Question

show that
\[\sum\limits_{k=1}^n e^{i2k\pi/n} = 0\]

w/o using geometric series

http://www.wolframalpha.com/input/?i=%5Csum%5Climits_%7Bk%3D1%7D%5En+e%5E%7Bi2k%5Cpi%2Fn%7D

Answer 1

They're the roots of \(x^n - 1\) so... using Vieta's Formulas?

Answer 2

Clever! that is still algebra...

Answer 3

this is also helpful :O

\(\Large e^{i\theta}=\cos \theta+i \sin \theta \)

Answer 4

It's pretty easy when you geometrically see it.

Answer 5

\[e^{i \pi.\frac{2k}{n}}=e^{i.n'.\pi}=\cos(n' \pi)+i.\sin(n' \pi)\]\[\implies \cos(\frac{2k}{n}\pi)+i.\sin(\frac{2k}{n}\pi)\]
can we use this?
or is the formula stricly for positive integers??

Answer 6

@Nishant_Garg au can and its also work

Answer 7

sure euler formula works for ALL \(\theta\), need not be an integer..

Answer 8

tell me about the geometric method @ParthKohli

Answer 9

Eh, so far, it's only about seeing. It's not a strict proof.