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Math Help - Residue calculation

  1. #1
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    Question Residue calculation

    Hello to everyone!
    Could someone explain to me why the residue of the complex function f(z)= \frac{e^{1/z}}{z^{2}+1} is a_{-1}=Res(f,0)=1-\frac{1}{3!}+\frac{1}{5!}-...= \sin{1}

    Thanks in advance!
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Residue calculation

    f(z)=e^{1/z}\cdot \dfrac{1}{1+z^2}=\sum_{n=0}^{+\infty}\dfrac{1}{z^n n!}\cdot \sum_{n=0}^{+\infty}(-1)^nz^{2n} for 0<|z|<1 . Now, find the coefficient of 1/z .
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  3. #3
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    Re: Residue calculation

    So it was so easy... I should have thought it! Thanks for the help!
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