# Find and classify the isolated singularities of $\frac{z}{e^z-1}$

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?

• 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? Commented May 3, 2020 at 13:23
• Sorry, were does the $-\frac{z^2 \log z}{e^{1/z}}$ come from? Commented May 3, 2020 at 13:34
• You have a sequence of poles whose locations tend to $\infty$. Commented May 3, 2020 at 13:39

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.

• I thought I proved that the singularity is non-isolated, is this correct? Commented May 5, 2020 at 11:15