# holomorphic function from punctured disc that misses a slit has an essential singularity at the hole

Let $D^\star=\{z\in\mathbb{C}:|z|<1\}\setminus\{0\}$.

Let $f:D^\star\to\mathbb{C}$ be a holomorphic function such that $f(D^\star)\subseteq\mathbb{C}\setminus[0,\infty)$.

Prove that $f$ has a removable singularity at the point $z=0$.

We can define $g(z)=\sqrt{-f(z)}$, so we get that $g:D^\star\to\mathbb{C}$ is a holomorphic function.

Observe that $g(D^\star)\subseteq \{z\in\mathbb{C}:\Re(z)>0\}$.

If $f$ has an essential singularity at $z=0$, then it is easy to see that $g$ also has an essential singularity at $z=0$. Therefore, $g(D^\star)$ is dense in $\mathbb{C}$, contradiction.

But, what about a pole at $z=0$.

You are almost there. Instead of $g(z)=\sqrt{f(-z)}$, let $g(z)=h(f(z))$ where $h:\Bbb C\setminus[0,\infty)\to D$ is a biholomorphic bijection. Then $g$ is holomorphic on $D^\star$ and bounded, it has a removable singularity at $z=0$ and so does $f=h^{-1}\circ g$.
• Oh, I see. You use Riemann mapping theorem to obtain $h$, right? – Don Fanucci Jun 13 '18 at 17:09
• No need for that; $h=\phi(\sqrt{-z})$ where $\phi$ is a Moebius map taking the upper half plane to the unit disk. – Julián Aguirre Jun 13 '18 at 17:11