# Uniform convergence of $f_n\rightarrow f$ and limit of zeroes to $f_n$

I'm having some doubts on a homework question:

Let $f_n\rightarrow f$ uniformly on compact subsets of an open connected set $\Omega \subset \mathbb{C}$, where $f_n$ is analytic, and $f$ is not identically equal to zero.

(a) Show that if $f(w)=0$ then we can write $w=\lim z_n$, where $f_n(z_n)=0$ for all $n$ sufficiently large.

(b) Does this result hold if we only assume $\Omega$ to be open?

I'm not too sure how to do (a)-- I think I might be able to do it just by using the definition of uniform convergence and the fact that $f_n$ has a zero at $z_n$, but this doesn't use the assumption that $f_n$ is analytic or that $\Omega$ is connected. I'm also guessing that the result doesn't hold if we only assume $\Omega$ to be open and not connected for obvious topological reasons, but not knowing exactly how to do (a), I'm not sure if I know how to prove this. Could anyone give me some pointers? Thanks in advance.

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How do you know $f_n$ has a zero at $z_n$? You should be able to come up with a counterexample to $(b)$ before you prove $(a)$: taking two connected components for $\Omega, f=0$ on one of them, and $f=1$ on the other, there are some readily available $f_n$ that would contradict the statement. – Kevin Carlson Sep 24 '12 at 7:37
@GeorgesElencwajg: To see that Kevin is right, consider $\Omega=B(0,1)\cup B(2,1)$, $f_n(z)=\begin{cases}\frac1n&z\in B(0,1)\\1&z\in B(2,1)\end{cases}$ and hence $f(z)=\begin{cases}0&z\in B(0,1)\\1&z\in B(2,1)\end{cases}$. The $f_n$ don't have any zeroes at all. – Hagen von Eitzen Sep 24 '12 at 10:40
Dear Hagen, you are absolutely right: I'll delete the relevant comments (and present my apologies) . Of course, the correct hypothesis is that $f$ is not identically zero in any connected component of $\Omega$. I appreciate the irony that your counter-example is a modification of the one I give in my answer:-) – Georges Elencwajg Sep 24 '12 at 12:48
Thanks everyone, your comments were very helpful. I ended up using Rouché's theorem to prove the result. – mathstudent12 Sep 24 '12 at 18:07
Just to remark that analyticity is important: Take $f(z)=|z|^2$ and $f_n(z)=|z|^2+1/n$. – timur Sep 25 '12 at 2:12

Take a small circle around $w$.
Then by Rouché's theorem $f_n$ has a zero $z_n$ inside the circle for $n$ large enough (and maybe several if $w$ is a multiple zero of $f$).
Now shrink the circle and repeat: you will obtain the convergent sequence $(z_n)$.
By the way, this sketch of proof shows why we must assume that $f$ is not identically zero near $w$: if $f\equiv 0$ just take $f_n=\frac {1}{n}$ (which has no zero at all!) to get a counter-example.