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

Question

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!

Answer 1

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 $n>1$.

Answer 2

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 $S=0$.

Answer 3

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:

• Intuitive understanding of why the sum of nth roots of unity is $0$
• Proof that sum of complex unit roots is zero
• Sum of n-th roots of unity
• Sum of nth roots of unity

(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.)