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Math Help - Laurent expansion

  1. #1
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    Laurent expansion

    Hey, I'm having a lot of trouble with this Laurent expansion because it's a bit more complicated than the ones I'm used to. Sorry it's not in LaTeX.

    f(z) = z / (z - 2i)(z + 1)

    valid for (i) 0 < |z + 1| < sqrt(5)
    (ii) |z| > 2


    Thanks for any help!
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Use partial fractions and then geometric series.
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  3. #3
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    how do you partial fractions when there's a complex number in the denominator?
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  4. #4
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by schrodingersdog View Post
    how do you partial fractions when there's a complex number in the denominator?
    You do just as you would if it were a real number! Give it a try, and post your calculations if you get stuck.
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  5. #5
    MHF Contributor chisigma's Avatar
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    For (i) You can use the substitution s= \frac{1}{1+z} that with some steps transforms the function...

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

    ... into ...

    \displaystyle f(s)= \frac {s\ (1-s)}{1- s (1+ 2 i)} (2)

    Now if |s|<\frac{1}{|1+2i|} = \frac{1}{\sqrt{5}} is...

    \displaystyle f(s)= \frac {1}{1- s (1+ 2 i)} = 1 + s\ (1+2 i) + s^{2}\ (1+2 i)^{2} + ...  (3)

    ... so that if You turn back to the z variable and consider that is...

    \displaystyle f(s)= \frac {z}{(1+ z)^{2}} = \frac{1}{1+z} - \frac{1}{(1+z)^{2}} (4)

    ... from (3) with some other steps you obtain finally the Laurent expansion...

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

    For (ii) You set s=\frac{1}{z} and proceed is similar way... the computation is left to You...

    Kind regards

    \chi \sigma
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