Results 1 to 2 of 2

Math Help - Cauchy Integral

  1. #1
    Member
    Joined
    Oct 2008
    Posts
    156
    Evaluate  \oint_{C} \left(\frac{1}{(z+i)^{3}}- \frac{5}{z+i}+8 \right) \ dz where  C is the circle  |z+i| = 1 with positive orientation.

    So we have  \int_{C} \frac{1}{(z+i)^{3}} \ dz + -5 \int_{C} \frac{1}{z+i} \ dz + \int_{C} 8 \ dz . The last integral is  0 by Cauchy-Goursat. For the two other integrals why can't we say that they are both  0 by Cauchy-Goursat? Because they both have singularities at  z = -i .

    Only the first one is  0 and the second one is  -5(2 \pi i) = -10 \pi i by Cauchy Integral Formula. But thet are both analytic within and on  C . So why are they not both  0 ?

    yeah you had to use Cauchy Integral formula for derivatives because the function was "too singular."
    Last edited by mr fantastic; March 11th 2009 at 06:14 AM. Reason: Merged posts
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by manjohn12 View Post
    Evaluate  \oint_{C} \left(\frac{1}{(z+i)^{3}}- \frac{5}{z+i}+8 \right) \ dz where  C is the circle  |z+i| = 1 with positive orientation.

    So we have  \int_{C} \frac{1}{(z+i)^{3}} \ dz + -5 \int_{C} \frac{1}{z+i} \ dz + \int_{C} 8 \ dz . The last integral is  0 by Cauchy-Goursat. For the two other integrals why can't we say that they are both  0 by Cauchy-Goursat? Because they both have singularities at  z = -i .

    Only the first one is  0 and the second one is  -5(2 \pi i) = -10 \pi i by Cauchy Integral Formula. But thet are both analytic within and on  C . So why are they not both  0 ?
    The integrands of the other two integrals are not analytic inside and on C (there is a singularity at z = -i) so the Cauchy-Goursat Theorem cannot be applied.


    You know that f^{(n)}(\alpha) = \frac{n!}{2 \pi i} \oint \frac{f(z)}{(z - \alpha)^{n+1}} \, dz. Therefore:


    1.  \oint_{C} \frac{1}{(z+i)^3} \ dz = 0 since n = 2 and f(z) = 1.


    2.  \oint_{C} \frac{1}{(z+i)} \ dz = 2 \pi i since n = 0 and f(z) = 1.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: August 31st 2010, 08:38 AM
  2. [SOLVED] Line integral, Cauchy's integral formula
    Posted in the Differential Geometry Forum
    Replies: 7
    Last Post: September 16th 2009, 12:50 PM
  3. [SOLVED] Line integral, Cauchy's integral formula
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: September 15th 2009, 02:28 PM
  4. Another Cauchy integral
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: July 12th 2009, 07:53 PM
  5. Cauchy Integral
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: July 10th 2009, 12:26 AM

Search Tags


/mathhelpforum @mathhelpforum