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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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There are branch cuts in the integrand, you have to be carefull. –  Pragabhava Oct 10 '12 at 19:45
    
...Why did you use the word 'residue'? Maybe because for this integral you want to use the residue theorem? Do you know binomial series (with noninteger exponents)? Also, do we know that $a,b\ne 0$? –  Berci Oct 10 '12 at 19:45
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Is this valid?: To find the residue of a function $f(z)$ at $z = a$, we can look at the coefficient of $1/(z-a)$ in the Laurent expansion of $f(z)$ about $z = a$. From WolframAlpha, we can find that the coefficient is $\sqrt{ab}$, and so by the Residue theorem: $$\oint_{|z|=\epsilon} z^{-1}[(z-a)(z-b)]^{1\over 2}\ dz = 2\pi i \sqrt{ab}.$$ Do branch cuts prevent use of the Residue theorem? –  Michael Zhao Oct 10 '12 at 19:54
    
@Berci: yes, thanks for correcting me. edited. –  Henry Oct 10 '12 at 19:58
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also, both a, b are greater than epsilon. –  Henry Oct 10 '12 at 20:18
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2 Answers

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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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Excellent graphics, thanks! –  Henry Oct 10 '12 at 23:29
    
@Henry I edited the answer for all cases. –  Pragabhava Oct 10 '12 at 23:36
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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.

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