Let $f$ be a holomorphic function on the punctured unit disk. If the imaginary part of $f$ is bounded, is it true that $f$ has a removable singularity at 0?
I see that $|e^{-if}|=e^{Im\;f}$ so $e^{-if}$ is a bounded holomorphic function on the punctured unit disk and it follows that $e^{-if}$ has a removable singularity at 0. The problem is how to say the same for $f$.
Best Answer
So you have observed that $g=e^{-if}$ is holomorphic on the open disk $D$. Therefore $g'=(-if')e^{-if}=-if'g$ is holomorphic on $D$. So $f'=ig'e^{if}$ is holomorphic on $D$. This implies that $f$ is holomorphic on $D$ (think of the Laurent series and its term by term derivative, for instance).