Assuming $ω_0,ω_1,...,ω_n$ are the $n^{th}$ roots of unity, I am asked to show $$(x−ω_0)(x−ω_1)···(x−ω_{n−1}) =x^n−1$$ and $$\sum^{n−1}_{a=0}ω_a= 0$$ I understand that by definition, the $n^{th}$ roots of unity are the roots of the polynomial $x^n-1$. I'm not sure I know the properties of the roots of unity well enough to even know where to begin.
Thank you.
On
Since the roots of unity are the roots of the polynomial $p(z)=z^n-1$, and since the coefficient of the $z$ term is $0$, it follows immediately that $\sum \omega_a=0$. And since the roots of unity split the polynomial completely over $\mathbb C$, it follows again at once that $p(z)=\prod (z-\omega_a)=z^n-1.$
On
If you pick a primitive root $\zeta$ for n>1 then we can represent all of the other roots as powers of it.
$$\sum_{a=0}^{n-1} \omega_a = \sum_{k=0}^{n-1} \zeta^k$$
This rearrangement makes all the difference since now it's the geometric series,
$$\sum_{k=0}^{n-1} \zeta^k = \frac{\zeta^n-1}{\zeta-1}$$
Because $n>1$ then $\zeta \ne 1$ and so there's no issue with the denominator. Since $\zeta$ is a root of $x^n-1$ the numerator is $0$ which is what you wanted to show.
Since omegas are the roots of $x^n-1$
You can factor $x^n-1$ to get $$x^n-1=(x−ω_0)(x−ω_1)···(x−ω_{n−1}) $$
The sum of roots of a monomial of degree $n$ is the opposite of the coefficient of $x^{n-1}$ which in this case is $0$