2Contour integration
IB Complex Analysis
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
i≥0
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
z∈∂T
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
z∈∂T
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.