How do I show that $z = -i$ is an essential singularity of $f(z) = \frac{1}{z} + \exp\left(\frac{z-i}{z^2+1} \right)$ and find its Residue $\operatorname{Res}(f,-i)$?
My idea was to show that the principle part of the Laurent series around $z = -i$ of $f$ has infinitely many terms but somehow I don't know how to deal with the $\frac{1}{z}$.
And of course I do know that \begin{align} \exp\left(\frac{z-i}{z^2+1} \right) = \exp\left(\frac{1}{z+i} \right) = \sum_{k=0}^{\infty} \frac{1}{k! (z+i)^k}. \end{align}
Would anyone be so kind to help me out?
It follows from your computations that the residue at $-i$ of $\exp\left(\frac{z-i}{z^2+1}\right)=1$. Since the residue at $-i$ of $\frac1z$ is $0$, $\operatorname{Res}(f,-i)=1$.