Winding number version of Cauchy integral formula

This article gives the statement, and possibly proof, of a basic fact in complex analysis.
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Statement

Suppose $U$ is an open subset of $\mathbb{C}$ and $f:U\to \mathbb{C}$ is a holomorphic function.

Suppose $c$ is a sum of oriented piecewise smooth loops $c_1,c_2,\dots ,c_r$, each completely inside $U$. In other words, when we use the symbol ${\oint }_{c}g(z)dz$, we actually mean $\sum _1^r{\int }_{c_r}g(z)dz$.

Denote by $n(c;z_0)$ the sum of the winding numbers of the $c_k$ about $z_0$. Suppose $c$ is zero-homologous, i.e. $n(c;z)=0$ for $z\in \mathbb{C}\setminus U$. Then we have, for any $z_0\in U$:

$n(c;z_0)f(z_0)=\frac{1}{2\pi i}{\oint }_c\frac{f(z)}{z-z_0}dz$

Proof

Proof outline

• We define two functions: one on the open set $U$, and the other on the open set $V$ of those points around which $c$ has winding number zero.

Proof details

The idea here is to define analytic functions on two open subsets, show that they agree on the overlap, and hence obtain an analytic function on the whole of $\mathbb{C}$.

Define first:

$\phi (z,w)=\frac{f(z)-f(w)}{z-w}$ if $z\ne w$, $f^{\prime }(z)$ if $z=w$

Observe that $\phi$ is analytic in each variable. To see this, note that for fixed $z$, $\phi$ is analytic in $w$ for $w$ away from $z$, and the power series shows that it is analytic in a neighborhood of $z$.

Now define an open subset:

$V:=\{z\in \mathbb{C}\setminus c\mid n(c;z)=0\}$

Then $U\cup V=\mathbb{C}$, and $V$ is an open subset. Consider the function $g_1,g_2:U,V\to \mathbb{C}$:

$g_1(z):={\oint }_c\phi (z,w)dw$

and:

$g_2(z):={\oint }_c\frac{f(w)}{w-z}dw$

To see that $g_1$ and $g_2$ agree on the intersection, observe that on the intersection $U\cap V$, the difference is given by:

${\oint }_c\frac{f(z)}{z-w}dz$

which is zero for points in the intersection, as they are points in $U$ about which the winding number is zero.

Thus, we can paste them together to get a single holomorphic function $g:\mathbb{C}\to \mathbb{C}$. It is clear from the expression that $g(z)\to 0$ as $\left|z\right|\to \mathrm{\infty }$; hence, using the fact that any bounded entire function is constant, we obtain that $g\equiv 0$. This shows that:

${\oint }_c\frac{f(z)}{z-w}dw={\oint }_c\frac{f(w)}{z-w}dw$

The left side evaluates to $n(c;z)f(z)$: precisely what we want.