicircle (which is not
closed).
γ
R
R R
In previous cases, we had
f
(
z
) =
O
(
R
2
), and then we can bound the
integral simply as
O
(
R
1
)
0. In this case, we only require
f
(
z
) =
O
(
R
1
).
The drawback is that the case
R
γ
R
f
(
z
) d
z
need not work it is possible that
this does not vanish. However, if we have the extra help from e
iαx
, then we do
get that the integral vanishes.
Proof. By assumption, we have
|f(z)|
M
|z|
for large |z| and some constant M > 0. We also have
|e
iαz
| = e
sin t
on
γ
R
. To avoid messing with
sin t
, we note that on (0
,
π
2
], the function
sin θ
θ
is
decreasing, since
d
dθ
sin θ
θ
=
θ cos θ sin θ
θ
2
0.
Then by consider the end points, we find
sin(t)
2t
π
for t [0,
π
2
]. This gives us the bound
|e
iαz
| = e
sin t
(
e
Ra2t/π
0 t
π
2
e
Ra2t
0
0 t
0
= π t
π
2
So we get
Z
π/2
0
e
iRαe
it
f(Re
it
)Re
it
dt
Z
2π
0
e
2αRt/π
· M dt
=
1
2R
(1 e
αR
)
0
as R .
The estimate for
Z
π
π/2
f(z)e
iαz
dz
is analogous.
Example. We want to show
Z
0
sin x
x
dx =
π
2
.
Note that
sin x
x
has a removable singularity at x = 0. So everything is fine.
Our first thought might be to use our usual semi-circular contour that looks
like this:
γ
R
R R
If we look at this and take the function
sin z
z
, then we get no control at
iR γ
R
.
So what we would like to do is to replace the sine with an exponential. If we let
f(z) =
e
iz
z
,
then we now have the problem that
f
has a simple pole at 0. So we consider a
modified contour
R
ε
ε
R
γ
R,ε
×
Now if
γ
R,ε
denotes the modified contour, then the singularity of
e
iz
z
lies outside
the contour, and Cauchy’s theorem says
Z
γ
R,ε
f(z) dz = 0.
Considering the
R
-semicircle
γ
R
, and using Jordan’s lemma with
α
= 1 and
1
z
as the function, we know
Z
γ
R
f(z) dz 0
as R .
Considering the
ε
-semicircle
γ
ε
, and using the first lemma, we get a contribu-
tion of
, where the sign comes from the orientation. Rearranging, and using
the fact that the function is even, we get the desired result.
Example. Suppose we want to evaluate
Z
−∞
e
ax
cosh x
dx,
where a (1, 1) is a real constant.
To do this, note that the function
f(z) =
e
az
cosh z
has simple poles where
z
=
n +
1
2
for
n Z
. So if we did as we have done
above, then we would run into infinitely many singularities, which is not fun.
Instead, we note that
cos(x + ) = cosh x.
Consider a rectangular contour
R
γ
0
R
γ
+
vert
γ
1
γ
vert
×
πi
2
πi
We now enclose only one singularity, namely ρ =
2
, where
Res(f, ρ) =
e
cosh
0
(ρ)
= ie
i/2
.
We first want to see what happens at the edges. We have
Z
γ
+
vert
f(z) dz =
Z
π
0
e
a(R+iy)
cosh(R + iy)
i dy.
hence we can bound this as
Z
γ
+
vert
f(z) dz
Z
π
0
2e
aR
e
R
e
R
dy 0 as R ,
since
a <
1. We can do a similar bound for
γ
vert
, where we use the fact that
a > 1.
Thus, letting R , we get
Z
R
e
ax
cosh x
dx +
Z
−∞
+
e
i
e
ax
cosh(x + )
dx = 2πi(ie
i/2
).
Using the fact that cosh(x + ) = cos(x), we get
Z
R
e
ax
cosh x
dx =
2πe
aiπ/2
1 + e
i
= π sec
πa
2
.
Example. We provide a(nother) proof that
X
n1
1
n
2
=
π
2
6
.
