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Thread: Laurent series of a function

  1. #1
    MHF Contributor arbolis's Avatar
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    Laurent series of a function

    I must find the Laurent series of the function $\displaystyle g(z)=\frac{1}{z(z^2+1)}$ for 3 different regions: $\displaystyle 0<|z-i|<1$, $\displaystyle 1<|z-i|<2$ and $\displaystyle 2<|z-i|<\infty$.
    My attempt: I've found out that g has 3 singularities (at $\displaystyle z=\pm i$ and $\displaystyle z=0$) and is I think otherwise analytic.
    So for the first region, there is 1 singularity inside it (just at the center of the region).
    So I think I should write the Laurent series centered at the singularity $\displaystyle z=i$.
    I've tried many things like rewriting $\displaystyle g(z)=\frac{1}{z(z+i)} \cdot \frac{1}{z-i}$ but I realized it was better to keep it as it was. I was thinking about writing $\displaystyle \frac{1}{z^2+1}$ as an infinite series (now I realize that as it should be centered in i, I should have factored out by (z-i)...) and then divide by z the whole series. I'm going into circles and I don't click yet on how to get it. I'd like a tip.
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  2. #2
    MHF Contributor chisigma's Avatar
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    A possible confortable way is the complex variable subsitution $\displaystyle p= \frac{1}{z-i}$ so that is...

    $\displaystyle \displaystyle \frac{1}{z} = \frac{p}{1+i p}$

    $\displaystyle \displaystyle \frac{1}{z+i} = \frac{p}{1+2 i p} $ (1)

    ... and we have...

    $\displaystyle \displaystyle \frac{1}{z}\ \frac{1}{z-i}\ \frac{1}{z+i} = \frac{p^{3}}{(1 + i p)\ (1 + 2 i p)} = p^{3}\ (\frac{1}{1 + 2 i p} - \frac{1}{1 + i p}) $ (2)

    Now for $\displaystyle |p|<\frac{1}{2}$ is...

    $\displaystyle \displaystyle \frac{1}{1 + i p} = 1 - i p - p^{2} + i p^{3} + ...$

    $\displaystyle \displaystyle \frac{1}{1 + 2 i p} = 1 - 2 i p - 4 p^{2} + 8 i p^{3} + ...$ (3)

    ... so that is...

    $\displaystyle \displaystyle \frac{p^{3}}{(1 + i p)\ (1 + 2 i p)} = -i p^{4} -3 p^{5} + 7 i p^{6} + ...$ (4)

    ... and therefore...

    $\displaystyle \displaystyle \frac{1}{z\ (1+z^{2})} = - \frac{i}{(z-i)^{4}} - \frac{3}{(z-i)^{5}} + \frac{7 i}{(z-i)^{6}} + ...$ (5)

    The series (5) converges for $\displaystyle |z-i|>2$ ...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  3. #3
    MHF Contributor arbolis's Avatar
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    Thank you. That was actually very hard. So the main idea was to make a change of variable. Also, how do you reach that the series (5) converges for |z-i|>2?
    Edit: Oh nevermind I get it! Because of line 6 of your post. Great, many thanks.
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