I was reading this proof here: holomorphic function with bounded real part on punctured neighborhood $\dot{D}_{\epsilon}(z_0)$, which was proving that
Let $f$ be holomorphic on a punctured disk $D_\epsilon(z_0)/\{z_0\}$ and $Re(f(z)) < M$ for all $z \in D_\epsilon(z_0)/\{z_0\}$. Then it implies that $f$ has a removable singularity at $z_0$.
In the answer to the original post, they supposed that $z_0$ is a pole for the sake of contradiction, but why $g = \frac{1}{f}$ holomorphic in a "possible smaller" punctured neighbourhood $D_\delta(z_0)/\{z_0\}$? Also, why is $g(D_\delta(z_0)/\{z_0\})$ an open neighbourhood of $0$? The open mapping theorem might have been used here but why is it a neighbourhood of $0$? Last, why is $\frac{1}{D_r(0)\setminus \{0\}} = \mathbb{C}\setminus \overline{D_{\frac{1}{r}}(0)}$?
Thank you .
Best Answer
The proof in the link given does not say $g(D(z_0)\setminus\{z_0\})$ is an open neighborhood of $0.$ That doesn't even make sense: $0\notin g(D(z_0)\setminus\{z_0\})!$ What is stated is that $g$ has a removable singularity at $z_0.$ Still using the notation $g$ for the extension, we have $g(z_0)=0.$ It is the extended $g$ that maps $D(z_0)$ onto a neighborhood of $0.$