Are all algebraic integers with absolute value 1 roots of unity? 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 $\alpha^{n^m - 1} = 1$.

Let me first mention an example in Character Theory. Let $G$ be a finite group of order $n$ and assume $\rho$ is a representation with character $\chi:=\chi_\rho$ which is defined by $\chi(g)=Tr(\rho(g))$. Since $G$ is a finite group then, by invoking facts from linear algebra, one can show $\chi(g)\in\mathbb{Z}[\zeta_n]$. For abelian groups, it is easy to see $\chi(g)$ is a root of unity, when $\chi$ is irreducible, but what about non-abelian groups? In other words let $|\chi(g)|=1$, what can we say about $\chi(g)$?
This relates to your question. Let assume $K/\mathbb{Q}$ be an abelian Galois extension inside $\mathbb{C}$, and take an algebraic integer $\alpha\in\mathcal{O}_K$ such that $|\alpha|=1$, then for any $\sigma\in Gal(K/\mathbb{Q})$ we have 
$$
|\sigma(\alpha)|^2=\sigma(\alpha)\overline{\sigma(\alpha)}
$$
Since $K/\mathbb{Q}$ is abelian then $\overline{\sigma(\alpha)}=\sigma(\overline{\alpha})$ so 
$$
|\sigma(\alpha)|^2=\sigma(|\alpha|)=1
$$
Then norm of all its conjugate is one so it must be a root of unity. This answer to the question was posed, therefore if $|\chi(g)|=1$ then $\chi(g)$ is root of unity.  
