Finding the Laurent Series of $f(z)=\frac{1}{(z^2+1)^2}$ I'm trying to determine the residue at $i$ of $f(z)=\frac{1}{(z^2+1)^2}$. My first attempt was to transform this into a Laurent Series:
$$f(z)=\frac{1}{(z^2+1)^2}=\frac{1}{(z+i)^2(z-i)^2}$$
individually analysing the roots (using the properties of the geometric series, for $|z|<1$):
$$\frac{1}{z+i}=\frac{i}{zi-1}=\frac{-i}{1-zi}=-i\sum_{n=0}^\infty(zi)^n=\sum_{n=0}^\infty (-1)\cdot i^{n+1}z^n$$
$$\frac{1}{z-i}=\frac{i}{zi+1}=\frac{i}{1-(-zi)}=i\sum_{n=0}^\infty(-zi)^n=\sum_{n=0}^\infty i(-i)^{n}z^n$$
But I can't really tell how to determine the initial function and find the residue with this approach.
Is there an alternative way of determining this?
 A: There are two approaches to this. One is to use the Laurent series as you have stated here, but we need to expand around $i$ instead of $0$. To do this, we know the series of
$$\frac{1}{z+i}=\frac{1}{(z-i)+2i}=\frac{1}{2i}\frac{1}{1+\frac{z-i}{2i}}=\frac{1}{2i}\sum_{n=0}^\infty\left(-\frac{z-i}{2i}\right)^n$$
The derivative is then
$$-\frac{1}{(z+i)^2}=\frac{1}{2i}\sum_{n=0}^\infty n\left(-\frac{z-i}{2i}\right)^{n-1}\left(-\frac{1}{2i}\right)=\frac{1}{4}\sum_{n=0}^\infty(n+1)\left(-\frac{z-i}{2i}\right)^n$$
This gives us
$$\frac{1}{(z^2+1)^2}=\frac{1}{(z-i)^2}\frac{1}{(z+i)^2}=-\frac{1}{4}\frac{1}{(z-i)^2}\sum_{n=0}^\infty(n+1)\left(-\frac{z-i}{2i}\right)^n$$
The residue is when $n=1$, as this gives the term where $z-i$ is to the power $-1$, and we have
$$-\frac{1}{4}(2)\left(-\frac{1}{2i}\right)=-\frac{i}{4}$$
The second approach is using derivatives. This function as a pole of order $2$ at $i$ and we can compute the residue using this fact. To find the residue of a pole of order $k$ we compute
$$Res_w(f)=\frac{1}{(k-1)!}\frac{d^{k-1}}{dz^{k-1}}(z-w)^kf(z)\Big|_{w}$$
In other words, we multiply by $(z-i)^2$, take a derivative, then evaluate at $i$. This gives
$$\frac{d}{dz}\frac{1}{(z+i)^2}\Big|_i=\frac{-2}{(z+i)^3}\Big|_i=\frac{-2}{-8i}=-\frac{i}{4}$$
A: This gives nice Laurent series
$$f(z)=\frac{1}{(z^2+1)^2}=-\frac{1}{4 (z-i)^2}-\frac{i}{4 (z-i)}+\frac{3}{16}+\frac{1}{16}\sum_{n=1}^\infty i^n \frac{(n+3)}{2^{n}} (z-i)^n$$
