If $f(x)$ is a monic polynomial in ${\mathbb Z}[x]$ and all roots have absolute value 1 (i.e. lie on the circle $S^1$ of radius 1), then all roots are roots of unity (satisfy $x^k=1$). Indeed, if $x_1,...,x_n$ are the roots of $f$, then for every integer $k$, $x_1^k,...,x_n^k$ are all roots of some monic polynomial $f_k$ of degree $n$ over $\mathbb Z$. The coefficients of $f_k$ are (by Vieta's formulas) $\pm$ elementary symmetric polynomials of $x_1^k,...,x_n^k$. Since the absolute values of $x_i^k$ are 1, the coefficients of $f_k$ are bounded independently of $k$. Hence there are finitely many different polynomials $f_k$, so $x_i^k=x_i^l$ for some $k\ne l$ and $i=1,2,...,n$, hence $x_i^{l-k}=1$. Now if you take any algebraic number on $S^1$ which is not a root of unity, some of its conjugates will not belong $S^1$.

If $f(x)$ is a monic polynomial in ${\mathbb Z}[x]$ and all roots have absolute value 1 (i.e. lie on the circle $S^1$ of radius 1), then all roots are roots of unity (satisfy $x^k=1$). Indeed, if $x_1,...,x_n$ are the roots of $f$, then for every integer $k$, $x_1^k,...,x_n^k$ are all roots of some monic polynomial $f_k$ of degree $n$ over $\mathbb Z$. The coefficients of $f_k$ are (by Vieta's formulas) $\pm$ elementary symmetric polynomials of $x_1^k,...,x_n^k$. Since the absolute values of $x_i^k$ are 1, the coefficients of $f_k$ are bounded independently of $k$. Since these coefficients are integers (these numbers are algebraic integers and they are obviously stable under any automorphism of the field extension), there are finitely many different polynomials $f_k$, so $x_i^k=x_i^l$ for some $k\ne l$ and $i=1,2,...,n$, hence $x_i^{l-k}=1$. Now if you take any algebraic integer on $S^1$ which is not a root of unity, some of its conjugates will not belong $S^1$.