Isolated singularities of holomorphic function $f$ with $\mathfrak{R} f \geq 0$ are removable

Question

I am trying to solve the following problem:

Let $V \subseteq \mathbb{C}$ be open, $a \in V$ and $f$ holomorphic on $V \setminus \{a\}$ with $\mathfrak{R} f \geq 0$ on $V \setminus\{a\}$. Prove that $a$ is a removable singularity of $f$.

My attempt of proof:

Taking the connected component of $a$ in $V$, we may assume that $V \setminus\{a\}$ is connected. Also wlog $f$ is not constant. By the open mapping theorem, $f(V\setminus\{a\})$ is open in $\mathbb{C}$, so $f(V\setminus\{a\}) \subseteq \{\mathfrak{R} z \geq 0\}$ implies $f(V\setminus\{a\}) \subseteq \{\mathfrak{R} z > 0\}$. Taking a biholomorphic map $g: \{\mathfrak{R}z >0\} \rightarrow U$, $U$ the unit disc, we obtain a holomorphic function $g \circ f$ whose image is actually bounded. Then $a$ is clearly a removable singularity of $g\circ f$ (if $a$ was a pole, then $\lim_{z \rightarrow a}g(f(z)) =\infty$, and if $a$ was an essential singularity, then $(g\circ f)(U\setminus \{a\})$ would be dense in $\mathbb{C}$), so it is a removable singularity of $f$ as well.

The questions are the usual ones:

1. Is the proof correct and complete? Is there some subtle point I am overlooking? (The only thing that comes to mind is that one should show that $|\lim_{z \rightarrow a}g(f(z))|< 1$. But this should be clear from the open mapping theorem for $g\circ f$.)

2. Assuming the proof is correct, is there a simpler argument? In particular, is the use of the Riemann's mapping theorem an overkill?

Thanks in advance for any help.

Answer 1

There's a small point: Why does the removability of the singularity of $g\circ f$ at $a$ imply the removability of the singularity of $f$ at $a$?

Of course it follows by writing $f = g^{-1}\circ (g\circ f)$. But we need $\lvert(g\circ f)(a)\rvert < 1$ for that. Which you have covered in the parenthetical remark.

Using the Riemann mapping theorem to get a biholomorphic map between a half-plane and a disk is kind of overkill, we can write down such maps explicitly.

A somewhat different argument is by considering the behaviour at non-removable isolated singularities. The Casorati-Weierstraß theorem rules out an essential singularity, for the right half-plane isn't dense in $\mathbb{C}$, and if $a$ were a pole, then $f(V\setminus \{a\})$ would be a punctured neighbourhood of $\infty$, hence contain points with negative real part.