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Thread: Laurent series expansion

  1. #1
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    Laurent series expansion

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

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


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


    I know how to do part (a) when $\displaystyle z_0=0$ but I do not see how to do it when $\displaystyle z_0=1$. 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 $\displaystyle \|z\|>\frac{1}{5}$
    $\displaystyle \frac{zz_{0}}{5z+1}=\frac{1}{z}\frac{z_{0}}{\frac{ 5}{z}+\frac{1}{z^{2}}}=\frac{z_{0}}{5+\frac{1}{z}} =\frac{z_{0}}{5}\frac{1}{1-\left(-\frac{1}{5z}\right)}=\frac{z_{0}}{5}\sum^{\infty}_ {j=0}\left(\frac{1}{5z}\right)^{j}$

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

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

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

    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:


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

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


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

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


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


    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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