[Math] Essential singularities of $z(\exp(1/z)-1)^{-1}$

complex-analysis

I'm studying for a qualifying exam, so I'm trying to find efficient ways to answer difficult questions. Here's an example:

Find and classify the singularities of

$$
f(z)=\frac{z}{e^{1/z}-1}
$$ Intuitively, $f$ should have an essential singularity at $z=0$ and $z=\infty$. Is there an easy way to justify this? Computing the Laurent series seems difficult at first glance – or am I just being a wuss?

Best Answer

To me it seems it is certainly continuous at each point except $z=0$ and $\{e^{1/z}=1\}$. $\{e^{1/z}=1\}$ will give you poles, not an essential singularities. It remains to show that $z=0$ gives an essential singularity. Certainly it is an isolated point of discontinuity, so one need only show it is neither a pole, nor a removable discontinuity. To do so, find points $z_n$ arbitrarily close to $0$ with $|f(z_n)|$ bounded, and separately, show $f$ cannot have a limit at zero.

Edit : an alert user has pointed out that $z=0$ is not even an isolated singularity! Therefore, a priori it cannot be an essential singularity. So there cannot be any essential singularities of this function.

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