Sum of n-th roots of unity

Question

I'm being asked to prove that
$$1 + \omega + \omega^2 + ... + \omega^{n-1} = 0$$ where $\omega \ne 1$ is an n-th root of unity, and I don't know where to start

I feel like there's something terribly obvious that I'm missing here.

Hint: sum of the first n terms of a geometric series – gammatester Apr 21 '16 at 11:51 — gammatester, Apr 21 '16 at 11:51
Do you know how to sum a finite geometric progression? – Apr 21 '16 at 11:51 — , Apr 21 '16 at 11:51

score 5 · Answer 1 · answered Apr 21 '16 at 11:51

5

Hint : multiply by $\omega - 1 $

answered Apr 21 '16 at 11:51

Vincent

2,318

score 3 · Accepted Answer · answered Apr 21 '16 at 11:52

3

$$x^n-1=0$$ $$(x-1)(x^{n-1}+x^{n-2}+\cdots+x+1)=0$$ Then, $x=1$ or $x^{n-1}+x^{n-2}+\cdots+x+1=0$.

The root of the second equation is what we denote by $\omega$.

answered Apr 21 '16 at 11:52

GoodDeeds

11,185
3
22
42

this helped me greatly thank you very much – davkav9 Apr 21 '16 at 12:15

score 1 · Answer 3 · answered Apr 21 '16 at 11:52

1

Multiply both sides by $(1-w)$, then you will come up with an expression $1-w^n$, which must be zero.

answered Apr 21 '16 at 11:52

Sanwar

360

Sum of n-th roots of unity

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