= Solution
Put
$$
s(t)=\sqrt{1-t^2}
$$
and take the <branch cut> $[-1,1]$. On $\mathbb C\setminus[-1,1]$, choose the <analytic branch of a square root> whose upper boundary value is
$$
s(x+i0)=+\sqrt{1-x^2},
\qquad -1<x<1.
$$
Its lower boundary value is $s(x-i0)=-\sqrt{1-x^2}$. This fixes the integrand to be $+1$ at $0^+$ as required.
For $0\leq\arg z\leq\pi$, start at $0^+$ and use any path to $z$ lying above the cut; path deformations within that region do not alter the integral by the <Cauchy integral theorem>. For $\pi<\arg z<2\pi$, start in the same way, continue counterclockwise around the branch point $-1$ from the upper bank to the lower bank, and then continue to $z$ through the lower half-plane. This second prescription is the continuation obtained as $arg z$ increases through $\pi$.
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