# Counting finite zeros among infinite zeros

Let $G$ be open in $\mathbb{C}$ and $f:G\rightarrow \mathbb{C}$ be an analytic function.

Let's denote $deg(f,z)$ to mean the multiplicity of a zero $z$ of $f$, and $Z(f)$ to mean the set of zeros of $f$.

Let $\gamma:[0,1]\rightarrow \mathbb{C}$ be a closed rectifiable curve which does not pass through zeros of $f$ and $Wnd(\gamma,z)=0$ for all $z\in \mathbb{C}\setminus G$.

If $f$ has finitely many zeros in $G$, then $\frac{1}{2\pi i} \int_\gamma \frac{f'(z)}{f(z)} dz = \sum_{z\in Z(f)} deg(f,z) Wnd(\gamma, z)$. It's easy to prove this.

However, I'm curious whether the above equality holds in general

. So let's assume that $f$ has infinitely many zeros in $G$.

Since the set $\{z\in G:Wnd(\gamma,z)\neq 0\}$ is relatively compact in $G$, $f$ has only finitely many zeros in $G$ whose winding numbers are nonzero. Hence, the term $\sum_{z\in Z(f)} deg(f,z)Wnd(\gamma,z)$ is well-defined.

So to prove the equality above, it suffices to prove the following statement:

Let $g:G\rightarrow \mathbb{C}$ be an anlytic function and $\gamma$ be a closed rectifiable curve in $G$ which does not pass through zeros of $g$ and $Wnd(\gamma,z)=0$ for all $z\in\mathbb{C}\setminus G$.

If any zero $z$ of $g$ satisfies $Wnd(\gamma,z)=0$, then $\int_\gamma \frac{g(z)}{g'(z)}dz= 0$

How do I prove this? If this is false, what would be a counterexample?

## 2 Answers

Yes, it holds in general. If you let $G_\epsilon$ be the open subset of $G$ which consists of the points $z$ with $|1/z|<\epsilon$ and $\textrm{dist}(z,\partial G) > \epsilon$, then for $\epsilon > 0$ sufficiently small, $\gamma \subset G_\epsilon$, and $f$ has finitely many zeros in $G_\epsilon$ because it is compactly contained in $G$. In order to apply the result about functions with finitely many zeros, you just have to check that the winding number of $\gamma$ about points in the complement of $G_\epsilon$ is still zero, which is straightforward since every point in the complement of $G_\epsilon$ can be connected to a point in the complement of $G$ (possibly $\infty$) without passing through $\gamma$.

• I'm not sure why $f$ jas finitely many zeros in $G_\epsilon$. Would you please give me some details there? – Rubertos Oct 4 '15 at 16:55
• @Rubertos That is because $\overline{G_\epsilon}$ is a compact subset of $G$. Hence, unless $f$ vanishes identically on some component of $G$, $f$ can have only finitely many zeros in $G_\epsilon$. And for small enough $\epsilon > 0$, the trace of $\gamma$ lies in $G_\epsilon$, and $\gamma$ doesn't wind around any point outside $G_\epsilon$. – Daniel Fischer Oct 4 '15 at 16:58
• @DanielFischer Perfect illustration. Thank you! – Rubertos Oct 4 '15 at 17:00

You can just replace $G$ by $G\setminus g^{-1}(\{0\})$, and now you've reduced to the case where $g$ has no zeroes in $G$.

• Oh, I feel foolish myself.. Thank you! – Rubertos Oct 4 '15 at 16:52