= Solution
The denominator vanishes at the four roots of $z^4=-1$, namely $z_k=e^{i(2k+1)\pi/4}$ for $k=0,1,2,3$. They are <simple poles>, since $(1+z^4)'=4z^3$ is nonzero at each root and the numerator does not vanish there. The simple-pole <residue> formula gives
$$
\boxed{\operatorname{Res}_{z=z_k}\frac{z^2}{1+z^4}=\frac{z_k^2}{4z_k^3}=\frac1{4z_k}.}
$$
Explicitly,
$$
\begin{array}{c|c}
\text{pole}&\text{residue}\\\hline
(1+i)/\sqrt2&(1-i)/(4\sqrt2)\\
(-1+i)/\sqrt2&(-1-i)/(4\sqrt2)\\
(-1-i)/\sqrt2&(-1+i)/(4\sqrt2)\\
(1-i)/\sqrt2&(1+i)/(4\sqrt2)
\end{array}.
$$
There are no branch points or other finite singularities. At infinity the function is $O(z^{-2})$, so infinity is removable rather than a pole.
Back to article page
