Residue Complex Function I've applied the formula  $$\lim_{z \to i} \frac{d}{dz}{(z-z_0)^2.f(z)}$$ and I keep getting a value of 0. Could someone prove my sanity by starting this off?

Here a solution with less calculations.


*

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

*$Res_{z=i}\frac{1}{z^2+1} = \lim_{z\rightarrow i}\frac{z-i}{(z-i)(z+i)}=\frac{1}{2i}$

*For $Res_{z=i}\left(- \frac{2}{(z^2+1)^2}\right) = -2 Res_{z=i} \frac{1}{(z^2+1)^2}$ you need to calculate $$\frac{d}{dz}\frac{(z-i)^2}{(z-i)^2(z+i)^2} = \frac{d}{dz}\frac{1}{(z+i)^2}= \frac{-2}{(z+i)^3} \stackrel{z\rightarrow 0}{\longrightarrow}\frac{1}{4i}$$

*All together: $$Res_{z=i}f(z) = \frac{1}{2i} - 2\cdot\frac{1}{4i} = 0$$


p.s.: You can check your residues with wolfram alpha. Here the : output for above residue
