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Find the residue for

  1. Apr 14, 2015 #1
    1. The problem statement, all variables and given/known data

    Find Res(0;f) for

    [itex] f(z) = \frac{e^{4z} - 1}{sin^2(z)}.[/itex]

    2. Relevant equations

    Residue Theorem

    3. The attempt at a solution

    If there's a nice (z-a)n singularity in the denominator, then I can simply use the Residue Theorem. However, I'm skeptical on what I'm doing:

    The singularity occurs at z = 0 and its order is two. Using the Residue Theorem, I can take [itex] \ d/dz(sin^2(z)f(z) = d/dz(e^{4z} - 1) [/itex] evaluated at z = 0 which is 4.

    Expressing the terms in a power series gets messy.
     
  2. jcsd
  3. Apr 14, 2015 #2

    Dick

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    I guess I'm not sure why you are skeptical of that. It looks fine to me.
     
  4. Apr 14, 2015 #3
    Okay. I wasn't sure if I could multiply the sin2z term because it's not of the form (z-a)n as given in the theorem.
     
  5. Apr 14, 2015 #4

    Dick

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    Actually, that is a correct concern. I was being sloppy. Since you know the pole is of order 2 you should be multiplying by ##z^2## and then taking the derivative and the limit. You'll get the same result.
     
  6. Apr 14, 2015 #5
    I thought about that but I'd still have a sin2z term in the denominator if I take the derivative of [itex] f(z) = \frac{z^2(e^{4z} - 1)}{sin^2(z)}[/itex], using the quotient rule.
     
  7. Apr 14, 2015 #6

    Dick

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    You are probably better off actually working on the series expansion. Remember you only need the coefficient of the 1/z term. ##e^{4z}-1## has the form ##4z f(z)## where ##f(z)## is analytic and ##f(0)## is not zero. ##sin(z)^2## has the form ##z^2 g(z)## where ##g(z)## is analytic and ##g(0)## is not zero. There may actually not be much series expansion to do.
     
  8. Apr 14, 2015 #7
    Hm. Expanding the numerator of [itex] f(z) = \frac{e^{4z} - 1}{sin^2(z)}[/itex] is easy, i.e. [itex] f(z) = \frac{1 + 4z + \frac{4^2z^2}{2} + ... \frac{4^nz^n}{n!} - 1}{sin^2(z)}.[/itex] I'm not sure how expanding sin2(z) in the denominator would allow me to find the z-1 coefficient.
     
  9. Apr 14, 2015 #8

    Dick

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    Ok, I'll spell it out a little more. I claim that your function can be written as ##\frac{4z f(x)}{z^2 g(z)}## where ##\frac{f(z)}{g(z)}## is analytic at ##z=0##. So ##\frac{f(z)}{g(z)}=a_0+a_1 z+...##. So the coefficient you are looking for is just ##4 a_0## and ##a_0=\frac{f(0)}{g(0)}##.
     
  10. Apr 14, 2015 #9
    Okay. I think I follow you now. Of course, multiplying the analytic quotient [itex]\frac{f(z)}{g(z)}[/itex] by [itex]\frac{4z}{z^2}[/itex] gives me the n = -1 or z-1 coefficient in the Laurent expansion, which should just be [itex]4\frac{f(0)}{g(0)} = 4\frac{1}{1} = 4. [/itex]
     
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