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Laurent series around a singular point

  1. May 8, 2013 #1
    Find the Laurent series of the following function in a neighborhood of the singularly indicated, and use it to classify the singularity.

    1. The problem statement, all variables and given/known data
    [itex]f(z) = \frac{1}{z^2-4} ; z_0=2 [/itex]

    2. Relevant equations
    Laurent series
    [itex]\sum_{-\infty}^{\infty} a_n (z-c)^n [/itex]

    3. The attempt at a solution
    I started by doing a partial fraction decomposition of [itex] f(z) = \frac{1}{z^2-4} ; z_0=2 [/itex] which I got to be [itex] f(z) = \frac{-1}{4} \cdot \frac{1}{z+2} + \frac{1}{4} \cdot \frac{1}{z-2} [/itex] .
    Then I used [itex] 0< \vert z+2 \vert < 2 \Rightarrow \vert \frac{z+2}{2} \vert < 1 [/itex] to apply the geometric series and get [itex] \frac{1}{4} \sum_{k=0}^{\infty} (\frac{2}{z})^k [/itex].

    I am not sure if I made a mistake somewhere or if this method is not applicable. Can somebody please provide a hint or point out an error on my behalf?
    Last edited: May 8, 2013
  2. jcsd
  3. May 8, 2013 #2


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    Staff Emeritus
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    Gold Member

    You need to use tex and /tex tags to do math typesetting on this forum.

    If you are going to apply the geometric series formula, you need to

    1) get a power series in z, not negative exponents
    2) Get something that has (z-2)s in it since you want the Laurent series around z=2 (which means you can't take the power series around z=0)
  4. May 10, 2013 #3
    Take note of the following.
    [tex]f(z) = \frac{1}{z^2 - 4} = \frac{1}{z-2} \cdot \frac{1}{z+2} = \frac{1}{z - 2} \cdot \frac{\frac{1}{4}}{1 - \left( - \frac{z - 2}{4} \right)} = \frac{1}{z-2} \cdot \frac{1}{4} \sum_{n=0}^\infty \left( - \frac{z - 2}{4} \right)^n = \cdots[/tex]
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