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Math Help - Find the residue

  1. #1
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    Find the residue

    How to find the residue of
    \frac{sinh(z)}{cosh(z)-1}
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  2. #2
    Super Member girdav's Avatar
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    Where to we have to find the residue?
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by hurz View Post
    How to find the residue of \frac{sinh(z)}{cosh(z)-1}

    If you mean the residue at z=0 then, \dfrac{\sinh z}{\cosh z -1}=\dfrac{z+z^3/3!+\ldots}{ z^2/2!+z^4/4!+\ldots}=\dfrac{2}{z}+\ldots . So, the residue is 2 .
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  4. #4
    Super Member Random Variable's Avatar
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    Re: Find the residue

    If you don't want to use long division, you could use L'Hospital's rule.

     \lim_{z \to 0} \ (z-0) \  \frac{\sinh z}{\cosh z - 1} = \lim_{z \to 0} \frac{\sinh z + z \cosh z}{\sinh z}

     = \lim_{z \to 0} \frac{\cosh z + \cosh z + z \sinh{z}}{\cosh z} = \frac{1+1+0}{1} = 2
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  5. #5
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    Re: Find the residue

    Quote Originally Posted by FernandoRevilla View Post
    If you mean the residue at z=0 then, \dfrac{\sinh z}{\cosh z -1}=\dfrac{z+z^3/3!+\ldots}{ z^2/2!+z^4/4!+\ldots}=\dfrac{2}{z}+\ldots . So, the residue is 2 .
    And about the other roots:
    i(2n\pi + \pi) , with n integer

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  6. #6
    MHF Contributor FernandoRevilla's Avatar
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    Re: Find the residue

    Quote Originally Posted by hurz View Post
    And about the other roots: i(2n\pi + \pi) , with n integer
    The roots of \cosh z -1=0 are z_n=2n\pi i with n\in\mathbb{Z} . If n\neq 0 is better to use the "standard" method.
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