if $\omega$ is a primitive cube root of unity then $-\omega$ is a primitive sixth root of unity.

I was given a statement that if $$\omega$$ is a primitive cube root of unity then $$-\omega$$ is a primitive sixth root of unity.

The roots of $$x^n−1$$ in $$\mathbb C$$ which are not also roots of $$x^m −1$$ for some $$1 ≤ m ≤ n$$ are called the primitive n’th complex roots of unity

So if this example works with $$n=6$$ I would get the following:

We have $$x^6−1 = (x−1)(x^5+x^4+x^3+x^2+x+1)$$, $$x^6−1 = (x^2−1)(x^4+x^2+1)$$ and $$x^6 −1 = (x^3 −1)(x^3 + 1)$$. The roots of $$x−1$$, $$x^2 −1$$, $$x^3 −1$$ are $$1,−1,\frac {−1} 2 ± \frac{{\sqrt3}} 2 i$$. Thus the remaining two roots of $$x^6−1$$, namely, $$ω^1 = \frac {1} 2 + \frac{{\sqrt3}} 2 i$$ and $$ω^5 = \frac {1} 2 - \frac{{\sqrt3}} 2 i$$ are the primitive 6’th complex roots of unity.

Is it correct to prove the above statement by just pointing out that if we set $$\omega = \frac {−1} 2 ± \frac{{\sqrt3}} 2 i$$ that $$-\omega = \frac {1} 2 - \frac{{\sqrt3}} 2 i$$ and $$-\omega = \frac {1} 2 + \frac{{\sqrt3}} 2 i$$ which are both sixth roots of unity.

• It's correct because, as you identified, they are primitive. It wouldn't be enough for them just to be sixth roots of unity, Primitive roots are special.
– user403337
Apr 1, 2020 at 3:43

$$\omega$$ a primitive third root of unity implies $$\omega=e^{(2\pi i k)/3},\,k=1,2$$. And $$-1=e^{\pi i}$$. Thus $$-\omega=e^{\pi i}e^{(2\pi i k)/3}=e^{(6+4k)\pi i/6}=e^{(2\pi i k)/6}\,,k=1,5$$.
Let $$\zeta$$ be a primitive 3rd root of unity. Then $$\zeta^3=1,\zeta\neq 1$$.
Now a 6th root of unity $$\omega$$ must satisfy $$\omega^6=1$$. Primitive means that $$\omega^2\neq 1$$ and $$\omega^3\neq 1$$. (It has order 6 and not a strict divisor of 6).