Factorise $z^3 - 1 $. If $z$ is one of the three cube roots of unity, find the two possible values of $z^2 + z + 1$.
Factorising gives you :
$(z - 1)(z^2 + z + 1) = 0$ since $z$ is one of the three cube roots of unity.
z is complex so $z \neq 1$ so $z-1 \neq 0$
Hence, $z^2 + z + 1 = 0$
Where do I go from here?
Any help is appreciated!!
You are done. The two possible values of $z^2+z+1$ are $0$ for $z$ a third root of unity different from $1$, as you have computed, and $1+1+1=3$ for $z=1$.
$$ 0 = z^2 + z + 1 = (z+1/2)^2 - (1/2)^2 + 1 = (z+1/2)^2 + 3/4 \iff \\ z + \frac{1}{2} = \pm i \frac{\sqrt{3}}{2} \iff \\ z = \frac{-1 \pm i \sqrt{3}}{2} $$ By the way: $$ 1 + z + z^2 = \frac{z^3-1}{z-1} $$ as a finite term geometric series.
More generally, you have $$(z-1)\sum_{j=0}^{N-1} z^j = z^N-1,$$ so for $z=1$ you have $\sum_{j=0}^{N-1} z^j = N$ and if $z\neq 1$ you get $$\sum_{j=0}^{N-1} z^j = \frac{z^N-1}{z-1}.$$ Hence, if $z$ is any $N^{th}$ root of unity not equal to 1, we have $$\sum_{j=0}^{N-1} z^j = 0.$$
