TITLE: What kind of singularity is possessed by $z^{-1/2}$?
QUESTION [6 upvotes]: I'm learning about singularities [Stein Sharkarchi Chapter 3], and this question popped up in my mind: What kind of singularity is $f(z):= z^{-\frac{1}{2}}$ at $0$? 
It doesn't seem to fall into either of these three categories:

Removable singularity: It's clearly not a removable singularity as $\lim_{z \rightarrow 0} |f(z)| \rightarrow \infty$
Essential singularity: By Casorati-Weierstrass it's not essential singularity either since the image of the unit disc minus the origin is very clearly not dense (it completely misses a ball of radious $.9999$ centered at the origin)
Pole: It's not a pole either since it contradicts this theorem because $z^nf(z)$ vanishes at $0$ for all positive integer $n$:


If $f$ has a pole at $0$ then for a neighbourhood of $0$ there's a positive unique  integer $n$ such that $f(z) = h(z) z^{-n}$ and $h(z)$ is holomorphic and nonvanishing in the neighbourhood 

So what's this singularity ? This is perplexing me. 
I feel this may have something to do with the nonuniqueness of $f(z)$ around $0$ but then I know nothing about branch cuts to add substance to my intuition.

REPLY [3 votes]: Standard definitions: $f$ has an isolated singularity at $a$ if $f$ is holomorphic in $\{z:0<|z-a|<r\}$ for some $r>0$. An isolated singularity is removable if  ..., it is a pole if ..., otherwise it is an essential singularity.
So your function does not have an essential singularity at $0$, simply because it does not have an isolated singularity there.