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Math Help - calculating a residue

  1. #1
    Super Member Random Variable's Avatar
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    calculating a residue

    I want to calculate the residue of  f(z) = \frac{z}{(1+z^{2}) \sinh \pi z} at  z = i .

     z = i is a pole of order 2 since  \sinh( \pi i) = 0

    So the standard approach is  Res[f,i] = \lim_{z \to i} \frac{d}{dz}  (z-i)^{2} \frac{z}{(1+z^{2})\sinh \pi z} = \lim_{z \to i} \frac{d}{dz}  (z-i) \frac{z}{(z+i)\sinh \pi z}

    After differentiating and applying L'Hospital's rule once, the limit is still indeterminate and very messy. Is there a better approach than repeated applications of L'Hospitals rule?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: calculating a residue

    Quote Originally Posted by Random Variable View Post
    I want to calculate the residue of  f(z) = \frac{z}{(1+z^{2}) \sinh \pi z} at  z = i .

     z = i is a pole of order 2 since  \sinh( \pi i) = 0

    So the standard approach is  Res[f,i] = \lim_{z \to i} \frac{d}{dz}  (z-i)^{2} \frac{z}{(1+z^{2})\sinh \pi z} = \lim_{z \to i} \frac{d}{dz}  (z-i) \frac{z}{(z+i)\sinh \pi z}

    After differentiating and applying L'Hospital's rule once, the limit is still indeterminate and very messy. Is there a better approach than repeated applications of L'Hospitals rule?
    Try expanding \sinh(\pi z) as a series around i, namely \displaystyle \sinh(\pi z)=-\pi(z-i)-\frac{\pi^3}{6}(z-i)^3+O(z) you should have an easy time then, no? By the way I hope you know..bam!
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  3. #3
    Super Member Random Variable's Avatar
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    Re: calculating a residue

    Here's what I did. It probably could have been done in less steps.

    I wrote the function as  \frac{ z  \ \text{csch} \ \pi z}{1+z^{2}} . Then I found the Taylor series of  \sinh \pi z centered at  z = i , and used synthetic division to find the Laurent series for  \text{csch} \ \pi z centered at  z = i . Next I found the Taylor series for  z and  1+z^{2} centered at  z= i, and again used synthetic division to find the Laurent series for  \frac{z}{1+z^{2}} centered at  i . Finally I multiplied the Laurent series of  \frac{z}{1+z^{2}} by the Laurent series for   \text{csch} \ \pi z .
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  4. #4
    MHF Contributor chisigma's Avatar
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    Re: calculating a residue

    Quote Originally Posted by Random Variable View Post
    I want to calculate the residue of  f(z) = \frac{z}{(1+z^{2}) \sinh \pi z} at  z = i .

     z = i is a pole of order 2 since  \sinh( \pi i) = 0

    So the standard approach is  Res[f,i] = \lim_{z \to i} \frac{d}{dz}  (z-i)^{2} \frac{z}{(1+z^{2})\sinh \pi z} = \lim_{z \to i} \frac{d}{dz}  (z-i) \frac{z}{(z+i)\sinh \pi z}

    After differentiating and applying L'Hospital's rule once, the limit is still indeterminate and very messy. Is there a better approach than repeated applications of L'Hospitals rule?
    With the substitution s=z-i the problem is finding the residue at s=0 of...

    f(s)= - \frac{s+i}{s\ (s+2 i)\ \sinh \pi s} = -  \frac{1}{(s+2 i)\ \sinh \pi s} -  \frac{i}{s\ (s+2 i)\ \sinh \pi s} =

    = f_{1} (s) + f_{2} (s)  (1)

    First we indagate on f_{2} (*) that can be written as...

    f_{2} (s)= - \frac{1}{s^{2}} \ \frac{i}{s+2i}\ \frac{s}{\sinh \pi s}=

     = -\frac{1}{2 \pi}\ \frac{1}{s^{2}}\  \frac{\pi s}{\sinh \pi s}\ (1+ \frac {i s}{2} + ...) (2)

    ... and from (2) You can derive that is...

    \text{Res}_{s=0} \{f_{2} (s)\} = - \frac{i}{4 \pi} (3)

    Now we indagate on f_{1}(*) that can be written as...

     f_{1}(s)= \frac{i}{2 \pi}\ \frac{1}{s}\ \frac{\pi s}{\sinh \pi s}\ (1+ \frac {i s}{2} + ...) (4)

    ... and from (4) You can derive that is...

    \text{Res}_{s=0} \{f_{1} (s)\} = \frac{i}{2 \pi} (5)

    ... so that is...

    \text{Res}_{s=0} \{f (s)\} = \frac{i}{4 \pi} (6)

    Kind regards

    \chi \sigma
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