Find $w^2$ and compute a quadratic equation satisfied by $w$ I'm given $$\zeta_7 = e^{\frac{2\pi i}{7}} = \cos\frac{2\pi}{7}+i\sin\frac{2\pi}{7}$$
and I know $$\sum_{k=0}^{6}\zeta_7^k=0 $$
I'm then given $$w=\zeta_7+\zeta_7^2+\zeta_7^4$$
and that $\zeta_7^7 = 1$.
I'm asked to find $w^2$ and I found that that's $\zeta^2_7+2\zeta_7^3+\zeta_7^4+2\zeta_7^5+2\zeta_7^6+\zeta_7^8$ so $w^2=\zeta_7^3+\zeta_7^5+\zeta_7^6-1$ and if we add w, we get $\zeta_7^0+\zeta_7+\zeta_7^2+\zeta_7^3+\zeta_7^4+\zeta_7^5+\zeta_7^6-2\zeta_7^0-w$ and that, I think, is just $-2-w$
So, if $w^2=-2-w$ then $w^2+w+2=0$ and w solves it, so that's the quadratic, right?

Here's a slightly simpler approach:
$$w^2=(\zeta_7+\zeta_7^2+\zeta_7^4)^2=\zeta_7^2+\zeta_7^4+\zeta_7^8+2\zeta_7^3+2\zeta_7^5+2\zeta_7^6,$$
so
$$w^2+w=2\zeta_7^2+2\zeta_7^4+2\zeta_7^1+2\zeta_7^3+2\zeta_7^5+2\zeta_7^6=2(0-\zeta_7^0)=2(-1),$$
hence
$w^2+w+2=0$.
