= Solution
The square root produces <branch points> at $z=\pm1$, with local behaviour proportional to $(z\mp1)^{-1/2}$. One <branch cut> can join them along $[-1,1]$. Another choice uses the two rays $(-\infty,-1]\cup[1,\infty)$. Each choice prevents a loop from enclosing just one <branch point> within a single-valued domain. For the finite-segment choice fix $\sqrt{z^2-1}\sim z$ at infinity. Infinity is then a <pole> of order three, not a <branch point>, since
$$
f(z)=z^3+\frac12z+\frac{3}{8z}+O(z^{-3}).
$$
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