Lecture 22 - Cauchy's theorem and Cauchy's Integral Formula

The content of Homotopies from the last lecture will be used in this lecture.

Cauchy’s Theorem

Let $f:D\to \mathbb{C}$ be holomorphic. Suppose ${\gamma }_1$ and ${\gamma }_2$ are two (parametrized) closed curves in $D$ such that ${\gamma }_1\sim {\gamma }_2$. Then we have
${\int }_{{\gamma }_1}f={\int }_{{\gamma }_2}f.$

Proof of Cauchy’s Theorem

Info

This is also called the Cauchy-Goursat theorem.
Cauchy proved the theorem with some extra assumptions, which was later shown to be unnecessary by Goursat.

We also assume the following:

• The derivative of $f^{\prime }$ is continuous
• The homotopy $h(t,s)$ has continuous second partial derivatives
  (The assumptions are not necessary, but makes the proof a lot simpler).

Fix $s\in [0,1]$ and let ${\gamma }_s$ be the closed curve $h_s(t)=h(t,s)$ with $0\le t\le 1$, where $h$ is the homotopy between ${\gamma }_1$ and ${\gamma }_2$.

We introduce a function $I:[0,1]\to \mathbb{C}$ by
$I(s)={\int }_{{\gamma }_s}f.$

Note

Note that $I(0)={\int }_{{\gamma }_1}f$ and $I(1)={\int }_{{\gamma }_2}f$.
Our main aim is to prove that $I(s)$ is constant, which in particular implies that $I(0)=I(1)$ and hence the result.

Since ${\gamma }_s$ is parametrized by $h_s(t)=h(s,t)$, we have by definition of complex integral
$I(s)={\int }_0^{1}f(h(t,s))\frac{\partial h}{\partial t}dt.$
We compute the derivative with respect to $s$, giving
$\frac{dI}{ds}=\frac{d}{ds}{\int }_0^{1}f(h(t,s))\frac{\partial h}{\partial t}dt={\int }_0^1\frac{\partial }{\partial s}\left(f(h(t,s))\frac{\partial h}{\partial t}\right)dt$
$={\int }_0^1\left(f^{\prime }(h(t,s))\frac{\partial h}{\partial s}\frac{\partial h}{\partial t}+f(h(t,s))\frac{{\partial }^{2}h}{\partial s\partial t}\right)dt.$
Since the second derivatives of $h$ are continuous, we can switch the order of the partial derivatives by Schwartz’s theorem. Then we can rewrite it as
$\frac{dI}{ds}={\int }_0^1\left(f^{\prime }(h(t,s))\frac{\partial h}{\partial t}\frac{\partial h}{\partial s}+f(h(t,s))\frac{{\partial }^{2}h}{\partial t\partial s}\right)dt={\int }_0^1\frac{\partial }{\partial t}\left(f(h(t,s))\frac{\partial h}{\partial s}\right)dt.$
Finally we use the fundamental theorem of calculus to get
$\frac{dI}{ds}=f(h(1,s))\frac{\partial h}{\partial s}(1,s)-f(h(0,s))\frac{\partial h}{\partial s}(0,s).$
Now since ${\gamma }_1$ and ${\gamma }_2$ are two homotopic closed curves, we have that $h(0,s)=h(1,s)$ for all $s\in [0,1]$.
Clearly this also implies that their derivatives coincide, that is $\frac{\partial h}{\partial s}(0,s)=\frac{\partial h}{\partial s}(1,s)$.
From this we conclude that $\frac{dI}{ds}=0$, that is $I(s)$ is constant.

Contractible Paths

A path $\gamma$ is called contractible if it is homotopic to a point (where a point is considered to be a closed constant path).

In this case we get an important corollary, which is very useful in practice.

Corollary

Let $f:D\to \mathbb{C}$ be holomorphic and $\gamma \subset D$ be contractible. Then
${\int }_{\gamma }f=0.$

Proof

We have that $\gamma$ can be modified into a point by some homotopy.
Then by Cauchy’s theorem:
${\int }_{\gamma }f={\int }_{pt}f.$
But the integral over a point is always zero.

Another Proof

We give another proof of Cauchy’s theorem (in the contractible setting) by using Green’s Theorem from vector calculus.

Let $C$ be a (simple, positively-oriented) closed curve in the plane and denoted by $R$ the region enclosed by $C$. Let $P(x,y)$ and $Q(x,y)$ be (real-valued) functions and assume that the first-derivatives are continuous.
Then we have the identity
${\int }_C(Pdx+Qdy)=\int {\int }_R\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dA.$
In this formula we use standard notations from vector calculus.