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 $\text{Re}(z)\in \mathbb{Q}$ ?
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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
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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
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1$\begingroup$ The fourth roots of unity $\{\pm 1,\pm i\}$ also have this property. $\endgroup$ – carmichael561 Jul 12 '16 at 16:52
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$\begingroup$ thanks! question updated to reflect this. $\endgroup$ – user3096876 Jul 13 '16 at 14:05
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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 $\pm i$.
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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
