Winding number version of Cauchy integral formula

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 $c_{1},c_{2},\ldots ,c_{r}$, 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

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:

$\varphi (z,w)={\frac {f(z)-f(w)}{z-w}}$ if $z\neq w$, $f'(z)$ if $z=w$

Observe that $\varphi$ is analytic in each variable. To see this, note that for fixed $z$, $\varphi$ 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 $g_{1},g_{2}:U,V\to \mathbb {C}$:

$g_{1}(z):=\oint _{c}\varphi (z,w)\,dw$

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 $|z|\to \infty$; 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.