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## Homework Statement

i) Define a path [itex]\gamma[/itex] whose image is the ellipse [tex]\frac{x^2}{a^2}[/tex] + [tex]\frac{y^2}{b^2}[/tex] = 1 traced counterclockwise.

ii) Show that [itex]\int[/itex] [itex]\frac{1}{z}[/itex] dz = [itex]\int[/itex] [itex]\frac{1}{z}[/itex] dz for a suitable circle [itex]\beta[/itex]

(NOTE: THE FIRST INTEGRAL IS OVER THE ELLIPSE [itex]\gamma[/itex], THE SECOND ONE IS OVER THE CIRCLE [itex]\beta[/itex]).

iii) Use the above result (part (ii)) to show that [itex]\int[/itex] [itex]\frac{1}{a^2 cos^2(t) + b^2 sin^2(t)}[/itex] dt = [itex]\frac{2\pi}{ab}[/itex] for a>0, b>0.

## Homework Equations

Green's Theorem, Cauchy's Theorem... Cauchy's formula too maybe?

## The Attempt at a Solution

Ok, so first of all I parametrized the ellipse, I got the following:

[itex]\gamma[/itex] (t) = t - i(b/a) [itex]\sqrt{a^2 - t^2}[/itex] for -a[itex]\leq[/itex] t [itex]\leq[/itex] a

[itex]\gamma[/itex] (t) = (2a-t) + i(b/a) [itex]\sqrt{a^2 - (2a-t)^2}[/itex] for a[itex]\leq[/itex] t [itex]\leq[/itex] 3a

I've checked and I'm pretty sure that is a correct parametrization for the ellipse given.

Now, my real problem is with ii) and iii). For ii) I'm thinking about using Cauchy's Theorem maybe... because the function 1/z is analytic everywhere except at the origin, so a line integral over an ellipse centered at the origin and a circle that's also centered at the origin would be the same (since the function is analytic everywhere in the space between the two paths). However, I'm not that sure on how to "show" that, mathematically. The corrector checks the homeworks very strictly so I need to show it very clearly. Any ideas? As for iii) well I need ii) first...

Any suggestions are greatly appreciated. Cheers.