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Sum of nth roots of unity

  1. Nov 12, 2012 #1
    i'm trying to prove the sum of nth roots of unity = 0, but I don't really know how to proceed:

    suppose z^n = 1 where z ε ℂ,

    suppose the roots of unity for z are 1, ω, ω^2, ω^3 .... ω^n

    the sum of these would be S = 1 + ω, ω^w, ω^3 +...+ ω^(n-1) + ω^n

    from here I had an idea to do some fancy manipulation of S, then show that S = 0, but if say I do ωS - S I don't get 0!

    I'm assuming I've made a very silly mistake or the way of approaches this is all wrong,

    does anyone have a better approach or can anyone spot my mistake?

    thanks,
     
  2. jcsd
  3. Nov 12, 2012 #2

    Mark44

    Staff: Mentor

    That's too many. Your roots should start at exponent 0 and end at exponent (n - 1).
    S = 1 + ω + ω2 + ... + ωn - 1

    Notice that the right side is a (finite) geometric series.
     
  4. Nov 12, 2012 #3

    Dick

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    Science Advisor
    Homework Helper

    And if you make the correction Mark44 gave, you will find ωS - S is zero.
     
  5. Nov 12, 2012 #4

    tiny-tim

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    alternatively, they're the n roots of the polynomial equation xn - 1 = 0 …

    so which coefficient is the sum of the roots? :wink:
     
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