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Integral where I need to classify the singularities

  1. May 7, 2007 #1
    Hi I'm just working on an integral where I need to classify the singularities of the integrand [tex]\frac{{\sin \left( {\pi z} \right)}}{{z^4 - 1}}[/tex].

    There are singularities at [tex]z = \pm 1, \pm i[/tex] but the ones I'm interested in are at [tex]z = \pm 1[/tex].

    [tex]
    \frac{{\sin \left( {\pi z} \right)}}{{z^4 - 1}} = \frac{{\sin \left( {\pi z} \right)}}{{\left( {z - 1} \right)\left( {z + 1} \right)\left( {z^2 + 1} \right)}}
    [/tex]

    My immediate thought was that z = 1 is a simple pole but if it is I should be able to write [tex]\frac{{\sin \left( {\pi z} \right)}}{{z^4 - 1}} = \frac{{h\left( z \right)}}{{\left( {z - 1} \right)}}[/tex] where h is analytic in a neighbourhood of z = 1 and non-zero at z = 1 but this doesn't seem to be the case here. The answer I have seems to be saying that z = 1 is a simple pole but at the moment I can't see how it can be.

    Any input would be great thanks.
     
  2. jcsd
  3. May 7, 2007 #2

    HallsofIvy

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    So the problem is that [itex]sin \pi z[/itex] has a zero at z= 1. How could you write [itex]sin \pi z[/itex] as a Taylor's series around z= 1?
     
  4. May 7, 2007 #3

    AKG

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    How did you figure out what the singularities are?
     
  5. May 8, 2007 #4
    I found the singularities by setting z^4 - 1 = 0.

    The first term in the series for the sin(pi*z), centred at 1, is zero but the second one (which is something multiplied by (z-1)) isn't. So the singularity should be removable but then it isn't a simple pole. It makes no difference to the integral that I was calculating but it's just something that's puzzling me.
     
  6. May 8, 2007 #5

    HallsofIvy

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    Yes, the singularity at z= 1 is removable, not a pole. What made you think it was a pole?
     
  7. May 10, 2007 #6
    The answer said it was a pole.:redface: I guess in future I should use my judgement when looking at the answers.

    Thanks for the help.
     
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