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Unit circle and the roots

  1. Dec 20, 2012 #1
    1. The problem statement, all variables and given/known data

    Calculate the integral ∫dθ/(1+acos(θ)) from 0 to 2∏ using residues.

    2. Relevant equations

    Res[itex]\underline{zo}[/itex](z)=lim[itex]\underline{z->zo}[/itex] (z-z0)f(zo)*2∏i


    3. The attempt at a solution

    To start I sub cos(θ)=1/2(e^(iθ)+e^(-iθ)) so that de^(iθ)=ie^(iθ)dθ

    Re-writing in terms of e^(iθ) and then subbing in z for e^(iθ) I get dz/(iz(1+a(1/2)(z+z*))

    This becomes -2idz/(az^2+2z+a). Solving for the pole I get two imaginary roots. The problem is this is all done on the unit circle and the roots I'm getting depend on a so it means the poles are not necessarily in the circle. One should be. 0 is another root but is trivial.

    Also I tried solving this integral numerically and it gave me 0 but the answer should be 2∏/√(1-a^2)
     
    Last edited by a moderator: Feb 4, 2013
  2. jcsd
  3. Dec 20, 2012 #2

    haruspex

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    Re: Residues

    I think the answer depends on whether a > 1. If it is, the roots are complex and both lie inside the unit circle, and they cancel. If < 1, only one root lies inside the circle, so you ignore the other.
     
  4. Dec 21, 2012 #3

    vela

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    Re: Residues

    When |a|>1, you should get two complex roots, namely ##z=-\frac{1}{a} \pm i\sqrt{1-\left(\frac{1}{a}\right)^2}##. You should be able to convince yourself that both roots lie on the unit circle.

    z=0 isn't a root.

    From the answer, you should see that you must have |a|<1, so that's the case you're primarily interested in.
     
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