# if $n$ is a Prime number and $\omega \neq 1$ then $1+\omega+\omega^2+...+\omega^{n-1}=0$

Let $n$ is a Prime number , $\omega \neq 1$ is a n-th root of unity , show that Other to form the n-th roots are $\omega ,\omega^2,...,\omega^{n-1}$ and we have $1+\omega+\omega^2+...+\omega^{n-1}=0$

• Hint: Factor $\omega^n - 1$. Commented Nov 3, 2014 at 19:44

Well here are a couple of hints:

$$(\omega^2)^n=\omega^{2n}=(\omega^n)^2$$

$$x^n-1=(x-1)(x^{n-1}+ \dots +1)$$

Note that $x^n-1=0$ has at most $n$ distinct roots in $\mathbb C$.

$0=w^n - 1=(w-1)(1 + w + w^2 + \cdots + w^{n-1})$

since $w- 1\neq 0$, we have $1 + w + w^2 + \cdots + w^{n-1} = 0$

Let $w = \exp(i\dfrac{2l\pi }{n})$, then $$w^k = \exp(i\dfrac{2lk\pi }{n}) = \exp(i\dfrac{2[lk]_n\pi }{n}) , k = 0,1,2,\cdots, n-1$$

Consider the set $\{[kl]_n\, k = 0,1,2,\cdots n-1\}$, we will see $[kl]_n \neq [k'l]_n$ if $k\neq k'$: because if $[kl]_n =[k'l]_n$, then since $gcd(l,n)= 1$, we have $[k]_n = [k']_n$. i.e. $k = k'$ since they are both less than $n$.

So $\{[kl]_n\, k = 0,1,2,\cdots n-1\}$ constain $n$ different elements, then it is equal to the set $\{0, 1,2,\cdots,n-1\}$.

So $\{w_k, k = 0,1,2,\cdots, n-1\}$ is the set of $n$ roots for $z^n -1 =0$