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Singularity in the infinity

  1. Oct 31, 2009 #1
    1. The problem statement, all variables and given/known data

    Find and classify the singularities in C* of [tex]f(z) = \frac{{\pi z - \pi {z^3}}}{{\sin (\pi z)}}[/tex], and give information about Res(f, 0) and Res(f, infinity)

    3. The attempt at a solution

    I found that the singularities in C are z = n, with n [tex]\in[/tex] Z, n[tex]\neq[/tex] 0, n[tex]\neq[/tex] 1. These are simple poles, while z=0 and z=1 are removable singularities (therefore, Res(f, 0)=0).

    Now, in C*: what I thought is that, since the poles tend to infinity when n tends to infinity, then there is a non-isolated singularity.

    But then I don't know how to calculate Res(f, infinity).

    Did I think that the right way or what am I missing?

    Thanks.
     
    Last edited: Oct 31, 2009
  2. jcsd
  3. Oct 31, 2009 #2
    Could somebody check this out, please?
     
  4. Nov 1, 2009 #3
    Did I write this correctly? Answer please, because I don't know if I'm translating my question properly into English.
     
  5. Nov 1, 2009 #4
    No one? Please, I need to know this urgently.
     
  6. Nov 2, 2009 #5

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    A singularity at infinity of f(z) has the same residue as the singularity at 0 of f(1/z).

    Look at
    [tex]f(\frac{1}{z}) = \frac{\frac{\pi}{z} - \frac{\pi}{z^3}}{\sin (\frac{\pi}{z})}[/tex]
     
  7. Nov 2, 2009 #6
    Thanks. You say the only way to find out is to find the a-1 coefficient (and then multiply by -1)?

    Was my assumption that it was a non-isolated singularity correct?
     
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