Question: Expand $\frac{1}{(z^2+1)^2}$ in a Laurent Series in the neighborhood of $z=i$.
My Thoughts: Let $f(z)=\frac{1}{(z^2+1)^2}=\frac{1}{(z+i)^2(z-i)^2}$. Since we want to find the Laurent series in a neighborhood of $z=i$, then let $w=z-i$. Thus, $f(w)=\frac{1}{w^2(w+2i)^2}$. Now I am just a bit stuck. I suppose we are really just playing with the definition of Laurent series of $$ f(z) = \begin{cases} \sum_{n=0}^\infty z^n & \text{if $|z|<1$} \\ -\sum_{n=1}^\infty\frac{1}{z^n} & \text{if $|z|>1$} & \end{cases} . $$ and I will just want to write $\frac{1}{w^2}$ and $\frac{1}{(w+2i)^2}$ in the above form, which would inevitably make the "bound" conditions on $z$ be $i$, thus getting us in a neighborhood of $z=i$, then replace $w$ with $z-1$ in the end. Or, maybe I could do partial fraction decomposition and go about it in the same way after I do that... or, maybe there is something I am not seeing here. Something just "feels off". Any help would be greatly appreciated! Thank you.
Let us start with $\frac1{z+i}$. We have, if $|z-i|<2$,\begin{align}\frac1{z+i}&=\frac1{2i+z-i}\\&=\frac1{2i}\times\frac1{1+\frac{z-i}{2i}}\\&=\frac1{2i}\sum_{n=0}^\infty\frac{(z-i)^i}{(2i)^n}\\&=\sum_{n=0}^\infty\frac{(z-i)^n}{(2i)^{n+1}}.\end{align}Therefore\begin{align}\frac1{(z+i)^2}&=-\left(\frac1{z+i}\right)'\\&=-\left(\sum_{n=0}^\infty\frac{(z-i)^n}{(2i)^{n+1}}\right)'\\&=-\left(\sum_{n=1}^\infty\frac{(z-i)^n}{(2i)^{n+1}}\right)'\\&=-\sum_{n=1}^\infty\frac n{(2i)^{n+1}}(z-i)^{n-1}\\&=-\sum_{n=0}^\infty\frac{n+1}{(2i)^{n+2}}(z-i)^n.\end{align}So\begin{align}\frac1{(z^2+1)^2}&=\frac1{(z-i)^2}\times\frac1{(z+i)^2}\\&=-\sum_{n=0}^\infty\frac{n+1}{(2i)^{n+2}}(z-i)^{n-2}\\&=-\sum_{n=-2}^\infty\frac{n+3}{(2i)^{n+4}}(z-i)^n.\end{align}