Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Tricky Identity

  1. Feb 8, 2014 #1
    I'd like to figure out why

    [tex] f(x) = x^n - r e^{i \theta} = \prod_{j = 0}^{n - 1} ( x - r^{1/n} e^{i \theta / n} e^{2 \pi i j / n} ) [/tex],

    as I've seen it used as an identity in a few courses but I cannot figure it out. Could somebody shed some light? I understand they are the roots of the function, but I'd like some kind of analytic description.

    Thanks a lot!
     
  2. jcsd
  3. Feb 9, 2014 #2
    Hi,
    I think you may have answered your own question: if you are willing to accept that the RHS above can be rewritten as
    [tex]
    \prod_{j = 0}^{n - 1} ( x - a_j )
    [/tex]
    where [itex]a_j[/itex] (with [itex]j=0,1,...,n-1[/itex]) are the n-th roots of the complex number in polar form [itex]r e^{i \theta}[/itex], then your question can be rephrased as: "can somebody point me to a proof of the Fundamental Theorem of Algebra"?

    My 2 cents.
     
    Last edited: Feb 9, 2014
  4. Feb 9, 2014 #3

    AlephZero

    User Avatar
    Science Advisor
    Homework Helper

    You have ##x^n = re^{i\theta}##, so taking the n'th root of each side you get
    $$x = \sqrt[n]{re^{i\theta}}\,\omega_k, \quad k = 0,\dots,n-1$$
    where the ##\omega_k## are the n'th roots of unity.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Tricky Identity
  1. A tricky one (Replies: 2)

  2. Tricky sequence (Replies: 1)

  3. Determinant identity (Replies: 4)

Loading...