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Laurent Expansion Problem (finding singularities)

  1. Dec 11, 2009 #1
    1. The problem statement, all variables and given/known data
    Find all Laurent expansion of the function f(z) = 1/(z(8(z^3)-1)) with centre z = 0.

    3. The attempt at a solution

    I tried to find all the singularities and came up with z = 0, z = 1/2, z = (1/2)exp((n*pi*i)/3)
    where n = +-2,+-4,+-6... . But according to the solution n can only be 2 and 4, what am I missing?
     
    Last edited: Dec 11, 2009
  2. jcsd
  3. Dec 11, 2009 #2
    n = 4 and n = -2 are the same (and n = -4 is the same as n = 2) n = 6 is the same as n = 0 which yields z = 1/2.
     
  4. Dec 11, 2009 #3
    exp((n*pi*i) = cos(n*pi) + i*sin(n*pi) = 1 + i*0 = 1 for n = +-2,+-4,+-6,+-8... isn't it? Wouldn't this make z = (1/2)exp((n*pi*i)/3) yield z = 1/2 for all n = +-2,+-4,+-6,+-8... ?
    (making all n = +-2,+-4,+-6,+-8... equal, hence my confusion over why only n=2 and n=4 were included in the solution).
     
  5. Dec 11, 2009 #4
    There is a factor 1/3 in the exponential.
     
  6. Dec 11, 2009 #5
    But that factor disappears because you take z to the power of three in the function
    [tex](f(z) = \frac{1}{z(8z^{3}-1)})[/tex]

    Edit: nvm this post, I got it. thx :)
     
    Last edited: Dec 11, 2009
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