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I am trying to find and classify the isolated singularities of $f(z) = \frac{z}{e^z-1}$.

So far I have found that: it has singularities for $e^z = 1$, so $z_k = 2k\pi i, k \in \mathbb{Z}$. First we analyze the point $z_0 = 0$. Then both the numerator and demominator have a root in $0$ of order $1$, since $(e^z-1)' = e^z$, in the point $z=0$, gives $1$. Thus, we have a removable singularity.

For $z_k = 2k\pi i, k \neq 0$, the denominator has a root $z_k$, and the numerator is non-zero in this point, so we have a pole of order 1.

Now I'm stuck at analyzing the point $z = \infty$. I wrote $$ f(1/z) = \frac{1/z}{e^{1/z}-1} = \frac{1}{ze^{1/z} -z}. $$ This has a singularity in the point $z = 0$ (and hence $f$ has one at the point $\infty$), but how do I determine its nature? Is it removable, a pole, or essential?

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  • $\begingroup$ By L'Hospital, we have $-\frac{z^2\log z}{e^{1/z}}$. As $z\to0^+$, the numerator limits to $0$, while the denominator limits to infinity. What does this tell you? $\endgroup$ May 3, 2020 at 13:23
  • $\begingroup$ Sorry, were does the $-\frac{z^2 \log z}{e^{1/z}}$ come from? $\endgroup$
    – Sigurd
    May 3, 2020 at 13:34
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    $\begingroup$ You have a sequence of poles whose locations tend to $\infty$. $\endgroup$ May 3, 2020 at 13:39

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Recall that $f$ has an essential singularity in $z=\infty$ if and only if the limit of $f(z)$ as $z\to\infty$ does not exist.

If we let $z\to\infty$ over the real axis, then clearly $\frac{z}{e^z-1}\to 0$. If we let $z\to\infty$ over the imaginary axis, prove that $f(z)\to\infty$ and conclude.

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  • $\begingroup$ I thought I proved that the singularity is non-isolated, is this correct? $\endgroup$
    – Sigurd
    May 5, 2020 at 11:15

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