Proving $0$ is a removable singularity

Question

let $f\in \mathbb{D}^* \rightarrow \mathbb{C} \backslash [0,\infty)$ be a holomorphic function ($\mathbb{D}^*=\{z\mid |z|<1 \}\backslash\{0\}$). Prove that $0$ is a removable singularity.

My try: I've tried to separate the problem into two cases and proving by contradiction: First I want to show that $0$ cannot be a pole, and then I'll show it cannot be an essential singularity. In the first case, there exist a holomorphic $\phi\in Hol(\mathbb{D})$ such that $f(z)=\frac{\phi(z)}{z^m}$. We notice that $g=\frac{1}{f}$ is a holomorphic function on the disk (0 is a removable singularity), however, I cannot get a contradiction... In the second case, I assume I should use Casorati-Weierstrass theorem, but again I cannot get a contradiction.

Also, I know that I can define a Logarithm branch Log(f(z)), but I'm not sure if it is necessary.

Thanks in advance.

Answer 1

We first construct an auxiliary function $g$ that maps $\Omega:={\mathbb C}\setminus{\mathbb R}_{\geq0}$ conformally onto $D$. Denote by $\sqrt{\cdot}$ the principal value of the square root function, which is defined on ${\mathbb C}\setminus{\mathbb R}_{\leq0}$. Then $w\mapsto\sqrt{-w}$ maps $\Omega$ conformally onto the right halfplane, and $$g(w):={\sqrt{-w}-1\over\sqrt{-w}+1}$$ maps $\Omega$ conformally onto $D$. It follows that the function $$h:=g\circ f:\quad D^*\to D,\qquad z\mapsto h(z):=g\bigl(f(z)\bigr)$$ is bounded in $D^*$, hence has a removable singularity at $z=0$. This allows to conclude that $$f:=h^{-1}\circ g:\quad D^*\to\Omega$$as a removable singularity at $z=0$ as well.