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.

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.
