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?
Best Answer
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.