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Laurent Expansion of sin(1-1/z)

  1. Mar 27, 2008 #1

    nicksauce

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    1. The problem statement, all variables and given/known data
    Find the Laurent expansion of [tex]f(z) = \sin(1-\frac{1}{z})[/tex] about z = 0, and state the annulus of convergence.


    2. Relevant equations



    3. The attempt at a solution
    I tried doing the regular expansion of sin(z), then applying the binomial expansion on the (1-1/z)^n terms, but I can't help but feel that there's a better way to approach the problem. Any thoughts?
     
  2. jcsd
  3. Mar 27, 2008 #2

    HallsofIvy

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    I would suggest that you use the sine sum formula to write it as
    [tex]sin(1-\frac{1}{z})= sin(1)cos(\frac{1}{z})- cos(1)sin(\frac{1}{z})[/tex]
    and then use the Taylor's series for sine and cosin.
     
  4. Mar 27, 2008 #3

    nicksauce

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    Boy don't I feel dumb. Thanks!
     
  5. Nov 18, 2011 #4
    is this sentence correct?
    sin(1/z)=1/z-1/z^3*3!+1/z^5*5!+-......
    for laurent series?
     
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