I'm trying to find the residue of the function :
$$ \frac{z^5-\mathrm iz^3}{\exp\left(\frac1{z-1}\right)-\mathrm e}$$
on the point $z=1$. In this point the function has essential singularity.
I either have to find the Laurent series of the function or try and calculate an integral of the function over a punctured circle, which I doubt it'll get me anywhere. I've tried calculating the Laurent series of the denominator, and I've got:
$$\frac1{\exp\left(\frac1{z-1}\right)-\mathrm e}=\frac1{1-\mathrm e+\frac1{z-1}+\frac1{2(z-1)^2}+\cdots} $$
However, I don't really think I'll progress much from this. I will gladly use some help to find out I'm missing something.
Best Answer
It may help you if I summarise the matter.
The denominator vanishes when
$$\frac{1}{z-1}=1+2kπi\quad k\in\mathbb{Z}$$
so
$$z_k=1+\frac{1}{1+2kπi}$$
are all isolated singularities (simple poles) of $f(z)$.
Since
$$\lim_{k\to \infty} z_k=1$$
$z=1$ is a non-isolated singularity, as pointed out by Maxim in the comments. $f(z)$ cannot be expanded as a Laurent series around $z=1$ as every neighbourhood of $z=1$ contains other singularities, so the residue you asked about does not exist.