Complete question: Give a proof or a counter-example. If $B^º$ is an open set of C and $f: B^º\rightarrow C$ is an analytic function, then f is bounded in $B^º$.
My attempt:
If $f$ is bounded, then $|f(z)|\leq M, M \in N$, $\forall z \in B^º$. Let $g(z)=e^z$, entire in C. Then, there exists an $M \in N$, such that, $|e^z|\leq M, \forall z \in B^º$.
$\rightarrow ln(|e^z|)\leq ln(M) \Rightarrow z\leq ln(M)$, but, as $B^º\subset C$ is any open set, we could choose it as $B^º\cup \{ z_0 \} $, where $z_0 > M$ which would led to a contradiction. Thus, $f$ being analytic does not imply being bounded in an open set.
Is this correct?
Your argument is not correct but it becomes correct if you just choose $z$ to be real. For complex $z$ it is not true that $\ln(|e^{z}|)=z$ and the inequality $z \leq \ln M$ does not make sense. Besides, you can take the open set $B^{0}$ to be $\mathbb C$ in your argument.