Normal family of holomorphic functions

Question

Suppose $V \subset \mathbb{C} $ is open, and we've a sequence $(f_n)$ of holomorphic functions on V such that $f_n$ converges uniformly on every compact subset of V to another function $f : V \rightarrow \mathbb{C}.$
(Clearly, $f$ is holomorphic: It follows from the uniform convergence theorem and Morera's theorem.) Now, let $f(z_0) = 0$ for some $ z_0 \in V.$ I want to show that there is sequence $(z_n)$ in $V$ such that $z_n$ converges to $z_0$ and $f_n(z_n) = 0$ for all $n \geq N$ for some natural number $N.$ I'm stuck for a while. I am thinking of applying Montel's theorem, but cannot find the right argument. Any help would be much appreciated.

Answer 1

This is a partial answer. Assuming that $f\neq0$, we find a radius $\delta>0$ so that $f(z)\neq0$ on the punctured disk $D(z_0,\delta)-\{z_0\}$.We have that the sequence $\{f_n\}_{n\geq1}$ converges locally uniformly to $f$ on the disk $D(z_0,\delta)$.

By Hurwitz's theorem, if $f_n(z)\neq0$ for all $n\geq 1$ and all $z\in D(z_0,\delta)$, then $f(z)\neq0$ for all $z\in D(z_0,\delta)$ a contradiction. So there exists $n_1\geq1$ and $z_1\in D(z_0,\delta)$ so that $f_{n_1}(z_1)=0$ and $|z_1-z_0|<\delta$.

The truncated sequence $\{f_n\}_{n>n_1}$ converges locally uniformly to $f$ on the disk $D(z_0,\frac{\delta}{2})$. By Hurwitz's theorem, if $f_n(z)\neq0$ for all $n>n_1$ and all $z\in D(z_0,\frac{\delta}{2})$, then $f(z)\neq0$ for all $z\in D(z_0,\frac{\delta}{2})$, a contradiction. So there exists $n_2>n_1$ and $z_2\in D(z_0,\frac{\delta}{2})$ such that $f_{n_2}(z_2)=0$ and $|z_2-z_0|<\frac{\delta}{2}$.

In this way one constructs a subsequence indexed by $n_1<n_2<n_3<\dots$ and a sequence $(z_n)$ so that $z_n\to z_0$ and $f_{n_k}(z_k)=0$ for all $k$.