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Suppose we had a function defined as $$ S(\zeta) = \zeta - \sqrt{\zeta^2 - c^2} $$

and we wish to evaluate $$ \oint_\gamma \frac{S(\zeta)}{z-\zeta}\, d\zeta, $$ where $\gamma$ is the (positively orientated) ellipse $$ \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1, \quad a > b > 0,\quad c^2= a^2-b^2 $$ and $z$ lies outside $\gamma$. First I notice that by Cauchy's theorem, $$ \oint_\gamma \frac{\zeta}{z-\zeta}\, d\zeta = 0 $$ since the integrand is holomorphic inside $\gamma$. So I am left with $$ -\oint_\gamma \frac{ \sqrt{\zeta^2 - c^2}}{z-\zeta}\, d\zeta $$ Is there a clever way to evaluate this without using a parameterisation? Letting $$ \zeta = a \cos t + ib\sin t, \quad 0 \leq t \leq 2\pi $$ leads me to $$ -\oint_\gamma \frac{ \sqrt{\zeta^2 - c^2}}{z-\zeta}\, d\zeta = \int_0^{2\pi} \frac{(b\cos t + ia\sin t)^2}{iz-ia\cos t + b\sin t} \, dt $$ which looks hard to evaluate.

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    $\begingroup$ $\sqrt{}$ is multivalued function. Should one choose a branch first? $\endgroup$ Nov 11, 2015 at 15:50
  • $\begingroup$ The chosen branch is such that for $\sqrt{\zeta^2 - c^2} > 0$ on the interval $(c, \infty)$ (I forgot to mention $c > 0$), with a branch cut $[-c,c]$. $\endgroup$ Nov 11, 2015 at 16:26
  • $\begingroup$ then contract your integral to the branch cut, and reverse the path of integration. your integral is then equal to the residues at $z$ and $\infty$ times a phase which is due to the discontinuity over the cut $\endgroup$
    – tired
    Nov 11, 2015 at 16:33

1 Answer 1

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First observe the the branch points corresponds to the two foci of the ellipse that constitutes your path of integration,therefore the branch cut is inside $\gamma$ . Now we do a little trick, instead of picking up the singularities inside of our contour, we pick up the ones at its outside (because it is easier to count residues then integrating around a branch cut). This yields

$$ I(z)=\oint_{\gamma}\frac{S(\zeta)}{z-\zeta}d\zeta=-2\pi i(\text{Res}(\zeta=\infty)+\text{Res}(\zeta=z)) $$

Note the minus sign, which stemms from the fact the we encircle the poles in clockwise direction now. The residues are given (choosing the correct branch of $\log$ which induces that $\pm i=e^{\pm i \pi/2}$)

$$ \text{Res}(\zeta=\infty)=0\\ \text{Res}(\zeta=z)=z-\sqrt{z^2-c^2} $$

and therefore

$$ I(z)=2\pi i(\sqrt{z^2-c^2}-z) $$

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  • $\begingroup$ OK! And if $z$ was inside $\gamma$, the same trick can be used, giving zero? $\endgroup$ Nov 11, 2015 at 20:27
  • $\begingroup$ it seems so, yes $\endgroup$
    – tired
    Nov 11, 2015 at 20:51

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