# If $\omega = e^{(\frac{2\pi i}{n})}$ why $1+ \omega + \omega^{2} + ... + \omega^{n-1} = 0$? [duplicate]

Let $\omega = e^{(\frac{2\pi i}{n})}$ why $1+ \omega + \omega^{2} + ... + \omega^{n-1} = 0$?

I saw this on a algebra PPT slice. However the teacher did not explain why this equation is correct, can somebody show me some clue?

Thank you for your honest help!

• Geometrically it is rather clear, and interesting.
– Vim
May 31, 2016 at 5:33

Note that $\omega^n=1$, so by summing the geometric series we get $$1+\omega+\omega^2+\dots+\omega^{n-1}=\frac{1-\omega^n}{1-\omega}=0$$ as long as $n>1$.

• Thank you carmichael. that's a very concise and beautiful answer!
– Long
May 31, 2016 at 7:09

Assume $n>1$ (for $n=1$ this is not true). Let $S=1+\omega+\omega^2+\dots+\omega^{n-1}$. Then $$\omega S=\omega+\omega^2+\omega^3+\dots+\omega^n=\omega+\omega^2+\omega^3+\dots+1=S.$$ Since $\omega\neq 1$ (since $n>1$), this implies $S=0$.

• Thank you Eric. I did not notice that N must be great than 1. your answer is beautiful.
– Long
May 31, 2016 at 7:11
• @Eric: Hey Eric, I would like to contact you via email. Can I request your email please?
– C.S.
May 31, 2016 at 16:12

You can find some explanation of this also in Wikipedia article on roots of unity. You can also find some further resources simply by googling for sum of roots of unity is zero

Geometrically, this corresponds to the fact that barycenter of regular $n$-gon is the center of escribed circle.

If you know Viete's formulas, you can also use the fact that these numbers are precisely the roots of the polynomial $x^n-1=0$. So their sum is the coefficient of $x^{n-1}$, which is zero.

Having a look at some related posts could also help:

(I have only noticed that one of similar posts has already been suggested as a duplicate after posting this answer. I have added it to this list for better visibility.)

• Thank you martin. I speak poor english so I failed to search my question exhaustively, Thank you for your abundant solution.
– Long
May 31, 2016 at 7:07