Compute the integral $\int_C (z^2-1)^\frac{1}{2} dz$ where $R>1$

Question

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

Answer 1

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.

Answer 2

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,

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

Answer 3

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\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}