The order of the pole of $z^{-\frac{|2k-3|}{k}}$

Question

I have a problem with determining the order of a pole. The function is $$f(z) = z^{-\frac{|2k-3|}{k}},$$ where $k$ is real and $0<k<3/2$. It seems that it has a pole at $z = 0$ where $f(z)$ is singular. However, what is the pole's order? When I look at the definition, the order of a pole is always an integer. But in this case, it seems to be unclear to me how to apply the definition.

Answer 1

Talking about a pole of a certain order only makes sense if the "pole" is an isolated singularity, that is the function is defined everywhere in a neighborhood of the singularity (except at the singularity itself). For function of the form $$f(z) = \frac1{z^a}$$ for $a \in \mathbb{R}$ this is only possible if $a$ is an integer. Else your function will have a brach cut going from zero to infinity. In that case $0$ won't be an isolated singularity.

So you need $n := \frac{2k - 3}{k} \in \mathbb{Z}$ and then $n$ is the order of your pole.