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Math Help - Laurent series expansion

  1. #1
    Junior Member
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    Laurent series expansion

    Write the Laurent series expansion for the following functions f(z) centered at z_0. State what are the possible domains.

    (a) f(z) = \frac{z}{5z+1} z_0=1


    (b) g(z) = \frac{z}{z^2-3iz-2} z_0=0


    I know how to do part (a) when z_0=0 but I do not see how to do it when z_0=1. For part (b), I used the quadratic formula, then I did not know how to finish the problem. I need a push in the right direction. Thanks in advance.
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  2. #2
    Member Mauritzvdworm's Avatar
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    Both problems depend on the region you will be working with, let us fist consider the first problem:

    We consider the region \|z\|>\frac{1}{5}
    \frac{zz_{0}}{5z+1}=\frac{1}{z}\frac{z_{0}}{\frac{  5}{z}+\frac{1}{z^{2}}}=\frac{z_{0}}{5+\frac{1}{z}}  =\frac{z_{0}}{5}\frac{1}{1-\left(-\frac{1}{5z}\right)}=\frac{z_{0}}{5}\sum^{\infty}_  {j=0}\left(\frac{1}{5z}\right)^{j}

    next consider the region \|z\|<\frac{1}{5}
    \frac{zz_{0}}{5z+1}=zz_{0}\frac{1}{1-\left(-5z\right)}=zz_{0}\sum^{\infty}_{j=0}\left(-5\right)^{j}z^{j}=z_{0}\sum^{\infty}_{j=0}\left(-5\right)^{j}z^{j+1}

    The second problem can be done in a similar fashion, just consider the regions of interest separately, namely
    i. \|z\|<1
    ii. 1<\|z\|<2
    iii. \|z\|>2

    you can then also make use of the following decomposition
    \frac{zz_{0}}{z^{2}-3iz-2}=\frac{zz_{0}}{\left(z-2i\right)\left(z-i\right)}=zz_{0}\left(\frac{1}{i\left(z-2i\right)}-\frac{1}{i\left(z-i\right)}\right)

    Now you will easily be able to find the solutions
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  3. #3
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    I am stuck on part (b). So far I have:


    \frac{z}{z^2-3iz-2}

    now consider the regions |z|<1, 1<|z|<2, and |z|>2. Now,


    \frac{z}{z^2-3iz-2}=\frac{z}{(z-2i)(z-i)}=z (\frac{1}{i(z-2i)}-\frac{1}{i(z-i)})<br />

    a) \frac{1}{i(z-2i)}=\frac{-1}{2}(\frac{1}{1-\frac{iz}{2}})


    b) \frac{1}{i(z-i)}=\frac{1}{1+iz}


    Now how do we get a Laurent series from all this work? I don't see how to use these intervals to get a Laurent series.
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