 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. ThereforeThe 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.