Contour Integral $\oint\limits_{C}\mathrm{d}z \,\sqrt{\frac{z-a}{z-b}}$

Question

$C$ is any closed curve encompassing the whole branch cut. The approach to this problem would involve using the residue theorem:

1) We first want to find the residues at infinity so we change the form to where we are able to perform a series expansion.

2) Obtain the coefficients of $\frac{1}{z}$.

3) Plug the residues back into the formula: $\oint\limits_{C}\mathrm{d}z \, f(z)=2\pi i \, \mathrm{Res}(f,\infty)$

So for the integral: $\oint\limits_{C}\mathrm{d}z \,\sqrt{\frac{z-a}{z-b}}$

I would like some help in obtaining the form which allows for an expansion. If this approach is bad for this integral, then please inform me of a better solution.

A similar problem has been posted before here, for which $\oint\limits_{C}\mathrm{d}z \,\sqrt{(z-a)(z-b)} = \dfrac{\pi i}{4}(a-b)^2$. I am just referencing this because the problem is so similar as with our approach utilizing residues at infinity.

Answer 1

More complete version of Pedrpan's answer... As $z \to \infty$, we have $$ \frac{z-a}{z-b} = \frac{1-a/z}{1-b/z} = \left(1-\frac{a}{z}\right)\left(1+\frac{b}{z}+O(z^{-2})\right) = 1 +\frac{b-a}{z}+O(z^{-2}) $$ then one branch of the square root is analytic near $z=\infty$ and $$ \left(\frac{z-a}{z-b}\right)^{1/2} = 1 +\frac{b-a}{2z}+O(z^{-2}) $$ We chose the branch of the square root so that the integrand $ \to 1$ at $z=\infty$.

The residue at $z=\infty$ is $\frac{b-a}{2}$ so the integral is $$ 2\pi i \frac{b-a}{2} = i\pi(b-a) . $$

Answer 2

We have that $$\sqrt{\frac{z-a}{z-b}} = 1 + \frac{b-a}{2}\frac{1}{z} + O\bigg(\frac{1}{z^2}\bigg)$$ therefore using Cauchy's integral formula we have that $$\oint \sqrt{\frac{z-a}{z-b}}dz = 2 \pi i \frac{b-a}{2} = (b-a)\pi i .$$