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Math Help - complex analysis

  1. #1
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    complex analysis

    I=\int^{2\pi}_{0}\frac{cos(3\theta)}{5-4\cos(\theta)}d\theta

    z=e^{i\theta},\frac{dz}{d\theta}=i\\e^{i\theta}=i\  \z\Rightarrow\\d\theta=\frac{dz}{i\\z}

    I=\oint_{C_{1}(0)}\frac{z^{3}+z^{-3}}{2\\i(-2z^{2}+5z-2)}dz

    factorizing gives

    I=\oint_{C_{1}(0)}\frac{z^{3}+z^{-3}}{2\\i(z-2)(z-\frac{1}{2})}dz

    now from here.. i need to do 2\pi\\i*\sum\\Res\\|_{z} but everytime i do it, i get dumb answers, im pretty sure the answer is \frac{\pi}{12}

    please can anyone help me.. i may have done something stupid
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  2. #2
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    I think the residues are at z=0,1/2 and 2... but not sure... and the contour only encloses z=0 and 1/2
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by fredrick08 View Post
    I=\int^{2\pi}_{0}\frac{cos(3\theta)}{5-4\cos(\theta)}d\theta
    Rewrite it as iI=\int_0^{2\pi}\frac{\left(e^{i\theta}\right)^2}{  5-4\frac{e^{i\theta}+\frac{1}{e^{i\theta}}}{2}}e^{i\  theta}d\theta but this is clearly \underset{|z|=1}{\oint}\frac{z^2}{5-2(z+\frac{1}{z})}dz=\underset{|z|=1}{\oint}\frac{-z^3}{2(z-2)(z-\tfrac{1}{2})}. But, clearly f(z)=\frac{-z^3}{2(z-2)} is analytic on \mathbb{D} and so Cauchy's integral formula tells us that iI=2\pi i\frac{-(\tfrac{1}{2})^3}{2(\tfrac{1}{2}-2)}=\frac{\pi i}{12}\implies I=\frac{\pi}{12}

    .
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  4. #4
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    ok yes, that makes perfect sense... but could please do it by residue theorem, coz thats how we are meant to atm.. srry but also.... how did u get the -z^3 on top.... where does the z^-3 go?

    because i know that z^{3}+z^{-3}=\frac{z^{6}+1}{z^{3}} can that work at all?
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by fredrick08 View Post
    ok yes, that makes perfect sense... but could please do it by residue theorem, coz thats how we are meant to atm
    .........

    Ok, well let f(z) be as above and note that by it's analyticity that f(z)=\sum_{n=0}^{\infty}\frac{f^{(n)}(\tfrac{1}{2}  )(z-\tfrac{1}{2})^n}{n!} and so \frac{f(z)}{z-\tfrac{1}{2}}=\frac{f(\tfrac{1}{2})}{z-\tfrac{1}{2}}+\frac{1}{2}f'(\tfrac{1}{2})(z-\tfrac{1}{2})+\cdots and so \underset{z=\tfrac{1}{2}}{\text{Res}}\text{ }\frac{f(z)}{z-\tfrac{1}{2}}=f(\tfrac{1}{2})..........
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by fredrick08 View Post
    -z^3 on top.... where does the z^-3 go?
    \frac{z^2}{5-2(z+\tfrac{1}{z})}=\frac{z^3}{5z-2(z^2+1)}=\frac{z^3}{-2z^2+5z-2}=\frac{-z^3}{2z^2-5z+2}
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  7. #7
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    Quote Originally Posted by Drexel28 View Post
    \frac{z^2}{5-2(z+\tfrac{1}{z})}=\frac{z^3}{5z-2(z^2+1)}=\frac{z^3}{-2z^2+5z-2}=\frac{-z^3}{2z^2-5z+2}
    srry i dont understand how u went from


    \int^{2\pi}_{0}\frac{cos(3\theta)}{5-4\cos(\theta)}d\theta


    iI=\int_0^{2\pi}\frac{\left(e^{i\theta}\right)^2}{  5-4\frac{e^{i\theta}+\frac{1}{e^{i\theta}}}{2}}e^{i\  theta}d\theta

    how can u change cos(3theta) which is 1/2(z^3+z^-3) to z^2???

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  8. #8
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by fredrick08 View Post
    srry i dont understand how u went from


    \int^{2\pi}_{0}\frac{cos(3\theta)}{5-4\cos(\theta)}d\theta


    iI=\int_0^{2\pi}\frac{\left(e^{i\theta}\right)^2}{  5-4\frac{e^{i\theta}+\frac{1}{e^{i\theta}}}{2}}e^{i\  theta}d\theta

    how can u change cos(3theta) which is 1/2(z^3+z^-3) to z^2???

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    Oops, forgot to explain that. I'm kinda used to just writing stuff. I'll let you explain, think about \int_0^{2\pi}\frac{e^{3i\theta}}{5-4\cos(\theta)}d\theta and how that relates to the real part of the residue? It just happens that the real part of the residue is the entire residue and so I just omitted it. My sloppy mistake
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  9. #9
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    Quote Originally Posted by Drexel28 View Post
    Oops, forgot to explain that. I'm kinda used to just writing stuff. I'll let you explain, think about

    \int_0^{2\pi}\frac{e^{3i\theta}}{5-4\cos(\theta)}d\theta

    and how that relates to the real part of the residue? It just happens that the real part of the residue is the entire residue and so I just omitted it. My sloppy mistake
    srry i dont know what you mean by real part of residue??

    are u saying that \cos(3\theta)=\frac{1}{2}(z^{3}+z^{-3})=z^{3}????

    na i definately dont understand that... and is that how it became z^2.. because u brought i out the front then cancelled the z form i*z?
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  10. #10
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by fredrick08 View Post
    srry i dont know what you mean by real part of residue??

    are u saying that \cos(3\theta)=\frac{1}{2}(z^{3}+z^{-3})=z^{3}????

    na i definately dont understand that... and is that how it became z^2.. because u brought i out the front then cancelled the z form i*z?
    Ok, well I'll just say it then.

    \cos(3\theta)=\text{Re }e^{3i\theta}...so?
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  11. #11
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    Quote Originally Posted by Drexel28 View Post
    Ok, well I'll just say it then.

    \cos(3\theta)=\text{Re }e^{3i\theta}...so?
    hmm ya i see how the real part of e^i3(theta)=cos(3(theta) but how do u get z^2 form there? like


    e^{3\\i\theta}=\cos(3\theta)+i\sin(3\theta)=z^{2}???? srry... i dont understand, my maths knowledge is not great..
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  12. #12
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by fredrick08 View Post
    hmm ya i see how the real part of e^i3(theta)=cos(3(theta) but how do u get z^2 form there? like


    e^{3\\i\theta}=\cos(3\theta)+i\sin(3\theta)=z^{2}???? srry... i dont understand, my maths knowledge is not great..
    In some sense what we're saying is that \int_0^{2\pi}\frac{\cos(3\theta)}{5-4\cos(\theta)}d\theta=\int_0^{2\pi}\text{Re }\frac{e^{3i\theta}}{5-4\cos(\theta)}d\theta=\text{Re }\int_0^{2\pi}\frac{e^{3i\theta}}{5-4\cos(\theta)}d\theta and the calculation I did shows that \int_0^{2\pi}\frac{e^{3i\theta}}{5-4\cos(\theta)}d\theta=\frac{\pi}{12} and so our integral is equal to \text{Re }\frac{\pi}{12}=\frac{\pi}{12}
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