Line Integrals (Complex Variables)

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SUMMARY

This discussion focuses on evaluating line integrals in complex variables, specifically using an ellipse and a circle as paths. The parametrization of the ellipse is defined as \(\gamma(t) = a \cos(t) + i b \sin(t)\) for \(t\) in the range \([0, 2\pi]\). The participants discuss applying Cauchy's Theorem to show that the integral \(\int \frac{1}{z} dz\) over the ellipse \(\gamma\) is equal to that over a circle \(\beta\). Additionally, they derive the integral \(\int \frac{1}{a^2 \cos^2(t) + b^2 \sin^2(t)} dt = \frac{2\pi}{ab}\) for \(a > 0\) and \(b > 0\).

PREREQUISITES
  • Understanding of complex variables and line integrals
  • Familiarity with Cauchy's Theorem and its applications
  • Knowledge of parametrization techniques for curves
  • Basic principles of Green's Theorem
NEXT STEPS
  • Study the application of Cauchy's Integral Theorem in complex analysis
  • Learn about parametrizing different curves in the complex plane
  • Explore the derivation and applications of Green's Theorem
  • Investigate the properties of elliptic coordinates and their transformations
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Students and professionals in mathematics, particularly those focusing on complex analysis, as well as educators preparing coursework on line integrals and related theorems.

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Homework Statement


i) Define a path \gamma whose image is the ellipse \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 traced counterclockwise.
ii) Show that \int \frac{1}{z} dz = \int \frac{1}{z} dz for a suitable circle \beta
(NOTE: THE FIRST INTEGRAL IS OVER THE ELLIPSE \gamma, THE SECOND ONE IS OVER THE CIRCLE \beta).
iii) Use the above result (part (ii)) to show that \int \frac{1}{a^2 cos^2(t) + b^2 sin^2(t)} dt = \frac{2\pi}{ab} 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:

\gamma (t) = t - i(b/a) \sqrt{a^2 - t^2} for -a\leq t \leq a
\gamma (t) = (2a-t) + i(b/a) \sqrt{a^2 - (2a-t)^2} for a\leq t \leq 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.
 
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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 \gamma(t)= a cos(t)+ i b sin(t). You should find that simpler. The dz/z becomes (-a sin(t)+ i b cos(t))dt/(a cos(t)+ i b sin(t)). Try multiplying both numerator and denominator by the complex conjugate of the denominator.
 
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 \mu cos v
y = a sinh \mu sin v "
Can you show me how to get to your parametrization?

Thanks again.
 
Last edited:
Those are ellipitic coordinates in a plane, not a parameterization of an ellipse. What does Wikipedia say \mu and \nu 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).
 
Last edited by a moderator:
How can I evaluate integral 1/z over the ellipse?
 

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