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Complex integral

  1. Dec 20, 2006 #1
    1. The problem statement, all variables and given/known data
    [tex]\oint _{|z+i|=1} \frac{e^z}{1+z^2} dz =?[/tex]

    3. The attempt at a solution

    I substituted z+i=z' and [itex]z'=e^{i\theta}[/tex] to arrive at

    [tex]e^{-i} \int _0 ^{2 \pi} \frac{e^{e^{i \theta}}}{-ie^{i \theta}-2} d \theta[/tex]

    I have no clue how to solve such an integral, any ideas??

    (I also did a similar excercise to arrive at the same integral but now [itex]sin(\pi/4 + exp(i \theta))[/tex] in the numerator. Are these kind of integrals analytically solvable??)
     
    Last edited: Dec 20, 2006
  2. jcsd
  3. Dec 20, 2006 #2

    TD

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    Factor the denominator: z²+1 = (z+i)(z-i), then partial fractions.
    Do you know Cauchy's integral formula? It states for a inside C:

    [tex]f(a) = {1 \over 2\pi i} \oint_C {f(z) \over z-a}\, dz [/tex]
     
  4. Dec 20, 2006 #3
    Thanks a lot! Should have thought of that of course, but now I know I can also make the others, great help!
     
  5. Dec 20, 2006 #4

    TD

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    No problem :smile:
     
  6. Dec 20, 2006 #5
    Well, maybe I can bother you with one more question? Most of em I can do, but there is this this one with a denominator 1+z^4 which I don't know how to seperate. I tried (z^2+i)(z^2-i) but then I can't seperate these...

    Do you maybe have an idea?
     
  7. Dec 20, 2006 #6

    TD

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    You need to find +/- sqrt(i) and +/- sqrt(-i). It factors like this:

    [tex]
    \left( {z + \frac{{\sqrt 2 + \sqrt 2 i}}{2}} \right)\left( {z + \frac{{\sqrt 2 - \sqrt 2 i}}{2}} \right)\left( {z - \frac{{\sqrt 2 + \sqrt 2 i}}{2}} \right)\left( {z - \frac{{\sqrt 2 - \sqrt 2 i}}{2}} \right)
    [/tex]
     
  8. Dec 20, 2006 #7
    Worked like a charm! Thanks a lot!
     
  9. Dec 20, 2006 #8

    TD

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    You're welcome :smile:
     
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