I'm suppose to check the singularities and behaviour at infinity. However, I've never seen that and couln't find something about it online. So i have a function $ f(z) = \frac{1}{\exp{(z)} -1)}-\frac{1}{z}$
My attempt: expand in the variable: $u = \frac{1}{z}$.
So we have:
$f(u) = \frac{1}{\exp(\frac{1}{u})-1}-u$
Now I think I am supposed to do an Laurent expansion around $u=0$? But I am not sure.
Would love your input!
The easiest way goes as follows:
First rewrite the function $f(z)=\frac{1}{\exp(z)-1}-\frac{1}{z}$ as a Laurent series.
Notice that the $\frac{1}{z}$ part is already in the form of a Laurent series so we just need to rewrite the $\frac{1}{\exp(z)-1}$ part. This is done by solving the equation: $\frac{1}{\exp(z)-1}=g(x)$ iff $g(x)\cdot(\exp(z)-1)=g(x)\cdot(1+z+\frac{z^2}{2}+O(z)-1)=1$. This part takes practice and you need to try it out for yourself but basically you write $g(x)$ term by term so that the terms cancel with the terms from the other factor. The first two terms in each should equal one and all others should equal $0$ since you have no $z$'s on the right hand side.
You should get: $(\frac{1}{z}-\frac{1}{2}+\frac{z}{4}+O(z^2))(z+\frac{z^2}{2!}+\frac{z^3}{3!}+O(z^4))=1$. Now you put this back into the function $f$ and you get: $f(z)=\frac{1}{z}-\frac{1}{2}+\frac{z}{4}+O(z^2)-\frac{1}{z}=-\frac{1}{2}+\frac{z}{4}+O(z^2)$. Now you see that you have no $z^k$ terms for some $k<0$. This tells you that you have a removable singularity. That is not always the case. The Laurent series form should tell you what type of singularity you are dealing with. It will not always be a removable singularity as you have found here and calculating the Laurent series will not always be the best way to tackle the problem. I hope this helps!