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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!

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    also, both a,$b$are greater than epsilon.2012-10-10

2 Answers 2

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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

enter image description here

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

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

$\hskip1.3in$enter image description here

and then see what happens with the small branch going from $a$ to $b$ by taking the contour

$\hskip1.3in$enter image description here

and make the gap of the external circle go to zero. Then the result will be the contribution of the branch plus the pole.

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    @Henry I edited the answer for all cases.2012-10-10
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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.