# Singularities/Laurent series of $(ze^{\frac1{z}})^n$?

I wish to find the singularities of  $$f(z)=(ze^{\frac1{z}})^n,$$ where $$n\in\mathbb{Z}^+$$. I suspected that the points of interest are $$z=0$$ and $$z=\infty$$, and attempted to analyse them using a Laurent expansion. However, I don't make much sense of what I've done. Here's what I understand so far:

For $$z=0$$, the expansion of $$e^{\frac1{z}}$$ is $$1+\frac1{z}+\frac1{2z^2}+...$$ and therefore, the expansion of $$ze^{\frac1{z}}$$ is $$z+1+\frac1{2z}+...$$. The presence of $$z$$ in front does not change the fact that $$z=0$$ is an essential singularity. However, what changes when we consider the $$n^\text{th}$$ power? I think $$(z+1+\frac1{2z}+...)^n$$ still has an infinite number of negative powers, implying $$z=0$$ is an essential singularity, is this correct? What is the best way to properly calculate the actual Laurent expansion about $$z=0$$?

As for $$z=\infty$$, I took the substitution $$\xi=1/z$$, such that $$f(z)$$ $$f\left(\frac1\xi\right)=\frac{e^{n\xi}}{\xi^n}$$ before taking the limit as $$\xi\to0$$, which should be the indeterminate form $$1/0$$. Does this imply $$\infty$$ is an essential singularity as well, or is there more to look at?

• Recall that $(ze^{\frac{1}{z}})^n = z^ne^{\frac{n}{z}}$ Feb 1, 2020 at 19:55
• The singularity at the infinity is a pole of order $n$ by your calculations. Feb 1, 2020 at 19:57

For $$z=0$$, you don't need to take the $$n$$-th power of the Laurent series: $$f(z) = z^n e^{n/z}$$, so the Laurent series around $$0$$ becomes $$z + n + \frac{n^2}{2z} + \frac{n^3}{6z^2} + \cdots$$ What kind of singularity is there at $$z=0$$?
For $$z = \infty$$, you are correct that you should consider the substitution $$\xi = z^{-1}$$: the Laurent series of $$f(\xi^{-1}) = \xi^{-n}e^{n\xi}$$ around $$\xi = 0$$ is $$\frac{1}{\xi^n}+\frac{n}{\xi^{n-1}} + \frac{n^2}{2\xi^{n-2}} + \frac{n^3}{6\xi^{n-3}} + \cdots + \frac{n^n}{n!} + \frac{n^{n+1}\xi}{(n+1)!} + \cdots$$ What kind of singularity is there at $$\xi = 0$$?