How can I write this function $\frac {1}{(z^2+1)^2}$ as a Laurent series around the point $z =i$?
I am struggling with the partial fractions part. It's unreducible. The only thing that came to my mind was: $(z^2+1) = (z-i)(z+i)$ and I remained stuck here. Every my attemt was a fail (I think they were a nonsense so I won't show it).
My textbook says, the solution is $ \sum _{n=-2} ^{\infty} \frac {(n+3)i^n(z-i)^n}{2^{n+4}}$ , where $z$ is a complex number
$$\frac{1}{z^2+1}=\frac{1}{(z-i)(z+i)} = \frac{1}{2i}\left[\frac{1}{z-i}-\frac{1}{z+i}\right]\tag{1} $$ now square both sides and exploit $(1)$ again: $$ \begin{eqnarray*}\frac{1}{(z^2+1)^2} &=& -\frac{1}{4}\left[\frac{1}{(z-i)^2}+\frac{1}{(z+i)^2}-\frac{2}{(z-i)(z+i)}\right]\\ &=& -\frac{1}{4}\left[\color{red}{\frac{1}{(z-i)^2}}\color{blue}{+\frac{1}{(z+i)^2}}+\color{red}{\frac{i}{z-i}}\color{blue}{-\frac{i}{z+i}}\right]\tag{2}\end{eqnarray*}$$ Now the red terms give the singular part and the blue terms give the regular part of the wanted Laurent series.