alleryIf $\omega$ is a non-real cube root of unity, find the value of $1 + \omega + \omega^2$.
A$0$
B$3$
C$-1$
D$1$
Answer & Solution
Correct answer: A. $0$
The three cube roots of unity are $1, \omega, \omega^2$, and they are roots of $z^3 - 1 = (z-1)(z^2 + z + 1) = 0$.
Vieta's formulas on $z^2 + z + 1$ give $\omega + \omega^2 = -1$, so $1 + \omega + \omega^2 = 0$.
Equivalently, the sum of *all* $n$-th roots of unity is zero for $n \ge 2$.
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