# What is the cardinality of the set of roots of unity?

Consider the geometric interpretation of "roots of unity":

My intuition says that you can place arbitrarily many equidistant points on the unit circle and catch every point that lies on it. Therefore every $z \in \mathbb{C}$ that lies on the unit circle should be a solution to $z^n = 1$ for some $n \in \mathbb{N}$.

But, if I understand correctly, for any $n$, $z^n = 1$ has exactly $n$ roots.

Therefore, we have a countable union of countable sets, and therefore the set of roots of unity is countable.

Does that mean that there are points on the unit circle that are not in the sets of roots of unity?

• The statement "Therefore every $z \in \mathbb{C}$ should be a solution to $z^n=1$ for some $n \in \mathbb{N}$" is incorrect unless you allow $n = 0$. Consider $z = 2$ (or any point $z$ not on the unit circle). Everything you have written after that statement is correct. – JimmyK4542 Jul 22 '15 at 4:20
• That same intuition would also lead you to believe every real number is rational, since we can (by your logic) subdivide the interval $[0,1]$ arbitrarily many times and "catch every point on it". – David Wheeler Jul 22 '15 at 4:22
• Corrected. Thank you. – Andrei Savin Jul 22 '15 at 4:24

For a concrete example, consider $z=e^i$.

Suppose for the sake of contradiction that $z^n=1$ for some $n$. Then

$$e^{in}=1\hspace{5mm}\implies\hspace{5mm}n=2m\pi$$ for some integer $m$. But this means $\pi=\dfrac{n}{2m}$ which is impossible because $\pi$ is irrational.

Your argument that the number of roots is countable is good. Your intuition that you "catch every point that lies on it" is not correct, and your other argument proves that. Every root of unity is an algebraic number, as it solves a polynomial $z^n-1=0$ The argument, which is essentially the same, that the algebraic numbers are countable applies here.

Does that mean that there are points on the unit circle that are not in the sets of roots of unity?

Yep!

My intuition says that you can place arbitrarily many equidistant points on the unit circle and catch every point that lies on it.

Your intuition about "catching" a point reflects the fact that the unit circle is the closure of the set of roots of unity. So this example proves that the closure of a countable set may be uncountable, which is very, very useful.

Yes, take any angle $\theta$ that is not commensurable with $\pi$, i.e., $\theta/\pi \notin \mathbb Q$. Then, applying DeMoivre's theorem:

$(\cos\theta+i\sin \theta)^n=\cos(n\theta)+i\sin(n\theta)=\cos(4k \pi)+i\sin(4k\pi)$, which forces $n\theta =2k\pi \implies\theta=2k\pi/n$. So $\theta, \pi$ must have this relation.