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Contour radical integral

Consider the branch of \displaystyle f(z) =\sqrt{z^2-1} which is defined outside the segment [-1, 1] and which coincides with the positive square root \sqrt{x^2-1} for x>1. Let R>1 then evaluate the contour integral:

    \[\ointctrclockwise \limits_{\left | z\right |=R}\frac{{\rm d}z}{\sqrt{z^2-1}}\]


It is a classic case of residue at infinity. Subbing z \mapsto \frac{1}{z} the counterclockwise contour integral rotates the northern pole of the Riemannian sphere to the southern one and the contour integral is transformed to a clockwise one. Hence:

    \begin{align*} \oint \limits_{\left | z \right |=R}\frac{{\rm d}z}{\sqrt{z^2-1}} &=-2\pi i \underset{z=\infty}{\mathfrak{Res}}\frac{1}{\sqrt{z^2-1}} \\ &=-2\pi i \underset{w=0}{\mathfrak{Res}} \frac{-1}{w^2\sqrt{w^{-2}-1}} \\ &=2\pi i \underset{w=0}{\mathfrak{Res}}\frac{1}{w\sqrt{1-w^2}} \\ &= 2\pi i \lim_{w\rightarrow 0}\frac{1}{\sqrt{1-w^2}}\\ &=2\pi i \end{align*}

The equality w\sqrt{w^{-2}-1}=\sqrt{1-w^2}does hold for all |w|<1 if we take the standard branch \sqrt{1-w^2}=\exp \left ( \frac{1}{2}\mathrm{Log} \left ( 1-w^2 \right ) \right ) , otherwise it is not that obvious why this holds, since we are dealing with a multi-valued function.

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