# Generalization of the argument principle

This exercise is from big Rudin:

Let $f \in H(U)$ and $D(a,r)\subset U$ be a disk s.t. $f$ has no zero on
the boundary of the disk.

Let $\gamma$ be a curve parametrizing the boundary of the disk and Compute: $$I =\frac{1}{2\pi i}\int_\gamma \frac{f’}{f}z^pdz \space \text{ where p\in \mathbb{N}}$$
Then replace $z^p$ with an arbitrary holomorphic function $\phi \in H(U)$.

My attempt:

Suppose $f$ has a zero of degree $m$ at the point $a$. By the local mapping theorem $f=\psi^m$ in some neighborhood $V$ and $\psi$ is a holomorphic function with $\psi'(a) \ne 0$. Let $\gamma$ be the path of a circle with center $a$ contained in $V$.

$$\frac{1}{2\pi i}\int_\gamma \frac{f’}{f}\phi=\frac{m}{2\pi i}\int_\gamma \frac{\psi’}{\psi}\phi= \frac{m}{2\pi i}\int_\gamma \frac{\phi}{(z-a)}=m\phi(a)$$

The third step is by observing that the $\frac{\psi’}{\psi}$ has only one pole of order one (with residue 1) an so that’s the only term in the Laurent series that contributes to the integral.

Now generalizing, the integration over a circle can be decomposed into an integration over a circle with circular punctures wherever there’s a zero of $f$ plus the sum of all the individual circles. Here’s a picture:

Thus,

$$\frac{1}{2\pi i}\int_\gamma \frac{f’}{f}\phi dz = \sum_{a \in Z(f)} m\phi(a)$$

That is, the sum of values of $\phi$ $\space$ on the zeros of $f$ weighted by their multiplicity.

Is that right? can it be done more elegantly?