1
$\begingroup$

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.

Thanks in advance.

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$
    – user403337
    Apr 1, 2020 at 3:43

2 Answers 2

0
$\begingroup$

$\omega$ a primitive third root of unity implies $\omega=e^{(2\pi i k)/3},\,k=1,2$. And $-1=e^{\pi i}$. Thus $-\omega=e^{\pi i}e^{(2\pi i k)/3}=e^{(6+4k)\pi i/6}=e^{(2\pi i k)/6}\,,k=1,5$.

$\endgroup$
0
$\begingroup$

Yes, that is correct.

Another way, without computing anything is this:

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).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .