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Partial fractions

  1. Dec 22, 2009 #1
    I'm trying to do a question that requires the expansion of the following using partial fractions:
    [tex]f(z)=\frac{1}{(1+z^3)^2}[/tex].
    The fact that the bottom is squared is throwing me off for some reason... I've factorized the bottom, but I'm not sure whether I should use the complex roots or not, or even if it's possible without using complex roots.
    Any help would be appreciated.
     
  2. jcsd
  3. Dec 22, 2009 #2
    Hey there,
    The trick is to use identity 1+Z^3=(1+z)(1-Z+Z^2) and then just square this expression and do the partial fraction as usual.
    Hope this helps:)))
     
  4. Dec 23, 2009 #3

    HallsofIvy

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    [tex]\frac{1}{(1+z^3)^2}= \frac{1}{(z+1)^2(z^2-z+1)^2}[/tex]
    and can be written as "partial fractions" as
    [tex]\frac{A}{z+1}+ \frac{B}{(z+1)^2}+ \frac{Cx+D}{z^2- z+1}+ \frac{Ex+F}{(z^2- z+1)^2}[/tex]

    There was an earlier question about the "principal part" or "Laurent series" for [itex]1/(z^2+1)^2[/itex] which I answered by reducing to partial fractions with complex coefficients. Is this related to that thread?
     
    Last edited: Dec 24, 2009
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