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Find laurent series about z=-2

  1. Jun 19, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the laurent series about z=-2 for:
    f(z) = 1/(z(z+2)3)


    2. Relevant equations



    3. The attempt at a solution
    Setting t = z+2 yields:
    f(t) = 1/(t3(t-2))
    = 1/t (-1/(2(1-t/2))) = (1/t)3 * (-1/2) * Ʃ(t/2)n which can be put together in a sum, but I can't be bothered due to my poor Latex skills.
    My question is however: In what region will this sum converge? Am I right at saying that the expansion will only be valid in the region described by the circle lz+2l<2? If not please tell me, because the term 1/(t-2) should wreak havoc according to me if we go outside this circle.
     
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  3. Jun 19, 2012 #2

    HallsofIvy

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    There's a typo to begin with:
    [tex]\frac{1}{t^3(t- 2)}= \frac{1}{t}\frac{-1}{2(2- t/2)}[/tex]
    when you surely mean
    [tex]\frac{1}{t^3(t- 2)}= \frac{1}{t^3}\frac{-1}{2(2- t/2)}[/tex]
    but the rest is correct.
    You sum becomes
    [tex]-\frac{1}{2}\sum \frac{1}{2^n} t^{n- 3}= -\frac{1}{2}\sum \frac{1}{2^n}(z+ 2)^{n-3}[/tex]
    You can find the radius of convergence of that using the "ratio test"
     
  4. Jun 19, 2012 #3
    Okay, using the ratio test I find that indeed we have R=2. However, I am a little unsure about where the circle of convergence is centered. Is it true that it is centered about z=-2? Because in many books I see figures where the domain of convergence for a laurent series about a pole z0 is an annullus like the red one on the attached picture. When does this type of convergence happens?
     
  5. Jun 19, 2012 #4

    vela

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    When you performed the ratio test, you should have found the series converges when |z+2|<2. You can write that more suggestively as |z-(-2)|<2, which tells you the series converges when z is within a distance of 2 of the point z0=-2.

    You get an annulus when you have more than one singularity. For example, suppose you wanted to find a Laurent series for
    $$f(z)=\frac{1}{(z-1)(z-2)} = \frac{1}{z-2}-\frac{1}{z-1}$$ expanded about the point z0=0. For the first term, you can find a series which converges for |z|<2 and one which converges for |z|>2. Similarly, for the second term, you can find one that converges for |z|<1 and another for |z|>1. The series for f(z) will consist of some combination of those four series. Depending on which combination you take, the series for f(z) will converge in different regions of the complex plane, where the region of convergence for the individual series overlap. This breaks the complex plane into three regions — |z|<1, 1<|z|<2, and |z|>2 — and the middle one is an annulus.
     
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