= Solution
Factor the polynomial denominator as $(z-i)^2(z+i)^2$. The numerator has values $i$ and $-i$ at the two roots, so neither cancels. Therefore
$$
\boxed{z=i\text{ and }z=-i\text{ are double poles}.}
$$
The factor $z^{-1/2}$ gives a <branch point> at zero, with local behaviour $f(z)\sim z^{-1/2}$. Infinity is also a <branch point>, since in the coordinate $w=1/z$ the leading behaviour is $f(1/w)\sim w^{5/2}$. A single-valued branch requires a <branch cut> joining these points. Away from that cut and the two <poles>, the selected branch is <holomorphic>. For the integral below, use $0<\arg z<2\pi$, putting the cut along the positive real axis.
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