I've seen here that a holomorphic function with bounded imaginary part on a punctured neighborhood of $0$ has a removable singularity at $0$.
I just wanted to know if this result could be also extended to get this:
Let $\epsilon >0$, $z_0 \in \mathbb{C}$ and $f$ be holomorphic on a punctured neighborhood $\dot{D_{\epsilon}}(z_0)$. Futhermore it holds for all $z \in \dot{D_{\epsilon}}(z_0)$ that $Re(f(z))< K \in \mathbb{R}$
This implies that $z_0$ is a removable singularity of $f$ $(|f(z)|$ is bounded ?$)$.
If the answer is yes I'm searching for a proof
Thanks for help !
Best Answer
Yes. $z_0$ is an isolated singularity of $f$, so there are three possibilities,
If $z_0$ is a pole of $f$, then $g = \dfrac{1}{f}$ is holomorphic in a possibly smaller punctured neighbourhood $\dot{D}_\delta(z_0)$ of $z_0$, and has a removable singularity in $z_0$, which becomes a zero of $g$ after removing it. Since $g$ is not identically $0$, $g(D_\delta(z_0))$ is an open neighbourhood of $0$, hence contains a disk $D_r(0)$ for some $r > 0$. Thus
$$f(\dot{D}_\delta(z_0)) = \frac{1}{g(\dot{D}_\delta(z_0))} \supset \frac{1}{\dot{D}_r(0)} = \mathbb{C}\setminus \overline{D_{1/r}(0)}.$$
That means that neither the real nor the imaginary part of $f$ is bounded on $\dot{D}_\delta(z_0)$.
If $z_0$ is an essential singularity of $f$, the Casorati-Weierstraß theorem asserts that $f(\dot{D}_\delta(z_0))$ is dense in $\mathbb{C}$ for all $0 < \delta \leqslant \varepsilon$, hence neither the real nor the imaginary part of $f$ is bounded there.
The only remaining possibility is that $z_0$ is a removable singularity.