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Series/Function Proof (Complex Numbers)

  1. Jan 25, 2007 #1
    Problem: (I dont have latex/mathtype for this sorry in advance)

    Let n,k both be in the Natural Numbers

    n does not divide k (I have already completed the case when it does)

    Show that the series:

    1 + e^i(2pik/n) + e^i(4pik/n) + .... + e^i[(2n-2)pik/n) = 0

    Using eulers formula e^ix = cos(x) + isin(x) I used to prove that the above series ='s n if n divides k. I'm having some trouble starting on if n does not divide k. I was wondering if any of you could point me in the right direction as to were to look. I've done a few examples (n=2) etc and shown that it will equal 0 for all odd k (as n cannot divide k) in such a case, though I'm not totally sure I should be using an induction argument for higher cases of n. (didnt use n=1 since it divides all k).

    I'm thinking the proof ties into roots of unity since you can use cos(2pi/n) + isin(2pi/n) to construct the n-th roots of unity. Anyways, thanks for any help you can provide :smile:
  2. jcsd
  3. Jan 25, 2007 #2


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    So, you just want to sum the series

    [tex]1 + \zeta + \zeta^2 + \zeta^3 + \cdots + \zeta^{n-1},[/tex]

    where [itex]\zeta[/itex] is an n-th root of unity?
  4. Jan 25, 2007 #3
    Yes, except each term is to the power k in the natural numbers as well. I have to show that for all k such that n does not divide k the series ='s 0 and for n dividing k it equals n. I have shown the latter and looking for help on the prior.

    I'm reading my book and it seems that the I must show that if n does not divide k then the arguments are just shifts of 2pi from the principal arguments between (-pi, pi]. Does this seem like a good way to tackle the problem? Only problem with this I forsee is that even if I show that if n does not divide k then every root of unity is summed together if I can just show that it must be 0 in this case. I'll think about it more.

    Edit: I just realized this forum is not for homework, if you could move this thread I would appreciate it. I misread the labels obviously. Sorry for the mishap. (This is a homework problem)
    Last edited: Jan 25, 2007
  5. Jan 25, 2007 #4


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    [tex]\zeta = e^{2 \pi i k / n}[/tex]

    is a perfectly good n-th root of unity.

    I don't follow the approach you suggest. I guess I'm confused about what you mean by "principal arguments". Arguments of what? And why are they shifts of 2 pi?

    I do think you're overthinking the problem, though.
    Last edited: Jan 25, 2007
  6. Jan 25, 2007 #5
    Well, I understand that no matter what natural number you multiply it will end up being a n-th root of unity. The problem I'm having is when you add each root of unity, how do I know they are going to sum to zero? When n divides k each term becomes the root of unity 1 and therefore the sum ='s n. But if n does not divide k the series becomes a summation over different (maybe even every) root of unity which I must prove to sum to zero.

    Arguments are just reference to the angle of the root of unity. Usually the principal arguments are in the interval (-pi,pi] though obviously you can add 2pi and still have the same argument. I'm thinking by showing that if n divides k then each n-th root of unity is distinct with arguments shifted by some integer K*2pi and by summing all possible n-th roots of unity you get 0.

    Overall, I'm a little lost on how to start the proof which may be the reason for my overthinking possible ways of approaching a method of showing the sum to be 0.
  7. Jan 25, 2007 #6


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    It is true that the terms in this sequence consist of all of the (n / gcd(k, n))-th roots of unity, each repeated gcd(k, n) times. That's actually a rather useful fact to know. But that's by far the hard way.

    Maybe it will help if you forget for a moment that you're trying to prove the sum is zero, and forget whatever context in which this question arose.

    Think of this question as an exercise unto itself: what is the sum of

    [tex]1 + \zeta + \zeta^2 + \zeta^3 + \cdots + \zeta^{n-1}?[/tex]
    Last edited: Jan 25, 2007
  8. Jan 28, 2007 #7
    K, I proved the sum of the roots of unity are 0.

    namely, the roots of unity are roots of x^n -1 = 0 and the sum of every root is equal to b/-a = 0/-1 = 0. (Polynomial theorem I made a lemma).

    Thus the sum of the above is 0 since those are the roots of unity.

    I'm still having trouble taking that case into the case where each root is to the k power.

    0th root + 1st*k root + 2nd*k root....
    0th root + 1st*k root (mod n) + 2nd*k root (mod n)....
    was how I was trying to do it, since by powering each term by k you simply change the order in steps of k instead of 1. I found that this doesnt actually hit every root though, since for example k = 6, n = 10

    You have the sum:

    0th root + 6th root + 2th root + 8th root + 4th root + 0th root...

    Thus I'm still unsure how to prove the overall problem, though I think i'm closer :P.
  9. Jan 28, 2007 #8


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    That sum is a geometric series.
  10. Jan 29, 2007 #9
    Thanks for the help, the geometric series gives far greater insight in the mechanics of the problem from a rigorous proof standpoint.

    The problem can be seen as a geometric series of the k-th(mod n) root of unity from 0 to n-1. So the limit is:

    0/(1-e^(2piK/n) hince equals 0 assuming n does not divide k. If n does divides k then the limit is 0 as proved prior and the above is undefined.

    Thanks again Hurkyl, I dont think I would have noticed the geometric series representation as I was more focused on the graphic obviousness when dealing with roots of unity that their sum must go to 0 even with steps of k.
    Last edited: Jan 29, 2007
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