Recall we just avoided having to encircle infinitely poles by picking a rectangular
contour. Here we do the opposite we encircle infinitely many poles, and then
we can use this to evaluate the infinite sum of residues using contour integrals.
We consider the function
f
(
z
) =
π cot(πz)
z
2
, which is holomorphic on
C
except
for simple poles at Z \ {0}, and a triple pole at 0.
We can check that at n Z \ {0}, we can write
f(z) =
π cos(πz)
z
2
·
1
sin(πz)
,
where the second term has a simple zero at
n
, and the first is non-vanishing at
n 6= 0. Then we have compute
Res(f; n) =
π cos(πn)
n
2
·
1
π cos(πn)
=
1
n
2
.
Note that the reason why we have those funny
π
’s all around the place is so that
we can get this nice expression for the residue.
At z = 0, we get
cot(z) =
1
z
2
2
+ O(z
4
)
z
z
3
3
+ O(z
5
)
1
=
1
z
z
3
+ O(z
2
).
So we get
π cot(πz)
z
2
=
1
z
3
π
2
3z
+ ···
So the residue is
π
2
3
. Now we consider the following square contour:
× × × × × × × × × × ×
γ
N
(N +
1
2
)i
(N +
1
2
)i
N +
1
2
(N +
1
2
)
Since we don’t want the contour itself to pass through singularities, we make
the square pass through ±
N +
1
2
. Then the residue theorem says
Z
γ
N
f(z) dz = 2πi
2
N
X
n=1
1
n
2
π
2
3
!
.
We can thus get the desired series if we can show that
Z
γ
N
f(z) dz 0 as n .
We first note that
Z
γ
N
f(z) dz
sup
γ
N
π cot πz
z
2
4(2N + 1)
sup
γ
N
|cot πz|
4(2N + 1)π
N +
1
2
2
= sup
γ
N
|cot πz|O(N
1
).
So everything is good if we can show sup
γ
N
|cot πz| is bounded as N .
On the vertical sides, we have
z = π
N +
1
2
+ iy,
and thus
|cot(πz)| = |tan(y)| = |tanh(πy)| 1,
while on the horizontal sides, we have
z = x ± i
N +
1
2
,
and
|cot(πz)|
e
π(N+1/2)
+ e
π(N+1/2)
e
π(N+1/2)
e
π(N+1/2)
= coth
N +
1
2
π.
While it is not clear at first sight that this is bounded, we notice
x 7→ coth x
is
decreasing and positive for x 0. So we win.
Example. Suppose we want to compute the integral
Z
0
log x
1 + x
2
dx.
The point is that to define
log z
, we have to cut the plane to avoid multi-
valuedness. In this case, we might choose to cut it along
iR
0, giving a branch
of
log
, for which
arg
(
z
)
π
2
,
3π
2
. We need to avoid running through zero. So
we might look at the following contour:
R
ε ε
R
×
i
On the large semicircular arc of radius R, the integrand
|f(z)||dz| = O
R ·
log R
R
2
= O
log R
R
0 as R .
On the small semicircular arc of radius ε, the integrand
|f(z)||dz| = O(ε log ε) 0 as ε 0.
Hence, as
ε
0 and
R
, we are left with the integral along the negative
real axis. Along the negative real axis, we have
log z = log |z| + .
So the residue theorem says
Z
0
log x
1 + x
2
dx +
Z
0
log |z| +
1 + x
2
(dx) = 2πi Res(f; i).
We can compute the residue as
Res(f, i) =
log i
2i
=
1
2
2i
=
π
4
.
So we find
2
Z
0
log x
1 + x
2
dx +
Z
0
1
1 + x
2
dx =
2
2
.
Taking the real part of this, we obtain
Z
0
log x
1 + x
2
dx = 0.
In this case, we had a branch cut, and we managed to avoid it by going
around our magic contour. Sometimes, it is helpful to run our integral along the
branch cut.
Example. We want to compute
Z
0
x
x
2
+ ax + b
dx,
where
a, b R
. To define
z
, we need to pick a branch cut. We pick it to lie
along the real line, and consider the keyhole contour
As usual this has a small circle of radius
ε
around the origin, and a large circle
of radius R. Note that these both avoid the branch cut.
