The Laurent series of $1/(z^2+1)^2$ in the annulus $0<|z-i|<2$ I can't figure it out how to solve this problem:

Find the Laurent Series of the function $$f(z)=\frac{1}{(z^2+1)^2}$$ valid in $A=\{z \in \mathbb{C} : 0 < |z-i|<2\}$

I think that it is impossible because we have the $-i$ singularity that restrict $A$.
 A: Hint: Your function can be written as: $f(z)=\displaystyle\frac{1}{(z-i)^2}\displaystyle\frac{1}{(z+i)^2}$.
The factor $g(z)=\displaystyle\frac{1}{(z+i)^2}=-\left(\displaystyle\frac{1}{z+i}\right)'$.
Now you have to compute the power series expansion of the function
$\displaystyle\frac{1}{z+i}=\displaystyle\frac{1}{z-i+2i}=\displaystyle\frac{1}{2i}\frac{1}{1+\frac{z-i}{2i}}$ in the form  $\sum_{n=0}^\infty a_n (z-i)^n$, and mention where this expansion is valid. Then you  differentiate  the series term by term, and change the sign.
Finally you multiply with the first factor of $f$, $\displaystyle\frac{1}{(z-i)^2}$.
A: Here is an approach.
$$ f(z)=\frac{1}{(z^2+1)^2} = \frac{1}{(z-i)^2} \frac{1}{((z-i)+2i)^2}= \frac{1}{(z-i)^2} \frac{1}{((z-i)+2i)^2} $$
$$ = \frac{1}{(z-i)^2} \frac{1}{(2i)^2(1+(z-i)/(2i))^2} $$
$$ = \frac{1}{(2i)^2(z-i)^2}\sum_{k=0}^{\infty} {-2\choose k} \frac{(z-i)^k}{(2i)^k}   $$
$$ =  \sum_{k=0}^{\infty} {-2\choose k} \frac{(z-i)^{k-2}}{(2i)^{k+2}}, \quad |z-i|< 2  $$
Note: This is a suggestion by robjohn

$$ \binom{-2}{k}=(-1)^k(k+1). $$ 

A: Hint: Notice that $$\frac{1}{(z^2+1)^2} = \frac{i}{4(z+i)} + \frac{1}{4(z+i)^2} - \frac{i}{4(z-i)} - \frac{1}{4(z -i)^2} $$
and $\frac{1}{4(z+1)^2} = -\frac{d}{dz}\bigg(\frac{1}{4(z+i)}\bigg)$.
