Line Integrals (Complex Variables)

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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.
 

Answers and Replies

  • #2
HallsofIvy
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I wouldn't have used that parameterization. I would have used x= a cos(t), y= b sin(t) so that [itex]\gamma(t)= a cos(t)+ i b sin(t)[/itex]. You should find that simpler. The [itex]dz/z[/itex] becomes [itex](-a sin(t)+ i b cos(t))dt/(a cos(t)+ i b sin(t))[/itex]. Try multiplying both numerator and denominator by the complex conjugate of the denominator.
 
  • #3
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Oh, thanks Halls, that was great help.
How did you get that parametrization, are they simlpy obtained from the the elliptic coordinate transformations?
I found this on wikipedia:
"The most common definition of elliptic coordinates (μ,ν) is
x = a cosh [itex]\mu[/itex] cos v
y = a sinh [itex]\mu[/itex] sin v "
Can you show me how to get to your parametrization?

Thanks again.
 
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  • #4
HallsofIvy
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Those are ellipitic coordinates in a plane, not a parameterization of an ellipse. What does Wikipedia say [itex]\mu[/itex] and [itex]\nu[/itex] represent there?

My derivation starts from cos2(t)+ sin2(t)= 1 Since the equation of the ellipse is x2/a2+ y2/b2= 1, letting x/a= cos(t) and y/b= sin(t) should be obvious and that gives x= a cos(t), y= b sin(t).
 
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  • #5
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How can I evaluate integral 1/z over the ellipse?
 

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