Types of singularity that f may have at $0$

Question

Suppose f is holomorphic on a punctured neighborhood $D^*=\{z: 0<|z|<1\}$ and $f(z) \notin [0, \infty) \subset \mathbb{R}$ $ \forall z \in D^*.$ What are the possible types of singularity that f may have at $0$ (removable singularity, a pole or essential singularity)?

Thank you for any help!

Answer 1

Define $G = \Bbb C \setminus [0, \infty)$. Let $\sqrt z$ be the analytic branch of the square root in $G$ which is determined by $$ \sqrt{re^{i \phi}} = \sqrt r e^{i \phi/2} \text{ with } 0 < \phi < 2 \pi \, . $$ $\sqrt z$ maps $G$ conformally onto the upper halfplane, so that $$ h(z) = \frac{\sqrt z - i}{\sqrt z + i} $$ maps $G$ conformally onto the unit disk.

It follows that $g = h \circ f$ is bounded in $D^*$, and therefore has a removable singularity at $z=0$.

Finally conclude that $f = h^{-1} \circ g$ also has a removable singularity at $z=0$.

Answer 2

Few hints:

• In any arbitrarily small nbd. of an essential singularity the function takes all complex values (infinitely often) with at most one possible exception.

• If $z_0$ is pole of $f(z)$ then $\lim_{z\rightarrow z_0}|f(z)|\rightarrow\infty$