<s>[INST] Finding two possible values of $z^2 + z + 1$ given that $z$ is one of the three cube roots of unity. 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!!
 [/INST] $$
0 = z^2 + z + 1 = (z+1/2)^2 - (1/2)^2 + 1 = (z+1/2)^2 + 3/4 \iff \\
z + \frac{1}{2} = \pm i \frac{\sqrt{3}}{2} \iff \\
z = \frac{-1 \pm i \sqrt{3}}{2}
$$
By the way:
$$
1 + z + z^2 = \frac{z^3-1}{z-1}
$$
as a finite term geometric series.
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