Compute the integral $\int_C (z^2-1)^\frac{1}{2} dz$ where $R>1$ Let C be the circle of Radius $R&gt;1$, 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} &gt; 0 $ on the line $y=0$, $x&gt;1$. 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 $|{z}|&gt; 1$ and integrate term by term
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,

$$ \int_{|z|=R} dz \, \sqrt{z^2-1}  = i 2 \int_{-1}^1 dx \, \sqrt{1-x^2} = i \pi$$
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.