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Help with complex integral

  1. Jun 21, 2011 #1
    1. The problem statement, all variables and given/known data

    Evaluate the integral along the path given:

    integral(along a(t) of (b^2-1)/(b^2+1) db ) where a(t)=2*e^(it) , 0 <= t <= 2*pi


    2. Relevant equations

    none

    3. The attempt at a solution

    I am thinking of using the Residue Theorem.

    I think there are poles at -i and +i.

    But not sure what to do to show this explicitly.
     
  2. jcsd
  3. Jun 22, 2011 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, the function
    [tex]\frac{z^2- 1}{z^2+ 1}= \frac{z^2- 1}{(z+ i)(z- i)}[/tex]
    has simple poles at z= i and z= -i.

    That is true because
    [tex]\lim_{z\to i}\frac{z^2- 1}{(z+i)(z-i)}[/tex]
    does not exist while
    [tex]\lim_{z\to i}(z- i)\frac{z^2- 1}{(z+i)(z-i)}= \lim_{z\to i|}\frac{z^2- 1}{z+ i}= i[/tex]
    and similarly at z= -i.
     
  4. Jun 22, 2011 #3
    How can I use these facts to evaluate the integral?
     
  5. Jun 22, 2011 #4

    hunt_mat

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    Homework Helper

    You do know Cauchy's residue theorem right?
    [tex]
    \oint_{\gamma}f(z)dz=2\pi i\sum\textrm{Res}(f(z);\gamma )
    [/tex]
    Calculate the residues and you have your answer.
     
  6. Jun 22, 2011 #5
    so the integral would be zero in this case since the residues are -i and i?
     
  7. Jun 22, 2011 #6

    hunt_mat

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    The contour is a circle of radius 2 centred on the origin, these poles certainly lie within that contour. Have you calculated the residues for the poles?
     
  8. Jun 22, 2011 #7
    No, not sure how to...
     
  9. Jun 22, 2011 #8

    hunt_mat

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    To calculate the reside at i for example, compute:
    [tex]
    \textrm{Res}(f(x);i)=\lim_{z\rightarrow i}(z-i)f(z)
    [/tex]
     
  10. Jun 22, 2011 #9
    Thanks....I think that helps!
     
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