= 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 are
$$
\boxed{z=0,\quad1,\quad-1\quad\text{(all branch points; no finite poles).}}
$$
Infinity 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>.
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