If we have an algebraic number $\alpha$ with (complex) absolute value $1$, it does not follow that $\alpha$ is a root of unity (i.e., that $\alpha^n = 1$ for some $n$).  For example, $(3/5 + 4/5 i)$ is not a root of unity.

But if we assume that $\alpha$ is an algebraic integer with absolute value $1$, does it follow that $\alpha$ is a root of unity?



I know that if all conjugates of $\alpha$ have absolute value $1$, then $\alpha$ is a root of unity by the argument below:

The minimal polynomial of $\alpha$ over $\mathbb{Z}$ is $\prod_{i=1}^d (x-\alpha_i)$, where the $\alpha_i$ are just the conjugates of $\alpha$.  Then $\prod_{i=1}^d (x-\alpha_i^n)$ is a polynomial over $\mathbb{Z}$ with $\alpha^n$ as a root.  It also has degree $d$, and all roots have absolute value $1$.  But there can only be finitely many such polynomials (since the coefficients are integers with bounded size), so we get that $\alpha^n=\sigma(\alpha)$ for some Galois conjugation $\sigma$. If $\sigma^m(\alpha) = \alpha$, then $\alpha^{n^m} = \alpha$.

Thus $\alpha^{n^m - 1} = 1$.

No. There are algebraic integers on the unit circle which aren't roots of unity.

This paper by Ryan Daileda provides some useful information and references.

Also see Salem numbers.
