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Complex Variables

  1. Apr 9, 2008 #1
    Can anyone help me with this pls?

    How can you prove that the integral of f(z) around the contour z= 1 is 0

    where f(z) is Log(z+5)


    I know Log(z) is ln r + i (theta). But i dont know how that applies to this situation.

    Also, do I solve it as a normal integral or use the Cauchy Goursat theorem to prove that its integral is zero?
  2. jcsd
  3. Apr 9, 2008 #2


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    Welcome to PF!

    Hi soulsearching! Welcome to PF! :smile:

    Do you mean |z| = 1?

    If so, then z + 5 is just the circle of radius 1 and centre at 5. It doesn't include the origin (z + 5 = 0), so yes … always do it the easy way … use Cauchy Goursat! :smile:
  4. Apr 9, 2008 #3

    So the point z = 5 is outside the contour, so the integral vanishes right? Thank you Tim.

    When trying to show the same for this equation here around the same contour

    f(Z) = (4z^2 -4z +5)^-1

    I found the singularities here to be (1/2 + i) and (1/2-i). Is the integral zero because both singularities are outside the circle of radius 1? Thank you.

    I have a major exam tmr and my professor is not being of much help, but I really appreciate your help. :-)
  5. Apr 9, 2008 #4

    And in this case your integrand is analytic inside and on the circle |z| = 1, so again it's 0 by cauchy's.
  6. Apr 9, 2008 #5
    Thanks Dan.
  7. Apr 9, 2008 #6


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    Looks good to me! :smile:
  8. Apr 9, 2008 #7
    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
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