Essential singularity question, showing $f(z)$ can not have an essential singularity at certain point.

Question

Good time of time! I am preparing for an exam and got stuck with one of the problems.

Show that if $f$ is holomorphic in a punctured neighborhood of $z = a$, and satisfies in this neighborhood $|f(z)| ≤ M|z − a|^{−n}$ for some $M > 0$ and $n > 0$, then $z = a$ is not an essential singularity of $f$.

First I have attempted to look at $g(z)=\frac{1}{f(z)}$ to show that $z=a$ is actually a removable singularity, but I couldn't bound it (to use Riemann's theorem). Then I've tried to use contradiction-- assume it is an essential singularity and get a contradiction from the fact that $f(0<|z − a|<\epsilon)$ is dense, but got stuck again...

Thank you!

Answer 1

Besides Casorati or Picard, you can also argue as follows: The function $g(z)=f(z)(z-a)^n$ is holomorphic in a punctured neighborhood of $z=a$ and satisfies $|g(z)|\leq M$ there. Hence, $g$ can be extended holomorphically to $z=a$ (Riemann's Theorem). But then $f(z)=g(z)/(z-a)^n$ has a pole of order at most $n$ at $z=a$.

Answer 2

By Picard's great theorem, a holomorphic function $f$ on any punctured neighberhood of an essential singularity takes on all possible values in $\mathbb{C}$ except maybe one. However, this function is bounded on an open punctured disk of radius $r>0$: $$|f(z)|\leq M|z-a|^{-n}\leq \frac{M}{r^n}$$