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Math Help - solving this integral

  1. #1
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    solving this integral

    \int_{|z|=3}^{nothing}\frac{dz}{z^3(z^{10}-2)}\\
    f=\frac{1}{z^3(z^{10}-2)}\\
    f(\frac{1}{z})=\frac{1}{(\frac{1}{z})^3((\frac{1}{  z})^{10}-2)}\frac{z^{13}}{1-2z^{10}}=\\
    res(f,\infty)= res(\frac{1}{z^2}f(\frac{1}{z}),0)=\frac{1}{z^2}\s  um_{n=0}^{\infty}(2z^{10})^n\\
    res(f,\infty)= res(\frac{1}{z^2}f(\frac{1}{z}),0)=-[res(f,inside|z|=3)+res(f,outside|z|=3)]

    from the sum i get that there is no z^{-1} member in the series
    so the coefficient of z^{-1} is zero

    so the residiu of infinity is zero
    but still all of my singular points are |z|=3
    so the integral equals zero
    ??

    did i solved it correctly
    did i written every formula regarding the laws of residue correctly here
    ?
    Last edited by transgalactic; July 15th 2010 at 11:06 AM.
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  2. #2
    MHF Contributor chiph588@'s Avatar
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  3. #3
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    didd i solved it correctly
    ??
    did i written the formulas correctly
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    I didn't follow what you did. Why are you messing with  f\left(\frac1z\right) ?
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  5. #5
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    residue of infinity
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  6. #6
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by transgalactic View Post
    residue of infinity
    Why? You just need to find the residue at  0 and two times the tenth roots of unity.
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  7. #7
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    its z in power of 10
    thats 10 roots

    look at my way is it correct
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