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Principal Part of Complex Function

  1. Dec 21, 2009 #1
    I'm trying to do a question about finding the principal part of a complex valued function:
    [tex] f(z)= \frac{1}{(1+z^3)^2} [/tex].
    I really don't know how to go about even starting it, any tips?

    (just in case my terminology is different to anyone elses, the principal part is the terms of the Laurent expansion that have negative powers)
  2. jcsd
  3. Dec 22, 2009 #2


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    I assume you mean around the three roots of [itex]1+ z^3= 0[/itex]. Except around those three points, the function is analytic and has no Laurent series.

    Start by factoring: [itex]z^3+ 1= (z+ 1)(z^2- z+ 1)[/itex]. You can also "factor" [itex]z^2- z+ 1[/itex], of course. By the quadratic formula, its zeroes are
    [tex]\frac{1\pm\sqrt{1- 3}}{2}= \frac{1}{2}\pm i\frac{\sqrt{3}}{2}[/tex]
    [tex]z^2- z+ 1= \left(z- \frac{1}{2}- i\frac{\sqrt{3}}{2}}\right)\left(z-\frac{1}{2}+ i\frac{\sqrt{3}}{2}\right)[/tex]

    And that tells us that
    [tex]\frac{1}{(z^3+1)^2}= \frac{1}{\left(z+1\right)^2\left(z- \frac{1}{2}- i\frac{\sqrt{3}}{2}\right)\left(z-\frac{1}{2}+ i\frac{\sqrt{3}}{2}\right)}[/tex]

    Using "partial fractions", we can write that as
    [tex]\frac{A}{x+1}+ \frac{B}{(x+1)^2}+ \frac{C}{z- \frac{1}{2}- i\frac{1}{\sqrt{3}}{2}}}+ \frac{D}{\left(z- \frac{1}{2}- i\frac{\sqrt{3}}{2}}\right)^2}+ \frac{E}{\frac{z- 1}{2}+ i\frac{\sqrt{3}}{2}}+ \frac{F}{\left(z-\right(\frac{1}{2}+ i\frac{\sqrt{3}}{2}\right)^2}[/tex]

    Now, around z= -1, the last four fractions are analytic and can be written in Power series in z+1. The first two are negative powers of z+1 and so are the "principal part" of the function around z= -1.

    Similarly, the third and fourth fractions are the principal part of the function around
    [tex]z= \frac{1}{2}+ i\frac{\sqrt{3}}{2}}[/tex]
    and the fifth and sixth fractions are the principal part of the function around
    [tex]z= \frac{1}{2}- i\frac{\sqrt{3}}{2}}[/tex]
    Last edited by a moderator: Dec 22, 2009
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