1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    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


    User Avatar
    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


    User Avatar
    Homework Helper

    You do know Cauchy's residue theorem right?
    \oint_{\gamma}f(z)dz=2\pi i\sum\textrm{Res}(f(z);\gamma )
    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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    To calculate the reside at i for example, compute:
    \textrm{Res}(f(x);i)=\lim_{z\rightarrow i}(z-i)f(z)
  10. Jun 22, 2011 #9
    Thanks....I think that helps!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook