Evaluate $\oint\limits_{C}\mathrm{d}z \,\sqrt{(z-a)(z-b)}$ The premise of this integral is that $a$ and $b$ can take on any complex value except ones in which $a=b$. A finite branch cut for the square root is given and the contour is any closed curve encompassing the whole branch cut.
The approach to this problem would involve using the residue theorem:
1) Change the form to: $z\sqrt{1-\frac{a+b}{z}+\frac{ab}{z^2}}$
2) Perform a series expansion $\sqrt{1-Z}$
3) Getting the coefficients of $\frac{1}{z}$
4) Plugging back in and evaluating the integral: $\oint\limits_{C}\mathrm{d}z \, f(z)=-2\pi \, \mathrm{Res}(f,\infty)$
I am not sure if this is the correct approach to solving this integral and am a bit lost when it comes to step 3 in particular.
A similar problem has been posted before here, which is where I am referencing my general approach to the problem. That thread is effectively dead and no definitive answer or proper solution seems to have come out of it.
 A: It is a correct approach. There is a branch of $f(z) = z \sqrt {1 - a/z} \hspace {1px} \sqrt {1 - b/z}$ which is analytic around infinity, and the integral for that branch can be evaluated as $-2 \pi i$ (if $C$ goes around the origin counterclockwise) times the residue at infinity. The assumption $a \neq b$ is not necessary.
The Laurent expansion around infinity (on $|z| > R$ for some $R$) can be found from the binomial theorem. $\sum_{k \geq 0} \binom {1/2} k x^k y^{1/2 - k}$ converges for $|x| < |y|$, so we take $y = 1$:
$$f(z) = z
 \sum_{j \geq 0} \binom {1/2} j \left( -\frac a z \right)^j
 \sum_{k \geq 0} \binom {1/2} k \left( -\frac b z \right)^k.$$
Note that this assumes that we chose a branch of the square root for which $\sqrt 1 = 1$. To find the coefficient $c_{-1}$ at $z^{-1}$, we need to take the terms with $j + k = 2$. The residue at infinity is $-c_{-1}$.
A: Hint: using an affine transformation, we can reduce this to the case $a=0, b=1$.  Then take a contour that goes from $0$ to $1$ just below the real line and returns just above. 
