Classifying singularities of $\frac {z^{1/2}-1}{\sin{\pi z}}$

Question

I am trying to classify the singularities of $$\frac {z^{1/2}-1}{\sin{\pi z}}$$ where $-\pi<\arg z<\pi$.

I am confused by this because of the branch cut of $\sqrt z$ but here is my (bad) attempt:

The singularities are at $z = n$ for $n \in \mathbb Z$. We have a branch point at $z=0$ so this is a non-isolated singularity (not a simple pole, as I almost said). When $z$ is a negative integer the function is not even defined so I suppose these are not considered singularities? For $n=1$ it is removable since $1$ is a simple zero of both the numerator and denominator - the derivative of $z^{1/2}-1$ is $z^{-1/2}/2$ which is non-zero at $z=1$ so it must be a simple zero. Is that right? Then for $n>1$ we just have simple poles.

score 4 · Accepted Answer · answered May 25 '18 at 16:43

4

Concerning the square root see What are the branches of square root function?

So let us consider the function $z^{1/2}$. Although $0^{1/2}$ is of course well-defined, we must exclude $z = 0$ because we want to obtain a holomorphic function (these are defined only on open subsets of $\mathbb{C}$).

This means that the (holomorphic) function

$$f(z) = \frac{z^{1/2}-1}{\sin\pi z}$$

is defined on $U = \mathbb{C} \backslash ((-\infty,0] \cup \lbrace 1,2, 3, ... \rbrace)$. The singularities are at the points $n = 1,2,3,...$. All your arguments concerning these points are correct.

answered May 25 '18 at 16:43

Paul Frost

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Last sentence: I think you want to say "isolated singularities" – zhw. May 26 '18 at 19:26
@zhw Yes, of course! This is implicit in my answer since $\lbrace 1,2,3,... \rbrace$ does not have a cluster point. – Paul Frost May 26 '18 at 23:23

Classifying singularities of $\frac {z^{1/2}-1}{\sin{\pi z}}$

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