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Laurent series for this function

  1. Jan 9, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the Laurent series of the function f(z) = Sin(1/(z^2-z)) in the region 0<|z|<infinity.


    3. The attempt at a solution

    Now sin(z) = [e^(iz) - e^(-iz)]/(2i)

    Shall we replace z by 1/(z^2-z) to obtain the Laurent series for f(z)?
    I tried this but it gets messy. Is there a clever method? or any other approach?
     
  2. jcsd
  3. Jan 9, 2009 #2
    I would try using the series expansion
    [tex]\sin(z) = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \cdots[/tex]
    without going through the exponentials. That should be cleaner, but I haven't checked and I don't know if it's the best way to go.
     
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