Contour integral

Complex Analysis
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Tolaso J Kos
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Contour integral

#1

Post by Tolaso J Kos »

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:

$$\oint \limits_{\left | z\right |=R}\frac{{\rm d}z}{\sqrt{z^2-1}}$$
Imagination is much more important than knowledge.
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Tolaso J Kos
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Re: Contour integral

#2

Post by Tolaso J Kos »

It is a classic case of residue at infinity. Subbing \( z \mapsto 1/z \) 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:


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


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 a multi-valued function.
Imagination is much more important than knowledge.
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