= Solution
The rational factors give four <simple poles> at $z=\pm i,\pm2i$; none is canceled by the numerator. The square root has <branch points> at $z=\pm1$, with local behaviour proportional to $(z\mp1)^{-1/2}$ in the integrand.
For this part, choose $h(z)=\sqrt{z^2-1}$ analytic off $[-1,1]$ and satisfying $h(z)\sim z$ at infinity. Its values are $h(x)=\sqrt{x^2-1}$ for $x>1$, $h(x)=-\sqrt{x^2-1}$ for $x<-1$, and
$$
h(x+i0)=i\sqrt{1-x^2},\qquad h(x-i0)=-i\sqrt{1-x^2}\quad(-1<x<1).
$$
Thus the chosen <branch cut> is $[-1,1]$, and the function is analytic elsewhere apart from the four poles. Infinity is not an intrinsic branch point: $g(z)=O(z^{-2})$ on this branch.
Another legitimate convention applies the positive-axis square root of part (a) to $w=z^2-1$; it places the cuts on $(-\infty,-1]$ and $[1,\infty)$. Its upper-half-plane values agree with the branch used here, so the contour calculation below and the positive real square root in the required integral are unchanged. The branch choice must be stated because a square root alone does not specify a single-valued complex function.
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