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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?

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

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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 $\xi = 0$?

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