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How do i get sin(z)/2i = 1/4(e - 1/e) ? Resides at Poles question?

  1. Jun 6, 2007 #1
    1. The problem statement, all variables and given/known data

    Find Poles & eval resides at those points for

    f(z) = sin(z)/(z^2+1)

    2. Relevant equations

    I'm using Res f(z) = [g^m-1 (z_0)] / (m-1)!

    3. The attempt at a solution

    So I worked out the f(z) = 1/(z-i) so g(z) = sin(z) /(z+i) for my first one did the sub for z=i and got sin(i)/2i

    But the working out (i'm working on past papers so I have the worked solutions, just not very detailed) The next line shows;

    = 1/4(e-(1/e))

    What one earth don't I know about sin that gets to this point??
    Last edited: Jun 6, 2007
  2. jcsd
  3. Jun 6, 2007 #2


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    Homework Helper

    For simple pole at [tex]z_0[/tex], the residue of f(z) is given by
    [tex]\text{Res}\left[f(z), z_0\right]= \lim_{z\rightarrow z_0} (z-z_0) f(z)[/tex]

    [tex]\text{Res}\left[\frac{1}{(z^2+1)}, i \right]= \lim_{z\rightarrow i} \frac{z-i}{z^2+1} =\lim_{z\rightarrow i} \frac{1}{z+i}=\frac{1}{2i}[/tex]

    now, looks like you have got it right so far with sin(i)/2i, all it is need to get that next line is just by using
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