If $f$ is holomorphic in $\{|z|<1\}$ \ $\{0\}$ and doesn't get values in $(-\infty ,0]$ then $0$ is a removable singular point

Question

Let $f$ b a holomorphic in $\{|z|<1\}$ \ $\{0\}$. I want to show that if $f$ doesn't get values in $(-\infty ,0]$ then $0$ is a removable singular point.

I am not sure where to start, but since $f$ is never equal to $0$ then I can probably work with $\frac{1}{f}$ which is also holomorphic in the same domain.

From Riemann's theorem, I know that if $f$ is bounded in a neighborhood of $0$ then it is a removable singular point.

However I don't think it gets me anywhere here.

Help would be appreciated

Answer 1

If $0$ is not a removable singularity, there are two possibilities:

1. It is a pole. Then, near $0$, $f(z)$ can be written as $\frac{g(z)}{z^n}$ for some analytic function $g$ such that $g(0)\neq0$ and some $n\in\mathbb N$. If $g(0)=re^{i\theta}$, then, for each $r'>0$, if $z=r'e^{i(\theta-\pi)/n}$, then $f(z)$ is a real negative number.
2. It is an essential singularity. For each $z\in\mathbb C\setminus(-\infty,0]$, let $\sqrt z$ be the number defined as follows: if $z=re^{i\theta}$, with $\theta\in(-\pi,\pi)$, then $\sqrt z=\sqrt r^{i\theta/2}$. This defines an analytic function and so $\sqrt f$ is analytic and $0$ is still an essential singularity of $\sqrt f$. But the image of $\sqrt f$ is contained in the half-plane $\{z\in\mathbb C\mid\operatorname{Re}z>0\}$. This is impossible since, by the Casorati-Weierstrass theorem, that image is dense.

Since we get a contradiction in each case, $0$ is a removable singularity of $f$.