 Solution

PutThis 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. 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.