Again, on the R circle, we have
|f(z)||dz| = O
1
R
0 as R .
On the ε-circle, we have
|f(z)||dz| = O(ε
3/2
) 0 as ε 0.
Viewing
z
=
e
1
2
log z
, on the two pieces of the contour along
R
0
,
log z
differs
by 2
πi
. So
z
changes sign. This cancels with the sign change arising from
going in the wrong direction. Therefore the residue theorem says
2πi
X
residues inside contour = 2
Z
0
x
x
2
+ ax + b
dx.
What the residues are depends on what the quadratic actually is, but we will
not go into details.
3.6 Rouces theorem
We now want to move away from computing integrals, and look at a different
application Rouch´es theorem. Recall one of the first applications of complex
analysis is to use Liouville’s theorem to prove the fundamental theorem of algebra,
and show that every polynomial has a root. One might wonder if we know
a bit more about the polynomial, can we say a bit more about how the roots
behave?
To do this, recall we said that if
f
:
B
(
a
;
r
)
C
is holomorphic, and
f
(
a
) = 0,
then f has a zero of order k if, locally,
f(z) = (z a)
k
g(z),
with g holomorphic and g(a) 6= 0.
Analogously, if
f
:
B
(
a, r
)
\ {a} C
is holomorphic, and
f
has at worst a
pole at a, we can again write
f(z) = (z a)
k
g(z),
where now
k Z
may be negative. Since we like numbers to be positive, we say
the order of the zero/pole is |k|.
It turns out we can use integrals to help count poles and zeroes.
Theorem
(Argument principle)
.
Let
U
be a simply connected domain, and let
f
be meromorphic on
U
. Suppose in fact
f
has finitely many zeroes
z
1
, ··· , z
k
and finitely many poles
w
1
, ··· , w
`
. Let
γ
be a piecewise-
C
1
closed curve such
that z
i
, w
j
6∈ image(γ) for all i, j. Then
I(f γ, 0) =
1
2πi
Z
γ
f
0
(z)
f(z)
dz =
k
X
i=1
ord(f; z
i
)I
γ
(z
i
)
`
X
j=1
ord(f, w
j
)I(γ, w
j
).
Note that the first equality comes from the fact that
I(f γ, 0) =
1
2πi
Z
fγ
dw
w
=
1
2πi
Z
γ
df
f(z)
=
1
2πi
Z
γ
f
0
(z)
f(z)
dz.
In particular, if
γ
is a simple closed curve, then all the winding numbers of
γ
about points
z
i
, w
j
lying in the region bound by
γ
are all +1 (with the right
choice of orientation). Then
number of zeroes number of poles =
1
2π
(change in argument of f along γ).
Proof. By the residue theorem, we have
1
2πi
Z
γ
f
0
(z)
f(z)
dz =
X
zU
Res
f
0
f
, z
I(γ, z),
where we sum over all zeroes and poles of
z
. Note that outside these zeroes and
poles, the function
f
0
(z)
f(z)
is holomorphic.
Now at each
z
i
, if
f
(
z
) = (
z z
j
)
k
g
(
z
), with
g
(
z
j
)
6
= 0, then by direct
computation, we get
f
0
(z)
f(z)
=
k
z z
j
+
g
0
(z)
g(z)
.
Since at
z
j
,
g
is holomorphic and non-zero, we know
g
0
(z)
g(z)
is holomorphic near
z
j
. So
Res
f
0
f
, z
j
= k = ord(f, z
j
).
Analogously, by the same proof, at the w
i
, we get
Res
f
0
f
, w
j
= ord(f; w
j
).
So done.
This might be the right place to put the following remark all the time, we
have assumed that a simple closed curve “bounds a region”, and then we talk
about which poles or zeroes are bounded by the curve. While this seems obvious,
it is not. This is given by the Jordan curve theorem, which is actually hard.
Instead of resorting to this theorem, we can instead define what it means to
bound a region in a more convenient way. One can say that for a domain
U
, a
closed curve γ U bounds a domain D U if
I(γ, z) =
(
+1 z D
0 z 6∈ D
,
for a particular choice of orientation on
γ
. However, we shall not worry ourselves
with this.
The main application of the argument principle is Rouch´es theorem.
Corollary
(Rouces theorem)
.
Let
U
be a domain and
γ
a closed curve which
bounds a domain in
U
(the key case is when
U
is simply connected and
γ
is a
simple closed curve). Let
f, g
be holomorphic on
U
, and suppose
|f| > |g|
for all
z image
(
γ
). Then
f
and
f
+
g
have the same number of zeroes in the domain
bound by γ, when counted with multiplicity.
Proof.
If
|f| > |g|
on
γ
, then
f
and
f
+
g
cannot have zeroes on the curve
γ
.
We let
h(z) =
f(z) + g(z)
f(z)
= 1 +
g(z)
f(z)
.
This is a natural thing to consider, since zeroes of
f
+
g
is zeroes of
h
, while
poles of h are zeroes of f . Note that by assumption, for all z γ, we have
h(z) B(1, 1) {z : Re z > 0}.
Therefore
hγ
is a closed curve in the half-plane
{z
:
Re z >
0
}
. So
I
(
hγ
; 0) = 0.
Then by the argument principle,
h
must have the same number of zeros as poles
in
D
, when counted with multiplicity (note that the winding numbers are all
+1).
Thus, as the zeroes of
h
are the zeroes of
f
+
g
, and the poles of
h
are the
poles of f, the result follows.
Example.
Consider the function
z
6
+ 6
z
+ 3. This has three roots (with
multiplicity) in {1 < |z| < 2}. To show this, note that on |z| = 2, we have
|z|
4
= 16 > 6|z| + 3 |6z + 3|.
So if we let
f
(
z
) =
z
4
and
g
(
z
) = 6
z
+ 3, then
f
and
f
+
g
have the same number
of roots in {|z| < 2}. Hence all four roots lie inside {|z| < 2}.
On the other hand, on |z| = 1, we have
|6z| = 6 > |z
4
+ 3|.
So 6
z
and
z
6
+ 6
z
+ 3 have the same number of roots in
{|z| <
1
}
. So there is
exactly one root in there, and the remaining three must lie in
{
1
< |z| <
2
}
(the
bounds above show that |z| cannot be exactly 1 or 2). So done.
Example. Let
P (x) = x
n
+ a
n1
x
n1
+ ··· + a
1
x + a
0
Z[x],
and suppose a
0
6= 0. If
|a
n1
| > 1 + |a
n2
| + ··· + |a
1
| + |a
0
|,
then
P
is irreducible over
Z
(and hence irreducible over
Q
, by Gauss’ lemma
from IB Groups, Rings and Modules).
To show this, we let
f(z) = a
n1
z
n1
,
g(z) = z
n
+ a
n2
z
n2
+ ··· + a
1
z + a
0
.
Then our hypothesis tells us |f| > |g| on |z| = 1.
So f and P = f + g both have n 1 roots in the open unit disc {|z| < 1}.
Now if we could factor
P
(
z
) =
Q
(
z
)
R
(
z
), where
Q, R Z
[
x
], then at least
one of
Q, R
must have all its roots inside the unit disk. Say all roots of
Q
are
inside the unit disk. But we assumed
a
0
6
= 0. So 0 is not a root of
P
. Hence it
is not a root of
Q
. But the product of the roots
Q
is a coefficient of
Q
, hence an
integer strictly between 0 and 1. This is a contradiction.
The argument principle and Rouch´es theorem tell us how many roots we
have got. However, we do not know if they are distinct or not. This information
is given to us via the local degree theorem. Before we can state it, we have to
define the local degree.
Definition
(Local degree)
.
Let
f
:
B
(
a, r
)
C
be holomorphic and non-
constant. Then the local degree of
f
at
a
, written
deg
(
f, a
) is the order of the
zero of f(z) f (a) at a.
If we take the Taylor expansion of
f
about
a
, then the local degree is the
degree of the first non-zero term after the constant term.
Lemma. The local degree is given by
deg(f, a) = I(f γ, f(a)),
where
γ(t) = a + re
it
,
with 0 t 2π, for r > 0 sufficiently small.
