An application of Hurwitz theorem from Conway's Complex Analysis Suppose that $f_{n}$ is a sequence in $H(G)$, $f$ is a non-constant function and $f_{n}\to f$ in $H(G)$. Let $a\in G$ and $f(a)=\alpha$. Show that there is a sequence $a_{n}$ in $G$, such that (i)$a=lim_{n \to \infty} a_{n}$ (ii) $f_{n}(a_{n})=\alpha$ for sufficiently large $n$.
I am stuck with this problem and I have no idea. It seem it is an application of Hurwitz's theorem but I am not able to apply it!
Thanks in advance!
 A: To simplify the notation we can (without loss of generality) assume that $a = \alpha = 0$, i.e. $f_n \to f$ locally uniformly in $G$, $f$ is not constant, and $f(0) = 0$.
The zeros of a non-constant function are isolated, so we can choose a $r > 0$ such that $\overline{B_r(0)} \subset G$ and $f(z) \ne 0$ for $0 < |z| \le r$. Now Hurwitz's theorem can be applied  on $\overline{B_r(0)}$, and it follows that there is a $N \in \Bbb N$ such that for every $n \ge N$, $f_n$ has the same number of zeros as $f$ in $B_r(0)$, counting with multiplicities. 
In particular, $f_n$ has at least one zero in $B_r(0)$ for $n \ge N$. The set
$$
 Z_n = \{ z \in \overline{B_r(0)} : f_n(z) = 0 \}
$$
is non-empty and closed, therefore there is a $a_n \in Z_n$ with
$$
 |a_n| = \min \{ |z| : z \in Z_n \}
$$
In other words $a_n$ is (as in Robert's answer) a zero of $f_n$ which is closest to the origin among all zeros of $f_n$.
So we have found a sequence $(a_n)_{n \ge N}$ in $G$ such that $f_n(a_n) = 0$, and it remains to show that $\lim_{n \to \infty} a_n = 0$.
Assume that this is not the case. Then there is a $0 < \rho < r$ and a subsequence $(a_{n_k})_k$ such that $|a_{n_k}| > \rho$ for all $k$. Now we apply Hurwitz's theorem to the sequence $(f_{n_k})_k$ on $\overline{B_\rho(0)}$. It follows that for sufficiently large $k$, $f_{n_k}$ has a zero in $B_\rho(0)$. This is a contradiction to our definition of $a_{n_k}$ being a zero of $f_{n_k}$ closest to the origin. This completes the proof.
A: Let $\Gamma$ be a positively oriented circle around $a$ such that $\Gamma$ and its interior are contained in $G$.  For any $g \in H(G)$ such that $g \ne \alpha$ on $\Gamma$, by the Residue Theorem
$$ M(g) = \dfrac{1}{2\pi i} \oint_\Gamma \dfrac{ g'(z)\; dz}{g(z) - \alpha}$$
is the number of roots of $g - \alpha$ inside $\Gamma$, counted by multiplicity.  In particular, if $g = f_n$ is sufficiently close to $f$ on $\Gamma$ this will be at least $1$.  You can simply define $a_n$ to be the closest root of $f_n - \alpha$ to $a$, breaking ties arbitrarily.
