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The exercise consists of determining the integral $\int_{\gamma} \frac{dz}{z}$ where $\gamma$ is the ellipse $x^2 + 2xy + 2y^2 = 1$ circulated once.

My approach was to first parametrise the ellipse: $x^2 + 2xy + 2y^2 = 1 \iff (x + y)^2 + y^2 = 1$ with $z(t) = \cos(t) - \sin(t) + i\sin(t)$. This yields the integral

$\int_0^{2\pi} \frac{- \sin{t} - \cos{t} + i\cos{t}}{\cos{t} - \sin{t} + i\sin{t}} dt = \int_0^{2\pi}\frac{ie^{it} - \cos{t}}{e^{it} - \sin{t}} dt,$

which I'm having a hard time solving.

I would greatly appreciate any suggestions on how to solve the integral, or if there is another more viable approach.

Thanks,

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    $\begingroup$ Try using Cauchy's Residue Theorem. $\endgroup$ – aleden May 11 '18 at 10:45
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Maybe this will help: $$\int_0^{2\pi}\frac{ie^{it} - \cos{t}}{e^{it} - \sin{t}} dt=\int_0^{2\pi}\frac{d( e^{it} - \sin{t})}{e^{it} - \sin{t}} =$$ $$=[\ln{e^{it} - \sin{t}}]_0^{2\pi}=(\ln{e^{2i\pi}-\sin{2\pi}})-(\ln{1}-\sin{0})=2i\pi$$

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  • $\begingroup$ I can't believe that I didn't see that. Very neat! $\endgroup$ – chrstnsn May 11 '18 at 12:04

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