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Complex analysis

  1. Apr 9, 2008 #1
    Also when trying to find the integral of (1/8z^3 -1) around the contour c=1.

    I found the singularities to be 1/2, 1/2exp(2pi/3), and 1/2exp(4pi/3)

    What is the next step here. Do I just assume the integral is 6pi(i) after using partial fractions to find the numerators of the 3 fractions.

    The answer is actually 0, but I dont understand how it was solved. I know it might have something to do with the rule that says, the sum of the integral of a C1, C2 and C3 is equal to the integral of the outside contour, but how do i know the orientation of the three contours inside the big contour C.

    Thanks, hope im not asking too many questions
     
  2. jcsd
  3. Apr 9, 2008 #2
    Find the function's residue at each of the poles inside the contour. Then use the residue theorem.
     
  4. Apr 9, 2008 #3


    We haven't covered residue theorem yet. Is there any other way to solve it?
     
  5. Apr 9, 2008 #4
    Of course there is. Find a parametrization of the contour and do it by hand.

    In general if you want to evaluate
    [tex]
    \int_C dz f(z)
    [/tex]

    where C is the contour, you can find a parametrization

    [tex]
    \gamma : [0,1]\to \mathbb{C}
    [/tex]
    such that [itex]\gamma([0,1])=C[/itex] (formalities omitted - you are certainly familiar with that :smile:)

    Then your integral becomes

    [tex]
    \int_0^1 dt f(\gamma(t))|\gamma'(t)|
    [/tex]

    This will give a not-too eays but doable integral. Note however, that normally it works the other way around. If you have some definite real integral, you can sometimes try to interpret it as a complex integral over a closed contour and use the residue theorem to evaluate it, but that's a different story.
     
    Last edited: Apr 9, 2008
  6. Apr 9, 2008 #5

    tiny-tim

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    Hi soulsearching! :smile:

    I'm a little confused (especially by the 6πi). :confused:

    Are you saying that you know how to integrate (1/2z - 1) round a contour containing z = 1/2?

    If so, just use the same contour for all three partition fractions! :smile:
     
  7. Apr 9, 2008 #6


    Like after breaking up the fraction into 3 fractions using partial fractions, what's the next step?

    Thanks... :)
     
    Last edited: Apr 9, 2008
  8. Apr 9, 2008 #7

    tiny-tim

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    erm … if you don't know how to integrate (1/2z - 1) round a contour containing z = 1/2, then there is no next step … you're stuck (or, rather, you have to follow Pere Callahan's method)!

    So … do you know ? … you didn't say. :smile:
     
  9. Apr 9, 2008 #8

    I think its going to be 2pi(i) because the singularity (1/2) is inside the contour. is this right?
     
  10. Apr 9, 2008 #9
    i did not get Pere Callahan's method. and why did you choose the contour as z=1/2? the question says integrate it around z=1
     
  11. Apr 9, 2008 #10

    tiny-tim

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    No, I said a contour containing z = 1/2.

    You're getting confused between z = 1/2 (a point) and |z| = 1/2 (a circle) … I did warn you about that in another thread :rolleyes: … you must write |z| when you mean it, or you'll lose track.

    The integral, of course, is the same for any contour that includes the same singularities. :smile:

    So you could, for example, choose the contour |z| = 100!

    And then integrate all three partial fractions round that.
    erm …
    :redface: I can't remember! …:redface:

    but I know it's 1/2 times something. :rolleyes:
     
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