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 $f=h^{-1}\circ g$.

  • $\begingroup$ Oh, I see. You use Riemann mapping theorem to obtain $h$, right? $\endgroup$ – Don Fanucci Jun 13 '18 at 17:09
  • 2
    $\begingroup$ No need for that; $h=\phi(\sqrt{-z})$ where $\phi$ is a Moebius map taking the upper half plane to the unit disk. $\endgroup$ – Julián Aguirre Jun 13 '18 at 17:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.