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How can I convert the integral $\int_0^{2\pi} (a^2 \cos^2 t +b^2\sin^2 t)^{-1} dt$ into an integral $\oint_\gamma z^{-1} dz$ where $z\in \mathbb C$ and $\gamma: {x^2\over a^2}+{y^2\over b^2}=1$? I can see that $|z|^2 = a^2 \cos^2 t +b^2\sin^2 t$ but the Jacobian seems very messy and I can't get the desired form.

Edit: Perhaps into a form not exactly $\oint_\gamma z^{-1} dz$ but up to a constant multiple?

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The point $(a\cos t, b\sin t)$ lies on the ellipse $\gamma$ in the $x$-$y$ plane. –  Dilip Sarwate Mar 6 '12 at 14:58
    
@DilipSarwate: Thanks. Actually I did see that, hence I suggested $|z|^2 = a^2 \cos^2 t +b^2\sin^2 t$, but I still am not getting the right form... –  Gore Mar 6 '12 at 15:27

1 Answer 1

up vote 4 down vote accepted

With

$$z=a\cos t+\mathrm ib\sin t\;,\\\bar z=a\cos t-\mathrm ib\sin t\;,$$

we have

$$\mathrm dz=(-a\sin t+\mathrm ib\cos t)\mathrm dt$$

and thus

$$\bar z\mathrm dz=\left(\mathrm i ab+(b^2-a^2)\sin t\cos t\right)\mathrm dt\;.$$

Thus

$$ \oint_\gamma z^{-1}\mathrm dz = \oint_\gamma\frac{\bar z\mathrm dz}{\bar zz} = \int_0^{2\pi}\frac{\mathrm i ab+(b^2-a^2)\sin t\cos t}{a^2\cos^2t+b^2\sin^2t}\mathrm dt\;. $$

The first term is $\mathrm iab$ times your integral, and the second vanishes since the integrand is an odd function of $t$ and the integral is over a full period.

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Thanks, joriki!!! –  Gore Mar 6 '12 at 18:05
    
@Gore: You're welcome. –  joriki Mar 6 '12 at 18:32

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