Why is the equation $\frac{(z-i)^2}{(z^2+1)^2}=\frac{1}{(z+i)^2} $ in the residue theorem accurate?

Question

I don't understand the reasoning here:

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

Answer 1

Well, $z² + 1 = (z-i)(z+i)$. As the latter is squared in the denominator, you can cancel out $(z-i)^2$.