Let $G\subseteq\mathbb C$ be a bounded domain and $f:G\to\mathbb C$ a function that satisfies the following condition: For each sequence $(z_n)\in G$ with $z_n\to z\in\partial G$ we have $|f(z_n)|\to\infty$.
Show that $f$ is not holomoprhic.
I want to solve this problem by looking at the following 3 cases:
a) $f$ has no zero in $G$.
b) $f$ has finitely many zeros in $G$.
c) $f$ has infinitely many zeros in $G$.
I have trouble though finding a way to start. What happens, for instance, if $f$ has no zero in $G$? Does it have something to do with the maximum principle?
On the other hand I would assume that the identity theorem is helpful for c): Let $N$ be the set of zeros of $f$ in $G$. If $|N|=\infty$ then $N$ cannot be discrete (not too sure about this) and thus has an accumulation point, so $f\equiv 0$ which contradicts the behaviour on $\partial G$.
Hints: in a) apply MMP to $\frac 1 f$. In b) you have to divide $f$ by $\prod (z-z_j)$ where $z_j$'s are the zeros. Case c) cannot occur. The hypothesis implies that if there are infinitely many zeros then they will have a limit point (because there are no zeros near the boundary) which gives the contradiction $f\equiv 0$.