11 - Homology Form of Cauchy's Theorem

Homology Form of Cauchy's Theorem

Let $f:U\to \mathbb{C}$ be a holomorphic function
Let $\Gamma$ be a cycle in $U$ whose inside lies entirely in $U$ so $I(\Gamma ,z)=0$ for all $z\notin U$

Then Cauchy’s Theorem and Integral Formula hold so
1)

$${\int }_{\Gamma }f(\zeta )d\zeta =0$$

$$\frac{1}{2\pi i}{\int }_{f_{\Gamma }}\frac{f(\zeta )}{\zeta -z}d\zeta =I(\Gamma ,z)f(z)\text{ for all }z\in U\setminus {\Gamma }^{\ast }$$

Proof

Let

$$\begin{array}{rl}G(z)& :=\frac{1}{2\pi i}{\int }_{\Gamma }\frac{f(\zeta )}{\zeta -z}d\zeta \text{ for }z\in C\setminus {\Gamma }^{\ast }\\ F(z)& :=G(z)-f(z)I(\Gamma ,z)\end{array}$$

For fixed $z\in U$ define $g_z:U\setminus \{z\}\to \mathbb{C}$ by

$$g_z(\zeta ):=\frac{f(\zeta )-f(z)}{\zeta -z}$$

By Cauchy integral formula for the winding number then

$$\begin{array}{rl}F(z)& =G(z)-f(z)\cdot \frac{1}{2\pi }{\int }_{\Gamma }\frac{1}{\zeta -z}d\zeta \\ & =\frac{1}{2\pi i}{\int }_{\Gamma }\frac{f(\zeta )-f(z)}{\zeta -z}d\zeta \\ & =\frac{1}{2\pi i}{\int }_{\Gamma }{g}_z(\zeta )d\zeta \text{ for }z\in U\setminus {\Gamma }^{\ast }\end{array}$$

By General Cauchy integral formula for the winding number then
Both $g(z)$ and $I(\Gamma ,z)$ are holomorphic for all $z\in \mathbb{C}\setminus {\Gamma }^{\ast }$ so $F(z)$ is holomorphic on $U\setminus {\Gamma }^{\ast }$

For fixed $z$ then $g_z\zeta$ is also clearly a holomorphic function of $\zeta$ on $U\setminus \{z\}$ and as

$$\underset{\zeta \to z}{\lim }{g}_z(\zeta )=f^{\prime }(z)$$

Then it extends to a continuous function on $U$ which also denote by $g_z$ where

$$g_z(z)=f^{\prime }(z)$$

Since it is continuous at $\zeta =z$ then $g_z$ is bounded near $z$

By Riemann’s Removable Singularities Theorem then
Extension is holomorphic as a function of $\zeta$ on all of $U$

But extending the definition of $F$ from $U\setminus {\Gamma }^{\ast }$ to all of $U$ by setting

$$F(z):={\int }_{\Gamma }{g}_z(\zeta )d\zeta$$

If $z_0\in {\Gamma }^{\ast }$ then since $g_z(\zeta )$ is holomorphic then by Removal of a point in a cycle
There is a cycle $\Sigma \subseteq U$ not containing $z_0$ such that

$${\int }_{\Sigma }{g}_z(\zeta )d\zeta ={\int }_{\Gamma }{g}_z(\zeta )d\zeta$$

So replacing $\Gamma$ by $\Sigma$ and applying the same reasoning then $F$ is holomorphic on $U\setminus {\Sigma }^{\ast }$
Most importantly it is holomorphic at $z_0$

Since $z_0$ is arbitrary then $F$ is holomorphic on all of $U$

Let $V=\mathbb{C}\setminus ({\Gamma }^{\ast }\cup \mathrm{ins}(\Gamma )))$
Since $\Gamma (z)$ is holomorphic on all of $\mathbb{C}\setminus {\Gamma }^{\ast }$, it restricts to holomorphic function on $V$
By assumption $\Gamma \cup \mathrm{ins}(\Gamma )\subseteq U$ hence $U\cup V=\mathbb{C}$

Moreover if $z\in U\cap V$ then $z\in U$ but is neither inside $\Gamma$ or on ${\Gamma }^{\ast }$ so by definition

$$F(z)=G(z)$$

So by defining

$$H:\mathbb{C}\to \mathbb{C},H(z):=\{\begin{array}{ll}F(z)& z\in U\\ & \\ G(z)& z\in V\end{array}$$

with $H$ being a well-defined entire function

Clearly $F(z)\equiv 0$ on $U$ if $H(z)\equiv 0$ on $\mathbb{C}$
But $H(z)=G(z)$ for large $z$ so
By General cauchy integral formula for the winding number then

$$G(z)\to 0\text{ as }z\to \infty$$

Hence $H$ defines a bounded entire function which is constant so is $0$ by Liouville’s Theorem