= Solution
Put
$$
D=\mathbb C\setminus\bigl((-\infty,-1]\cup[1,\infty)\bigr).
$$
This slit <complex plane> is a <simply connected domain>. On $D$ there is one branch of $\sqrt{1-z^2}$ that equals $1$ at zero, and its reciprocal is <holomorphic function>[holomorphic]. The <primitive of a holomorphic function on a simply connected domain> therefore makes the integral from zero path-independent, so $G$ is single-valued.
For unrestricted paths, going once around either <branch point> changes the sign of the square root. The resulting <analytic continuation> therefore need not return to the original germ, and the value of $g(z)$ can depend on the path.
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