= Square root of z squared minus one
{title2=$f(z)=\sqrt{z^2-1}$}
The <branch points> are $\pm1$; infinity has two unbranched meromorphic choices. On the <complex plane> slit along $[-1,1]$, normalization $f(z)\sim z$ at infinity selects a <holomorphic square root> with boundary values $f(x+i0)=i\sqrt{1-x^2}$ and $f(x-i0)=-i\sqrt{1-x^2}$ for $-1<x<1$. On the exterior real intervals, $f(x)=\operatorname{sgn}(x)\sqrt{x^2-1}$. A counterclockwise <contour integral> around the finite <branch cut> satisfies $\oint d\zeta/f(\zeta)=2\pi i$: the upper lip runs right to left, the lower lip left to right, and the endpoint arcs contribute $O(\sqrt\varepsilon)$. The opposite global branch reverses these signs.
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