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Thread: residue

  1. #1
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    residue

    I have an integral to evaluate via contour integration:
    $\displaystyle ({1 \over 1+cos^2\theta}) \ wrt d\theta$
    Using the substitution $\displaystyle z =e^{i\theta}$I find the integral becomes,with $\displaystyle d\theta \ becoming \ {dz \over iz}$
    $\displaystyle {4z^2 \over z^4 + 6z^2 + 1}$ is that correct?
    Also how would I evaluate this via the residue theorem?
    Last edited by bigdoggy; Dec 3rd 2009 at 02:08 PM.
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  2. #2
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    That depends what you're integrating. For example, suppose:

    $\displaystyle \int_0^{2\pi}\frac{1}{1+\cos^2(t)}dt$

    and I let $\displaystyle z=e^{it}$, then I'd get:

    $\displaystyle -i\mathop\oint\limits_{|z|=1}\frac{4z}{z^4+6z^2+1}d z=(-i) 2\pi i \sum\mathop\text{Res}\frac{4z}{z^4+6z^2+1}$

    where the sum is over the residues of the enclosed poles.
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  3. #3
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    I've written it differently:

    $\displaystyle {1\over i} \oint ({3\over2z}+{z\over4}+{1\over4z^3})^{-1}dz$

    how can I pick the relevant term (the cofefficient of 1/z) to give the residue? The inverse is complicating matters....
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  4. #4
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    I don't see why you'd want to write it that way. Just factor the denominator of the expression I got, get the enclosed zeros, bingo-bango.
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  5. #5
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    Quote Originally Posted by shawsend View Post
    I don't see why you'd want to write it that way. Just factor the denominator of the expression I got, get the enclosed zeros, bingo-bango.
    Thanks.
    So solve $\displaystyle z^4 + 6z^2 +1 =0$ to begin with...? But how would I?
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by bigdoggy View Post
    Thanks.
    So solve $\displaystyle z^4 + 6z^2 +1 =0$ to begin with...? But how would I?
    Let $\displaystyle z^2=k$..."bingo-bango"!
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    Quote Originally Posted by Drexel28 View Post
    Let $\displaystyle z^2=k$..."bingo-bango"!
    Hi, which gives $\displaystyle k=-3 \pm2\sqrt2 =z^2$
    Am I doing what you're implying...!? But how can I get the poles here...?
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  8. #8
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by bigdoggy View Post
    Hi, which gives $\displaystyle k=-3 \pm2\sqrt2 =z^2$
    Am I doing what you're implying...!? But how can I get the poles here...?
    The poles are merely the zeros of the denominator. You know that $\displaystyle z^2=-3\pm \sqrt{2}\implies\text{Denominator equals zero}$ therefore $\displaystyle z=?\implies\text{Denominator equals zero}$?
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  9. #9
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    Quote Originally Posted by Drexel28 View Post
    The poles are merely the zeros of the denominator. You know that $\displaystyle z^2=-3\pm \sqrt{2}\implies\text{Denominator equals zero}$ therefore $\displaystyle z=?\implies\text{Denominator equals zero}$?
    I can't see it being $\displaystyle z= \pm\sqrt{-3\pm \sqrt{2}}$
    I'm missing something here ...
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  10. #10
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by bigdoggy View Post
    I can't see it being $\displaystyle z= \pm\sqrt{-3\pm \sqrt{2}}$
    I'm missing something here ...
    You screwed up the quadratic equation.

    - Wolfram|Alpha -Step 1

    - Wolfram|Alpha -Step 2...validattion.
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  11. #11
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    Thankyou.

    So if the integrand is around |z|=1, the solution is $\displaystyle z=\pm i\sqrt{3 -2\sqrt2}$
    I can't see how this would give a 'nice' (multiple of pi) answer (which I was told it should!)....

    I find the residue = $\displaystyle 1 \over \sqrt2$ correct??
    Last edited by bigdoggy; Dec 3rd 2009 at 05:47 PM.
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  12. #12
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    The zeros are $\displaystyle z_0=i\sqrt{3+2\sqrt{2}},z_1=i\sqrt{3-2\sqrt{2}}, z_2=-i\sqrt{3+2\sqrt{2}},z_3=-i\sqrt{3-2\sqrt{2}}$. Now, which ones are in the unit circle? Isn't it the $\displaystyle 3-2\sqrt{2}$ ones? So then we have:

    $\displaystyle \int_0^{2\pi} \frac{1}{1+\cos^2(t)}dt=2\pi \mathop\text{Res}\limits_{z_1,z_3} \left\{\frac{4z}{z^4+6z^2+1}\right\}$.

    I'll do one of the residues:

    $\displaystyle r_1=\frac{4z}{(z-i\sqrt{3+2\sqrt{2}})(z+i\sqrt{3+2\sqrt{2}})(z+i\sq rt{3-2\sqrt{2}})}\biggr|_{z=i\sqrt{3-2\sqrt{2}}}=\frac{1}{2\sqrt{2}}$

    Edit: Also, forgot to mention, if it's a simple pole and the derivative of the denominator is not zero there then:

    $\displaystyle \mathop\text{Res}\limits_{z=z_0} \frac{f}{g}=\frac{f(z_0)}{g'(z_0)}$

    May be easier to calculate these residues this way.
    Last edited by shawsend; Dec 4th 2009 at 09:27 AM.
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