Complex Analysis: Integration

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
 

gabbagabbahey

Homework Helper
Gold Member
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6
[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
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:

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