Contour integral How does one find find $$\oint_{|z|=\epsilon} z^{-1}[(z-a)(z-b)]^{1\over 2}\,\,\,dz$$ where $\epsilon>0$ is small and $a,b>\epsilon$ and real.
My initial thought was to write it as $z=\epsilon\exp(i\theta)$, but then it doesn't work because we don't get small terms so we can't expand it. I also want to use the the residue theorem. Please help!
 A: The validity of the result holds as long as you take your branches correctly. For example, if $\epsilon < a < b$, you can take the branches (blue for $a$ and green for $b$) as

and if $a < \epsilon < b$, then you can take the branches as

In the thick part, the branches cancel, the path is well defined and you can use Cauchy's integral formula without problems,
$$
\oint_{|z| = \epsilon} \frac{\sqrt{(z-a)(z-b)}}{z} dz = 2\pi i \sqrt{(z-a)(z-b)}\Big|_{z = 0} = 2 \pi i \sqrt{ab}
$$
If both $a$ and $b$ are negative, the same argument applies.
Case $a < \epsilon$ and $b < \epsilon$
In this case, you have to take
$\hskip1.3in$
and then see what happens with the small branch going from $a$ to $b$ by taking the contour 
$\hskip1.3in$
and make the gap of the external circle go to zero. Then the result will be the contribution of the branch plus the pole.
A: Since $a,b \geq \epsilon$, you can assume that $((z-a)(z-b))^{\frac{1}{2}}$ is holomophic on $\{z \in \mathbb{C} : |z| \leq \epsilon\}$. This works because $x^\frac{1}{2}$ can be made holomorphic on $\mathbb{C} \setminus \{\lambda c :\lambda \in \mathbb{R}, \lambda \geq 0\}$ for every $c$.
The integrand thus has just one pole at zero. Note that the coefficient of $z^{-1}$ in the laurent series expansion of the integrand around that pole is the coefficient of $z^0$ in the taylor expansion of $((z-a)(z-b))^{\frac{1}{2}}$, also around that pole, which is $(ab)^{\frac{1}{2}}$.
Thus, by the residue theorem, the value of the contour integral is $2\pi i (ab)^{\frac{1}{2}}$.
The whole thing is a bit dubios, however, because the square root isn't a well-defined function - it's essentially a family of many functions $f$ which share the property that $f(x)^2=x$. Here, we have picked the most convenient of these functions, and computed a value of the contour integral. Whether or not another choice would have led to a different value we haven't answered.
