I understand that if $f$ is holomorphic on $\Omega$ except possibly on $z_0$ and is bounded on $\Omega – \{z_0\}$ then $z_0$ is a removable singularity. But if we know that if $z_0$ is a removable singularity of $f$ does this imply that there is a neighborhood of $z_0$ on which $f$ is bounded? Can $f$ be holomorphic at a nearby point for which it tends to infinity?
[Math] Does removable singularity imply bounded
complex-analysis
Best Answer
Since the singularity is removable there is a neighborhood $U$ of $z_0$ and a holomorphic function $g$ on $U$ that coincides with $f$ on $U \setminus\{z_0\}$.
The function $g$ is in particular continuous. You can now, for instance, take a closed disk $D$ around $z_0$ with $D \subset U$ and note that $g$ is bounded on $D$ (a continuous function on a compact set is bounded) and thus $f$ is bounded on $D \setminus \{z_0\}$