Problem: (I dont have latex/mathtype for this sorry in advance)(adsbygoogle = window.adsbygoogle || []).push({});

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

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# Homework Help: Series/Function Proof (Complex Numbers)

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