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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

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Consider the contour $\Gamma$ below.

dogbone

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,

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

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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.

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  • $\begingroup$ So would the integral just be zero via Cauchy Integral Theorem? Or is there some aspect of the question I'm not grasping. $\endgroup$ – ktgster Mar 7 '15 at 3:48
  • $\begingroup$ No, you need a slightly better estimate as $R\to \infty$ (you need to take possible $1/z$ terms into account). A third option is to shrink the curve to the interval $[-1,1]$ and compute the resulting integral using calculus techniques. $\endgroup$ – mrf Mar 7 '15 at 8:48
  • $\begingroup$ Ok, so I have computed a Laurent series in the form $z - \frac{1}{2z} - \frac{1}{8z^3} - \frac{1}{16z^5} - \frac{5}{128z^7} + .... $ that converges for $R \rightarrow \infty$ and the by the Cauchy Integral Formula computed the integral to be $2 \pi i * (-1/2) = -\pi i$ $\endgroup$ – ktgster Mar 7 '15 at 14:52

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