# Complex roots and singularities

Maybe the question is trivial, but I cannot find an answer according to the standard books of Complex Analysis. We have three kinds of singularities: isolated, pole, and essential.

What kind of singularities are the following ones?

1) $\frac{1}{\sqrt z} \text{ in } z=0$

2) $\frac{1}{\sqrt z-1} \text{ in } z=1$

One could make up a "fractionary order" pole, but they are not provided in usual terminology, since the Laurent series is on integer numbers... The problem is that they actually do not admit any Laurent expansion in the foregoing points, and thus one cannot say that they are essential singularities and apply the Picard theorem....

So what's the truth?

• Those functions do not have isolated singularities but rather algebraic branch points. Removable singularity, pole, and essential singularity are all isolated singularities. – sharding4 Jun 15 '17 at 0:24
• I tried to edit in mathjax. Did you mean $\frac{1}{\sqrt{z}-1}$ or $\frac{1}{\sqrt{z-1}}$ in (2)? – Dando18 Jun 15 '17 at 0:30
• In order to make (1) into a single-valued function in the first place, you'll first have to specify some branch cut. Once you do that, the singularity becomes essential. – Daniel Schepler Jun 15 '17 at 0:42
• Of course, in (2), if you branch cut along the negative real axis for example, then $z=1$ is a simple pole because $(z-1) f(z) = \sqrt{z} + 1$ is holomorphic on a neighborhood of 1. – Daniel Schepler Jun 15 '17 at 1:10
• Yes. That is correct. Sorry about my confusion. I thought you had $1/\sqrt{z-1}$ for 2) – sharding4 Jun 15 '17 at 1:18