1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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 ?

  2. jcsd
  3. Aug 21, 2010 #2


    User Avatar
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook