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Complex Analysis: Integration

  1. Mar 6, 2009 #1
    1. The problem statement, all variables and given/known data

    Evaluate the following integral for 0<r<1 by writing [tex]\cos\theta = \frac{1}{2}(e^{i\theta} + e^{-i\theta})[/tex] reducing the given integral to a complex integral over the unit circle.

    [tex]Evaluate: \displaystyle{\frac{1}{2\pi}\int_0^{2\pi}\frac{1}{1-2r\cos\theta + r^2}\,d\theta}[/tex]


    2. Relevant equations

    none

    3. The attempt at a solution

    [tex]\displaystyle\cos\theta = \frac{1}{2}(e^{i\theta} + e^{-i\theta})}[/tex]

    [tex]\displaystyle{z=e^{i\theta}}[/tex]

    [tex]\displaystyle{ \cos t= \frac{1}{2}(z+ \frac{1}{z})}[/tex]

    [tex]\displaystyle{\frac{dz}{iz}= dt}[/tex]

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

    [tex]\displaystyle{\frac{1}{2\pi}\oint \frac{1}{1-2r[\frac{1}{2}(z+\frac{1}{z})]+r^2}\,\frac{dz}{iz}}[/tex]

    [tex]\displaystyle{\frac{1}{2\pi}\oint \frac{1}{z-2rz[\frac{1}{2}(z+\frac{1}{z})]+r^2}\,\frac{dz}{i}}[/tex]

    [tex]\displaystyle{\frac{-i}{2\pi}\oint \frac{1}{z-2rz[\frac{1}{2}(z+\frac{1}{z})]+r^2}\,dz}[/tex]

    [tex]\displaystyle{\frac{-i}{2\pi}\oint \frac{1}{z-r^2z^2-r+r^2}\,dz}[/tex]

    But I get stuck here. What do I do with the "r"? Should I factor it out, and if yes then how?

    Thanks
     
  2. jcsd
  3. Mar 6, 2009 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Leave the "r" where it is, the location of the poles will depend on its value...find those poles by solving the quadratic [itex]z-r^2z^2-r+r^2=0[/itex] for [itex]z[/itex]...

    Edit: You've also got a couple of algebra errors, double check your last 4 steps
     
    Last edited: Mar 6, 2009
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