Proof.
Note that by the identity theorem, we know that,
f
(
z
)
f
(
a
) has an
isolated zero at
a
(since
f
is non-constant). So for sufficiently small
r
, the
function
f
(
z
)
f
(
a
) does not vanish on
B(a, r) \{a}
. If we use this
r
, then
f γ
never hits f (a), and the winding number is well-defined.
The result then follows directly from the argument principle.
Proposition
(Local degree theorem)
.
Let
f
:
B
(
a, r
)
C
be holomorphic and
non-constant. Then for
r >
0 sufficiently small, there is
ε >
0 such that for any
w B
(
f
(
a
)
, ε
)
\ {f
(
a
)
}
, the equation
f
(
z
) =
w
has exactly
deg
(
f, a
) distinct
solutions in B(a, r).
Proof.
We pick
r >
0 such that
f
(
z
)
f
(
a
) and
f
0
(
z
) don’t vanish on
B
(
a, r
)
\{a}
.
We let
γ
(
t
) =
a
+
re
it
. Then
f
(
a
)
6∈ image
(
f γ
). So there is some
ε >
0 such
that
B(f(a), ε) image(f γ) = .
We now let
w B
(
f
(
a
)
, ε
). Then the number of zeros of
f
(
z
)
w
in
B
(
a, r
) is
just
I
(
f γ, w
), by the argument principle. This is just equal to
I
(
f γ, f
(
a
)) =
deg(f, a), by the invariance of I, ) as we move in a component C \ Γ.
Now if
w 6
=
f
(
a
), since
f
0
(
z
)
6
= 0 on
B
(
a, r
)
\{a}
, all roots of
f
(
z
)
w
must
be simple. So there are exactly deg(f; a) distinct zeros.
The local degree theorem says the equation
f
(
z
) =
w
has
deg
(
f, a
) roots for
w
sufficiently close to
f
(
a
). In particular, we know there are some roots. So
B(f(a), ε) is contained in the image of f . So we get the following result:
Corollary
(Open mapping theorem)
.
Let
U
be a domain and
f
:
U C
is
holomorphic and non-constant, then
f
is an open map, i.e. for all open
V U
,
we get that f(V ) is open.
Proof.
This is an immediate consequence of the local degree theorem. It suffices
to prove that for every
z U
and
r >
0 sufficiently small, we can find
ε >
0
such that
B
(
f
(
a
)
, ε
)
f
(
B
(
a, r
)). This is true by the local degree theorem.
Recall that Liouville’s theorem says every holomorphic
f
:
C B
(0
,
1) is
constant. However, for any other simply connected domain, we know there are
some interesting functions we can write down.
Corollary.
Let
U C
be a simply connected domain, and
U 6
=
C
. Then there
is a non-constant holomorphic function U B(0, 1).
This is a weak form of the Riemann mapping theorem, which says that there
is a conformal equivalence to
B
(0
,
1). This just says there is a map that is not
boring.
Proof.
We let
q C \ U
, and let
φ
(
z
) =
z q
. So
φ
:
U C
is non-vanishing.
It is also clearly holomorphic and non-constant. By an exercise (possibly on the
example sheet), there is a holomorphic function
g
:
U C
such that
φ
(
z
) =
e
g(z)
for all
z
. In particular, our function
φ
(
z
) =
z q
:
U C
can be written as
φ(z) = h(z)
2
, for some function h : U C
(by letting h(z) = e
1
2
g(z)
).
We let
y h
(
U
), and then the open mapping theorem says there is some
r >
0 with
B
(
y, r
)
h
(
U
). But notice
φ
is injective by observation, and that
h
(
z
1
) =
±h
(
z
2
) implies
φ
(
z
1
) =
φ
(
z
2
). So we deduce that
B
(
y, r
)
h
(
U
) =
(note that since y 6= 0, we have B(y, r) B(y, r) = for sufficiently small r).
Now define
f : z 7→
r
2(h(z) + y)
.
This is a holomorphic function f : U B(0, 1), and is non-constant.
This shows the amazing difference between C and C \ {0}.