How to expand out negative powers for complex numbers I have the following expansions but I don't know how my teacher gets them. Apparently there is a formula for it (though the guy who told me didn't know it well), but I cannot find it in my notes.
For example, let $z$ be a complex number. Then,
$$\frac{1}{z^2+1}=\frac{i/2}{z+i}-\frac{i/2}{z-i}$$
It makes total sense to me in this case, but I get completely confused in this second case:
$$\frac{1}{(z^2+1)^2}=\frac{i/4}{z+i}-\frac{1/4}{(z+i)^2}-\frac{i/4}{z-i}-\frac{1/4}{(z-i)^2}$$.
What is/are the formulas that I need to compute this? Thanks! 

Start from the original equation
$$
\frac{1}{z^2+1} = \frac{i/2}{z+i} - \frac{i/2}{z-i}
$$
And just square it.
$$
\frac{1}{(z^2+1)^2} = \left(\frac{i/2}{z+i}\right)^2 + \left(\frac{i/2}{z-i}\right)^2 - 2\frac{i/2}{z+i}\frac{i/2}{z-i}
= -\frac{1/4}{(z+i)^2} - \frac{1/4}{(z-i)^2} + \frac{1/2}{z^2+1}
$$
Reuse the first equation to get:
$$
\frac{1}{(z^2+1)^2} = -\frac{1/4}{(z+i)^2} - \frac{1/4}{(z-i)^2} + \frac{i/4}{z+i} - \frac{i/4}{z-i}
$$
