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Cauchy Integral Theorem!

  1. Feb 16, 2012 #1
    1. The problem statement, all variables and given/known data

    ∮ dz/(2 - z*) over Curve |z|=1?

    where z* = conjugate of z

    How to solve this?

    2. Relevant equations

    I tried doing by taking z*=e^(-iθ) , the answer was zero
    Then i did it by taking z*=1/z which gives ∏i/2.


    3. The attempt at a solution
     
  2. jcsd
  3. Feb 16, 2012 #2

    HallsofIvy

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    Why would you take z*= 1/z??
     
  4. Feb 16, 2012 #3

    Dick

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    So you are taking z=e^(iθ), I hope. Both of those should give you the same answer and i*pi/2 looks like the right one. What did you do to get zero the first time?
     
  5. Feb 16, 2012 #4
    Yes, i did take z*=e^(-iθ).
    then ∫dz/(2-e^(-iθ)) = ∫ e^(iθ)/( 2e^(iθ) - 1)dz

    taking 2e^(iθ) -1 = t
    2ie^(iθ)dθ = dt

    But limits 0 →2π change to 1→1
    Therefore ans is zero!

    What i think is that the function is not analytic, therefore we cannot
    integrate over 0→2π directly. We must break the limits. For eg: if you
    take 0→π and multiply the Whole Integral by 2, you will get the ans as
    ∏i/2. But i cannot see any reason as to why the function ceases to be
    "non analytic".

    Thankyou. :)
     
  6. Feb 16, 2012 #5
    ceases to be "analytic" *
     
  7. Feb 16, 2012 #6
    Yes because we are integrating the function along the curve
    i.e the line integral. So because |z|=1 and since z=re^(iθ)
    hence r=1.
     
  8. Feb 16, 2012 #7

    Dick

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    The big problem is that doing the integral gives you a log(t) term. log is not a single valued function. You could assign log(1)=0, or 2*pi*i, or 4*pi*i or any other multiple of 2pi. So just because t goes from 1 to 1 you can't say the integral is zero. You can't do the complex integral that way. The safest way to do it would be to break it up into real integrals and do them.

    On the other hand using the Cauchy theorem is gives you the correct answer much more easily.

    BTW z* is not analytic. But z*=1/z on the unit circle. So the contour integral of both should come out the same.
     
    Last edited: Feb 16, 2012
  9. Feb 16, 2012 #8
    Thanx alot! But can you suggest some good mathematical books which
    explains these advanced concepts in proper detail.?

    Thanx again! :)
     
  10. Feb 16, 2012 #9

    Dick

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    You are welcome. Here's a link to some peoples opinions on good texts. https://www.physicsforums.com/showthread.php?t=243903 I used the Conway book, but I didn't comparison shop a lot.
     
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