Consider the branch of $ 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:
$$\oint_{\left | z\right |=R}\frac{{\rm d}z}{\sqrt{z^2-1}}$$
Solution:
It is a classic case of residue at infinity. Subbing $z \mapsto 1/z$ the counterclockwise contour integral rotates the northern pole of the Rimannian sphere to the southern one and the contour integral is transformed to a clockwise integral. Hence:
$$\begin{aligned}
\oint_{\left | z \right |=R}\frac{{\rm d}z}{\sqrt{z^2-1}} &=-2\pi i \underset{z=\infty}{\mathfrak{Res}}\frac{1}{\sqrt{z^2-1}} \\
&=-2\pi i \underset{w=0}{\mathfrak{Res}} \frac{-1}{w^2\sqrt{w^{-2}-1}} \\
&=2\pi i \underset{w=0}{\mathfrak{Res}}\frac{1}{w\sqrt{1-w^2}} \\
&= 2\pi i \lim_{w\rightarrow 0}\frac{1}{\sqrt{1-w^2}}\\
&=2\pi i
\end{aligned}$$
The equality $ w\sqrt{w^{-2}-1}=\sqrt{1-w^2} $ does hold for all $|w|<1$ if we take the standard branch $\displaystyle \sqrt{1-w^2}=\exp \left ( \frac{1}{2}{\rm Log}\left ( 1-w^2 \right ) \right ) $, otherwise it is not that obvious why this holds, since we are dealing with multi-valued function.
$$\oint_{\left | z\right |=R}\frac{{\rm d}z}{\sqrt{z^2-1}}$$
Solution:
It is a classic case of residue at infinity. Subbing $z \mapsto 1/z$ the counterclockwise contour integral rotates the northern pole of the Rimannian sphere to the southern one and the contour integral is transformed to a clockwise integral. Hence:
$$\begin{aligned}
\oint_{\left | z \right |=R}\frac{{\rm d}z}{\sqrt{z^2-1}} &=-2\pi i \underset{z=\infty}{\mathfrak{Res}}\frac{1}{\sqrt{z^2-1}} \\
&=-2\pi i \underset{w=0}{\mathfrak{Res}} \frac{-1}{w^2\sqrt{w^{-2}-1}} \\
&=2\pi i \underset{w=0}{\mathfrak{Res}}\frac{1}{w\sqrt{1-w^2}} \\
&= 2\pi i \lim_{w\rightarrow 0}\frac{1}{\sqrt{1-w^2}}\\
&=2\pi i
\end{aligned}$$
The equality $ w\sqrt{w^{-2}-1}=\sqrt{1-w^2} $ does hold for all $|w|<1$ if we take the standard branch $\displaystyle \sqrt{1-w^2}=\exp \left ( \frac{1}{2}{\rm Log}\left ( 1-w^2 \right ) \right ) $, otherwise it is not that obvious why this holds, since we are dealing with multi-valued function.
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