How can I determine the type of a singularity of a function $$f(z)={e^{1/z} \over z-1}+{\pi z \over 2\sin(\pi z)}$$
at $z_0=0$?
I don't see an easy way to represent it using Laurent series, neither I don't see how I can find the limit of the function with $z\to 0$.
On
It is essentially because the part $e^{\frac{1}{z}}=\sum_{n=0}^{\infty}\frac{\frac{1}{z^n}}{n!}=\sum_{n=0}^{\infty}\frac{1}{z^n\cdot n!}$ attains infinity times often addends in the form $\frac{a_k}{z^k}$ (an the other parts are holomorphic or liftable in $0$, in a neighborhood of $0$) so the principal part has infinity many addeds. Thus $z_0=0$ as an essentially singularity.
Look at $e^{1/z}$ around $z = 0$. This function approaches every point in the complex plane (take the limit along the real axis, positive or negative or along the imaginary axis). Dividing by and adding a meromorphic function does not change this, so there is no singularity that can be characterized in the normal way with Laurent series, residues, etc. This is called an essential singularity.