4
$\begingroup$

All of the $4^{\text{th}}$ and $6^{\text{th}}$ roots of unity have real parts that are rational numbers. Are these the only roots of unity $z$ such that $\text{Re}(z)\in \mathbb{Q}$ ?

$\endgroup$
4
  • 1
    $\begingroup$ Re(z) is not rational for all of the $12^{th}$ roots of unity. $e^{\frac{i\pi}{6}}$ has real part $\frac{\sqrt{3}}{2}$. $\endgroup$ – Christian Jul 12 '16 at 16:36
  • 1
    $\begingroup$ Sorry about that, I should have said 6th roots of unity, question has been edited to reflect that. $\endgroup$ – user3096876 Jul 12 '16 at 16:43
  • 1
    $\begingroup$ The fourth roots of unity $\{\pm 1,\pm i\}$ also have this property. $\endgroup$ – carmichael561 Jul 12 '16 at 16:52
  • $\begingroup$ thanks! question updated to reflect this. $\endgroup$ – user3096876 Jul 13 '16 at 14:05
4
$\begingroup$

Suppose $z^n=1$ and $\Re(z)\in \mathbb{Q}$. Then also $\left(\overline{z}\right)^n=1$, so $z$ and $\overline{z}$ are both algebraic integers. This means that $z+ \overline{z} = 2 \Re(z)$ is also an algebraic integer. The only algebraic integers that are rational are integers, so $2 \Re(z)$ must be an integer. Since $z$ lies on the unit circle we find that the only possible values for $2\Re(z)$ are $\{-2,1,0,1,2\}$. These values indeed correspond to the 6th roots of unity together with $\pm i$.

$\endgroup$
1
$\begingroup$

Let us take the $n$th root of unity. We want to find when $\cos{2\pi m/n}$ is rational for integers $m$. We know this is true for $n = 1,2,3,4,6$, just by substitution.

Now we prove that these are the only results. Take $m=1$. We will show that $\cos{2\pi/n}$ is irrational for all other values of $n$. My first instinct here was to use Galois theory, like the construction of numbers using a straightedge and compass, but a quick google search shows that this has already been done, for example, see https://wilsonong.wordpress.com/2010/06/25/when-is-cos2pin-irrational/

I'll attempt to summarize the argument, and introduce some of the concepts of Galois theory.

Start with the rational numbers $\mathbb{Q}$, which are a field. We perform what is called a field extension by including $\zeta = e^{2\pi i/n}$, $\mathbb{Q}[\zeta]$ which is the smallest field containing both the rationals and $\zeta$. When studying Galois theory, we are interested in automorphisms of $\mathbb{Q}[\zeta]$ that leave the base field $\mathbb{Q}$ fixed. Under such an automorphism $\cos(2\pi/n)$ is mapped to $\cos(2\pi j/n)$. If $\cos(2\pi/n)$ is rational, it is fixed under the automorphism, forcing $j=1$ or $j=n-1$. This means the Galois group has at most two elements, only possible for $n=1,2,3,4,6$. A nice summary of relevant facts about cyclotomic extensions are given here http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/cyclotomic.pdf

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.