$$f(z) = \frac{1}{z^4+4}$$
the roots of this are:
$z^2=\pm i\sqrt{2} \implies z=\pm\sqrt{i\sqrt{2}}$ and $z=\pm i\sqrt{i\sqrt{2}}$
i.e. $$f(z) = \frac{1}{(z\pm\sqrt{i\sqrt{2}})(\pm i\sqrt{i\sqrt{2}})}$$
i would like to know if there is a better simplification for this pole, since the answer looks quiet tedious when i do it this way, i notice here that there is a special case for $\dfrac{1}{z^4+1}$, is there perhaps something similar for this -Thanks.
The roots of $z^4+4=0$ are $4^{1/4}e^{ik\pi/4}$ for odd $k$. This turns out to be the four numbers $\pm1\pm i$