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

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Series/Function Proof (Complex Numbers)

**Physics Forums | Science Articles, Homework Help, Discussion**