Results 1 to 2 of 2

Thread: Complex Integration

  1. #1
    Super Member Deadstar's Avatar
    Joined
    Oct 2007
    Posts
    722

    Complex Integration

    Right isn't so much a question as an 'am i doing this right'.

    Evaluate $\displaystyle \int_\delta \frac{z^2}{z - 1} dz$ where $\displaystyle \delta$ is the circle with centre 0 and radius 2, traversed in the usual anti-clockwise sense.

    And my answer...

    The function $\displaystyle \frac{z^2}{z - 1}$ is holomorphic on the complex plane except at z=1.

    So, using Cauchys integral formula we get $\displaystyle f(1) = \frac{1}{2 \pi i} \int_\delta \frac{f(z)}{z - 1} dz$ with $\displaystyle f(z) = z^2$.

    Hence, $\displaystyle \int_\delta \frac{f(z)}{z - 1} = 2 \pi i f(1) = 2 \pi i 1^2 = 2 \pi i$.

    This right?

    If so, how come the answer doesnt depend on $\displaystyle \delta$? Does it not matter what the radius of the circle is?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,
    Quote Originally Posted by Deadstar View Post
    Right isn't so much a question as an 'am i doing this right'.

    Evaluate $\displaystyle \int_\delta \frac{z^2}{z - 1} dz$ where $\displaystyle \delta$ is the circle with centre 0 and radius 2, traversed in the usual anti-clockwise sense.

    And my answer...

    The function $\displaystyle \frac{z^2}{z - 1}$ is holomorphic on the complex plane except at z=1.

    So, using Cauchys integral formula we get $\displaystyle f(1) = \frac{1}{2 \pi i} \int_\delta \frac{f(z)}{z - 1} dz$ with $\displaystyle f(z) = z^2$.

    Hence, $\displaystyle \int_\delta \frac{f(z)}{z - 1} = 2 \pi i f(1) = 2 \pi i 1^2 = 2 \pi i$.

    This right?
    Yes

    If so, how come the answer doesnt depend on $\displaystyle \delta$? Does it not matter what the radius of the circle is?
    It does depend on it.
    Because the pole z=1 is within the circle, you get the result.
    If it wasn't (say you considered $\displaystyle \delta$ as the circle of center 5 and radius 1), then the integral would have been 0.
    Look here : Cauchy's integral formula - Wikipedia, the free encyclopedia (they say 'Then for every a in the interior of D')
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Complex Integration II
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 12th 2010, 09:52 AM
  2. Complex Integration
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: Nov 9th 2009, 02:35 PM
  3. complex integration
    Posted in the Advanced Applied Math Forum
    Replies: 3
    Last Post: Oct 28th 2009, 03:36 AM
  4. More Complex Integration Please Help
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Oct 4th 2009, 09:45 PM
  5. Integration and complex nos
    Posted in the Calculus Forum
    Replies: 0
    Last Post: Dec 1st 2006, 05:45 AM

Search Tags


/mathhelpforum @mathhelpforum