Complex Analysis – Contour Integral of sqrt(z/(z-1))

Question

Which Laurent series could be used to solve
$$\oint_{|z|=2}\sqrt{\dfrac{z}{z-1}}dz$$
if it has a branch cut at $y=0$, $x\in(0,1)$?

I thought maybe it couldn't be solved by residue theorem because there is a branch cut inside the contour.

Answer 1

Set for all complex number $z$ such that $|z|>1$ $$\sqrt{\frac{z}{z-1}} = \Big(1-\frac{1}{z}\Big)^{-1/2} = \sum_{n=0}^{+\infty} (-1)^n {-1/2 \choose n}\frac{1}{z^n}.$$
Set $\gamma(t) = 2e^{it}$ for $t \in [0,2\pi]$. The series giving $\sqrt{\gamma(t)/(\gamma(t)-1)}\gamma'(t)$ is normally convergent, so we derive $$\oint_{|z|=2}\sqrt{\dfrac{z}{z-1}}dz = \sum_{n=0}^{+\infty} (-1)^n {-1/2 \choose n}\oint_{|z|=2}\frac{dz}{z^n} = i2\pi \times (-1)^1{-1/2 \choose 1} = i\pi.$$