Find the residues of $\frac {z^2} {(z^4+1)^2}$ I know that $f(z)=\frac{z^2}{(z^4+1)^2}$ has four poles at -1 .$z_1= e^{i\frac{\pi}{4}}$, $z_2= e^{i\frac{3\pi}{4}}$, $z_3= e^{i\frac{5\pi}{4}}$,$z_4= e^{i\frac{7\pi}{4}}$.But how do I find the residue at one of these poles?

There are simple formulae to find the residue, but I would like to illustrate how to compute from the definition using a Laurent series.  Let's choose the residue about $z=z_1 = e^{i \pi/4}$.  Let $\zeta=z-z_1$ and rewrite $f$ as
$$\begin{align}f(\zeta+z_1) &= \frac{(\zeta+z_1)^2}{\zeta^2 (\zeta+z_1-z_2)^2 (\zeta+z_1-z_3)^2(\zeta+z_1-z_4)^2} \\&= \frac{(z_1+\zeta)^2}{\zeta^2 (z_1-z_2)^2 (z_1-z_3)^2 (z_1-z_4)^2} \left (1+\frac{\zeta}{z_1-z_2} \right )^{-2} \left (1+\frac{\zeta}{z_1-z_3} \right )^{-2}\left (1+\frac{\zeta}{z_1-z_4} \right )^{-2}\end{align}$$
Now expand $f(z_1+\zeta)$ for small $\zeta$. Note that because we have a factor of $\zeta^2$ in the denominator, we are looking for the coefficient of $\zeta$ in the numerator as our residue.  This we can write down readily:
$$\frac{2 z_1}{ (z_1-z_2)^2 (z_1-z_3)^2(z_1-z_4)^2}\left ( 1 - \frac{z_1}{z_1-z_2}- \frac{z_1}{z_1-z_3}- \frac{z_1}{z_1-z_4} \right ) $$
Plugging in the numbers, we find that the residue of $f$ at $z_1=e^{i \pi/4}$ is
$$\frac{2 e^{i \pi/4}}{(\sqrt{2})^2 (2 e^{i \pi/4})^2 (i \sqrt{2})^2} \left [ 1 - \left (\frac1{\sqrt{2}} + \frac1{2 e^{i \pi/4}} + \frac1{i \sqrt{2}} \right ) e^{i \pi/4}\right ] = \frac1{16} e^{-i \pi/4}$$
You may check this against the simple formula given for the residue of a double pole:
$$\operatorname*{Res}_{z=z_1} [f(z)] = \left [\frac{d}{dz} \frac{z^2 (z-z_1)^2}{(z^4+1)^2} \right ]_{z=z_1} $$
