 Solution

The square-root factor gives an algebraic branch point at $z=0$. The logarithm is locally analytic there but has value an odd multiple of $i\pi$, which is nonzero, so it cannot cancel that branching. The zero and pole of $(z-1)/(z+1)$ give logarithmic branch points at $z=1$ and $z=-1$, respectively. Thus the finite singularities areInfinity is also a branch point. On the logarithm branch tending to zero there, $\log [(z-1)/(z+1)]=-2/z+O(z^{-3})$, so the product behaves as $-2z^{-1/2}$. Other logarithm branches add $2\pi ik\sqrt{z}$ and retain branching. A chosen branch cut joins or separates these branch points; its ordinary interior points are not additional branch points of the underlying multivalued function.