# Roots of Unity for Precalculus

(a): What is the smallest positive integer $n$ such that all the roots of $z^4 + z^2 + 1 = 0$ are $n^{\text{th}}$ roots of unity?

(b) What is the smallest positive integer $n$ such that all the roots of $z^4 - z^2 + 1 = 0$ are $n^{\text{th}}$ roots of unity?

I am trying to incorporate the roots of unity in here, but I am having trouble so far. I know that the 2nd roots of unity are inside the fourth units of unity, and I am stuck there. Any help is appreciated.

• Welcome to math stack exchange – Peter Apr 12 '15 at 23:29
• It might be useful to note that $x^4+x^2+1=(x^2+x+1)(x^2-x+1)$. – Lubin Apr 12 '15 at 23:38

1) Roots of $p(z)=z^2+z+1$ are $\{j, j^2\}\subset \:3^{th}$ roots of the unity $(z^2+z+1= z^3-1/z-1)$. The roots of $z^4+z^2+1=p(z^2)$ are square roots of the roots of $z^2+z+1$. Thefore $n=6$.
2) $z^4-z^2+1=p(iz^2)$. I let u guess n.
(a) $z^6-1=(z^2-1)(z^4+z^2+1)$, so the roots of $z^4+z^2+1$ are all the $\xi\in\overline{\mathbb{Q}}$ such that $\xi^6=1$, except $1$ and $-1$. This means that $n=6$
(b) $z^6+1=(z^2+1)(z^4-z^2+1)$, and moreover $z^{12}-1=(z^6-1)(z^6+1)$, therefore $\xi^4-\xi^2+1=0\longleftrightarrow(\xi^{12}=1\wedge\xi^6\neq1\wedge\xi\neq\pm i$). Which means $n=12$.