Results 1 to 3 of 3
Like Tree1Thanks
  • 1 Post By HallsofIvy

Math Help - Complex integral theorem

  1. #1
    Member
    Joined
    Oct 2009
    Posts
    195
    Thanks
    1

    Complex integral theorem

    This is a theorem from my complex analysis book, and one of the exercises is to prove it.

    When n = 1, the result follows directly from Cauchys integral formula. When n is 0 or negative, the integral is just a regular polynomial. A regular polymoial is analytical everywhere, and an integral around a close curve around it is zero.

    These two cases I get, but why is this integral 0 when n is a positive integer greater than or equal to 2?

    Thanks!
    Attached Thumbnails Attached Thumbnails Complex integral theorem-complexintegral.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    14,973
    Thanks
    1121

    Re: Complex integral theorem

    First, your two cases are "n= 1" and " n\ne 0" which does not make sense. I assume you meant " n\ne 1".


    You also don't state what the curve is. Obviously, if the curve does not have z_0 in its interior, the the integrand is analytic there so the integral is 0. So I presume that C is a curve having z_0 in its interior. In that case, we can transform the curve into a circle with center at z_0 without changing the integral because the integrand is analytic at every point between the given curve and such a circle.

    So we can assume that C is the circle with center at z_0 and radius R. In that case z= z_0+ Re^{i\theta} where \theta goes from 0 to 2\pi. Of course then dz= Ri e^{i\theta}d\theta, z- z_0= Re^{i\theta}, and the integral becomes \int_0^{2\pi}\frac{Ri e^{i\theta}d\theta}{R^ne^{ni\theta}}= R^{1-n}i\int_0^{2\pi}e^{i(1-n}\theta}d\theta.

    IF n\ne 1, then the integral is \frac{R^{1- n}}{1-n}\left[e^{i(1-n)\theta\right]_0^{2\pi}. But the exponential is 1 at both 0 and 2\pi i so the integral is 0.

    IF n= 1,then the integral is i\int_0^{2\pi} d\theta= 2\pi i
    Thanks from gralla55
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2009
    Posts
    195
    Thanks
    1

    Re: Complex integral theorem

    Your assumptions are correct, I should learn to be more presise. Thank you very much, I didn't think of it that way!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: November 27th 2012, 12:51 PM
  2. Complex analysis, Cauchy's integral theorem
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: August 21st 2011, 12:06 AM
  3. Complex variables(Cauchy Integral Theorem)
    Posted in the Advanced Applied Math Forum
    Replies: 0
    Last Post: October 18th 2010, 06:04 PM
  4. Volume integral to a surface integral ... divergence theorem
    Posted in the Advanced Applied Math Forum
    Replies: 3
    Last Post: October 7th 2010, 06:11 PM
  5. Replies: 2
    Last Post: August 31st 2010, 07:38 AM

Search Tags


/mathhelpforum @mathhelpforum