Compute the integral $\int_C (z^2-1)^\frac{1}{2} dz$ where $R>1$ Let C be the circle of Radius $R>1$, centered at the origin, in the complex plane. Compute the integral
$\int_C (z^2-1)^\frac{1}{2} dz$
where we employ a branch of the integrand defined by a straight branch cut connecting $z=1$ and $z=-1$, and $(z^2-1)^\frac{1}{2} > 0 $ on the line $y=0$, $x>1$. Note that the singularities are not isolated in this case. One way to do this integral is to expand the integrand in a Laurent series valid for $|{z}|> 1$ and integrate term by term
 A: Consider the contour $\Gamma$ below.

This consists of two pieces: the circle $z=R$, $R \gt 1$, minus the dogbone inside the circle, consisting of two small circular arcs at $z=\pm 1$ of radius $\epsilon$, and joined by line segments above and below the real axis.
Now consider $f(z) = \sqrt{(z-1)(z+1)}$.  On the upper line segment, $\arg{(z+1)} = 0$ and $\arg{(z-1)} = \pi$.  On the lower line segment, $\arg{(z+1)} = 0$, but $\arg{(z-1)} = -\pi$.  Thus, the contour integral about $\Gamma$ is, in the limit as $\epsilon \to 0$,
$$\oint_{\Gamma} dz \, \sqrt{z^2-1} = \int_{|z|=R} dz \, \sqrt{z^2-1} + i \int_1^{-1} dx \, \sqrt{1-x^2} - i \int_{-1}^1 dx \, \sqrt{1-x^2} $$
By Cauchy's theorem, the integral about $\Gamma$ is zero.  Thus,
$$ \int_{|z|=R} dz \, \sqrt{z^2-1}  = i 2 \int_{-1}^1 dx \, \sqrt{1-x^2} = i \pi$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
&\bbox[5px,#ffd]{\oint_{\verts{z}\ =\ R\ >\ 1}\,\,\,
\pars{z^{2} - 1}^{1/2}\,\,\dd z}\ =\
\overbrace{\oint_{\verts{z}\ =\ R\ >\ 1}\,\,\root{z + 1}\root{z - 1}
\,\,\dd z}^{\substack{\ds{Both\ root\ are\ de\!fined\ as} \\[1mm]
\ds{Principal\ Branchs}}}
\\[5mm] = &\
-\int_{-R}^{-1}\root{-x - 1}\expo{\ic\pi/2}\root{-x + 1}\expo{\ic\pi/2}
\,\dd x
\\[2mm] &\
-\int_{-1}^{1}\root{x + 1}\root{-x + 1}\expo{\ic\pi/2}\,\dd x
\\[2mm] &\
-\int_{1}^{-1}\root{x + 1}\root{-x + 1}\expo{-\ic\pi/2}\,\dd x
\\[2mm] &\
-\int_{-1}^{-R}\root{-x - 1}\expo{-\ic\pi/2}\root{-x + 1}\expo{-\ic\pi/2}
\,\dd x
\\[5mm] = &\
-2\ic\int_{-1}^{1}\root{1 - x^{2}}\,\dd x =
-4\ic\int_{0}^{\pi/2}\cos^{2}\pars{\theta}\,\dd\theta
\\[5mm] = &\
\bbx{-\pi\,\ic} \\ &
\end{align}
A: Hint: The value is independent of $R$ (by Cauchy's integral theorem). Let $R \to \infty$ and use the "residue at $\infty$" or compute the limit in another way (for example using that $(z^2-1)^{1/2} \approx z$ for $|z|$ large). You need to precise that estimate of course.
