I had to solve the following question in an exam:
Prove that there is not a holomorphic function on a bounded domain $\Omega$, $f:\Omega \to \mathbb{C}$ such that $\lim_{z \to w} f(z)= \infty$ for every $w \in Fr(\Omega)$ where Fr denotes the boundary of $\Omega$.
The exam suggested that one should proof that in this situation $f$ could not have zeros in the domain. I did the exercise in this case, but how can I proof that $f$ has no zeros in $\Omega$?
The function $f$ has only finitely many zeros in $\Omega$. Indeed, if $f$ would have infinitely many zeros, then there would be a converging sequence $(z_n)_n$ of zeros by boundedness of $\Omega$. The sequence $(z_n)$ cannot converge in $\Omega$ since $f$ is not identically zero. But then $z=\lim z_n\in\mathrm{Fr}(\Omega)$ and $0=f(z_n)\rightarrow 0$, contradicting the assumption. This proves that $f$ only has finitely many zeros $a_1,\ldots,a_n$, enumerated with multiplicities. Then $f$ divided by the product $\prod(z-a_i)$ is a holomorphic function on $\Omega$ having the same property as $f$ but without zeros in $\Omega$.