2Contour integration
IB Complex Analysis
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)
w−z
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 −
z−z
0
w−z
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
2πiθ
, we get
=
Z
1
0
f(z + ρe
2πiθ
) dθ
≤ sup
|z−w|=ρ
|f(w)|
≤ f(z).
So we must have equality throughout. When we proved the supremum bound
for the integral, we showed equality can happen only if the integrand is constant.
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
(w−z
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)
w−z
is holomorphic on sufficiently small open neighbourhoods of
the half-contours indicated. The area enclosed by the contours might not be
star-shaped, but we can definitely divide it once more so that it is. Hence the
integral of
f(w)
w−z
around the half-contour vanishes by Cauchy’s theorem. Adding
these together, we get
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
i−1
, x
i
], both
φ
(
t
) and
ψ
(
t
) belong
to C
i
.
φ
ψ
φ(x
i−1
)
φ(x
i
)
φ(x
i−1
)
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
|w−z
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
n−1
z
n−1
+ ··· + 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.