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Math Help - Laurent Series/ Laurent Series Expansion

  1. #1
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    Laurent Series/ Laurent Series Expansion

    Hello everyone,

    I just had a couple questions about Laurent series.

    How do I find the Laurent series of:
    z^{2}cos(\frac{1}{3z}) in the region \left | z \right | > 0

    How do I find the Laurent series expansion of (z^{2} - 1)^{-2} valid in the regions:
    a) 0 < \left | z-1 \right |< 2

    b) \left | z+1 \right |> 2

    Any help would be great. Thanks for your time!
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  2. #2
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    Quote Originally Posted by FatherMike View Post
    Hello everyone,

    I just had a couple questions about Laurent series.

    How do I find the Laurent series of:
    z^{2}cos(\frac{1}{3z}) in the region \left | z \right | > 0

    How do I find the Laurent series expansion of (z^{2} - 1)^{-2} valid in the regions:
    a) 0 < \left | z-1 \right |< 2

    b) \left | z+1 \right |> 2

    Any help would be great. Thanks for your time!
    For the first one just use the Taylor series for cosine centered at zero.

    \displaystyle g(z)=\cos(z)=\sum_{n=0}^{\infty}\frac{(-1)^nz^{2n}}{(2n)!}

    Then your series is

    \displaystyle z^{2}cos\left(\frac{1}{3z}\right)=z^2g\left( \frac{1}{3z}\right)

    For the 2nd part this will get you started

    First note that

    \displaystyle \frac{1}{(1-z^2)^2}=\frac{1}{(1-z)^2(1+z)^2}

    lets focus on the 0 < |z-1|<2

    let \displaystyle h(z)=\frac{1}{1+z}=\frac{1}{2+(z-1)}=\frac{1}{2}\left(\frac{1}{1+\frac{z-1}{2}} \right) =\sum_{n=0}^{\infty}\frac{(-1)^n}{2^{n+1}}(z-1)^n

    Then

    \displaystyle h'(z)=\frac{-1}{(1+z)^2}=\sum_{n=1}^{\infty}\frac{n(-1)^n}{2^{n+1}}(z-1)^{n-1} = \sum_{n=0}^{\infty}\frac{(n+1)(-1)^{n+1}}{2^{n+2}}(z-1)^n

    Can you finish from here?
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