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Finding Laurent Series

  1. Nov 8, 2013 #1
    1. The problem statement, all variables and given/known data

    Specifically, I'm trying to find the laurent series for [itex]f(z)=\frac{z^2}{z+1}[/itex] around the point [itex]z=-1[/itex]. My real problem is my procedure in general though. I'm not sure what I'm doing wrong on a lot of these Laurent Series but for some reason I'm struggling with them.

    (Even more specifically, I'm trying to find the type of singularity and the residue at z=-1.)

    2. Relevant equations

    [itex]\sum^{\infty}_{n=-\infty}{c_n (z-z_0)^n}[/itex] where
    [itex]c_n=\frac{1}{2\pi i}\int_{C}{\frac{f(z)dz}{(z-z_0)^{n+1}}}[/itex]

    3. The attempt at a solution

    What I did is take the [itex]z^2[/itex] out and set [itex]z_0=-1[/itex] because that's the point I want to expand around. Then I set C such that [itex]z=-1+e^{i\theta}[/itex] where [itex]-\pi < \theta < \pi[/itex] and integrated.

    I found the integral, which was [itex]\frac{sin(n+1)\pi}{(n+1)\pi}[/itex] which equals zero for all values of n besides [itex]n=-1[/itex].

    I could probably find out more about the coefficient at -1 if I evaluated the integral for n=-1, but at this point I realized I still had the [itex]x^2[/itex] in there that would throw off the degree of my Laurent Series anyways...

    So I basically feel like I have spent way more time on this problem than I should have and have almost no results to show for it. Clearly, my procedure is not spot on :\. Can anybody help me out? How should I start out on a problem like this?

    Normally, when I try to find a Laurent Series, this is what I do:
    1. Decide where it should be centered.
    2. Think about other series representations that I've memorized to maybe do a quick easy substitution or break up the function into multiple parts.
    3. If I have no success this far, I'll usually just result to the laurent series formula that I have posted above, but clearly, I'm running into problems with it.
     
  2. jcsd
  3. Nov 8, 2013 #2
    You have a simple pole at z = -1. That means you will have one negative exponent in your Laurent expansion. Start with the Taylor's series for ##z^2## at the point -1 (i.e. you are expanding in powers of z+1. You will get

    ##z^2 = \sum [stuff]##. This series converges throughout the complex plane. Now divide it by z+1 to get

    ##z^2 = \sum [stuff]/(z+1) ##. This gives you the entire Laurent expansion.

    In general finding Laurent expansions via the integral is a fraught process. Sometimes it's the only way, but if you can think of something simpler, use it. The Laurent series is unique, so if you have a series that converges to your function in a domain around a singularity, it has to be the Laurent series. (If there is no singularity it has to be the Taylor's series).
     
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