Given a meromorphic function, $D_{r+\delta}(w)\backslash D_r(w)$ has no poles or zeros.

Question

Given a meromorphic function $f$ on $\Omega$ that is not identically zero, and $f$ has no poles and never vanishes on $\partial \overline{D_r(w)}\subseteq \Omega$. I'm trying to prove that there exists $\delta>0$ such that $D_{r+\delta}(w)\backslash D_r(w)$ has no poles or zeros.

Given the pieces in the problem, I think I need to use identity principle to prove by contradiction. But I'm not sure how to go about this. Any input is appreciated.

Answer 1

Suppose that no such $\delta$ exists. Then, for each $n\in\mathbb N$, there is a $z_n\in D_{r+\frac1n}(w)\setminus D_r(w)$ which is either a zero or pole of $f$. Passing to a subsequence, if needed, we can assume without loss of generality that each $z_n$ is a zero or that each $z_n$ is a pole. Since the sequence $(z_n)_{n\in\mathbb N}$ is bounded, it has a convergente subsequence; let $z_0$ be its limit. Then $z_0\in\partial\overline{D_r(w)}$. We can assume, again without loss of generality, that $\lim_{n\to\infty}z_n=z_0$. Now:

• if each $z_n$ is a zero of $f$, so is $z_0$, which is impossible, since $f$ has no zeros in $\overline{D_r(w)}$;
• if each $z_n$ is a pole of $f$, then, since $f$ is analytic and it has no pole in $\overline{D_r(w)}$, there is a neighborhhod of $z_0$ such that $f$ has no poles there, which contradicts the fact that $\lim_{n\to\infty}z_n=z_0$ and that each $z_n$ is a pole.