Laurent Series for this function

Question

$f(z) = \frac{1}{(z^2 + 1)^2}$

in the domain $\{z: 0<|z-i|<2\}$

but I'm struggling to get to the result, thanks in advance for your time

Answer 1

Hint:

$$\frac1{z^2+1}=\frac1{2i}\left(\frac1{z-i}-\frac1{z+i}\right)$$

and then, for example:

$$z+i=2i+(z-i)=2i\left(1+\frac{z-i}{2i}\right)$$

and you may want to check that

$$|z-i|<2\implies \left|\frac{z-i}{2i}\right|<1\;\;\;\text{and etc.}$$