What kind of singularity is possessed by $z^{-1/2}$?

Question

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.

Answer 1

The Casorati–Weierstrass theorem applies only when the function is holomorphic in a punctured disk around the singularity, which isn't the case here -- you need at least one branch cut where you switch to the other branch of the square root.

This is a branch point.

Answer 2

For most purposes, such a "singularity" does not fall into the "removable, pole, or essential" classification, because it cannot be defined as a holomorphic function on any punctured neighborhood of $0$.

A classical way to think about this is that $\sqrt{z}$ does admit analytic continuation along every path that does not go through $0$, ... but the monodromy is non-trivial, so there's no single "global" holomorphic function...

Essentially equivalently, and incorporating somewhat more modern things (covering spaces, from early 20th centurey?!?! :) we could say that there simply is no globally defined $\sqrt{z}$ (with or without $-1$) on a punctured neighborhood of $0$, but there is one on a two-fold (ramified) cover... that "unwinds" the square root...

Also, by the way, the notion of "branch cut" is not an intrinsic thing. It's just a way of reducing the "multi-connectedness" of a set so that the monodromy group homomorphism restricted to $\pi_1$ of the cut-up region is trivial, so that there does exist a globally defined holomorphic function off the "cuts", and so on.

Answer 3

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.