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Contour Integral: how to get the order of those poles?

  1. Aug 21, 2010 #1
    1. The problem statement, all variables and given/known data

    The following function :

    a) [tex] f(z) = \frac{1}{z^6 + 1} [/tex]

    has simple poles on :

    [tex] z_1 = e^{pi/6 i}, z_2 = e^{3pi/6 i}, z_3 = e^{5pi/6 i} [/tex]

    I know how to get the poles, but how could I demonstrate they are simple (order 1) ?
    I tried to write the Laurent series centered in the pole, but I got nowhere...
    Is there any general way to determine the order of a pole without using Laurent ?

    Thanks.
     
  2. jcsd
  3. Aug 21, 2010 #2

    lanedance

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    Homework Helper

    expanding the denominator could help, not also that theta = 7 pi/6, 9 pi/6 and 11 pi/6 are also poles
     
  4. Aug 21, 2010 #3
    Your function always could be written as 1 / [(z-a)(z-b)(z-c)(z-d)(z-e)(z-f)]
    where a,b,c,d,e,f are the roots of the polynomial z^6 - 1 = 0.

    The poles are simple poles, because there are no repeated roots.
    you have 6 complex roots to the polynomial z^6 - 1 = 0.
    I assume you want to do an integral using the "residue theorem"
    so you correctly chose 3 of then that are in the positive imaginary direction.

    *if your function was 1 / [(z-2)(z-3)^3] then you've had one simple pole at z=2 and one pole of order 3 at z=3.

    this is a general explanation.
     
  5. Aug 21, 2010 #4
    Thanks! I think I got it now.
     
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