I'm fearful about putting this forward, because it seems the answer should be elementary. Certainly, the Weak Approximation Theorem allows every system of simultaneous inequalities among archimedean absolute values to be satisfied. But equality combined with inequality?



Can an algebraic number on the unit circle have a conjugate with absolute value different from 1?
Yes. Take
$$
\alpha=\sqrt{2-\sqrt{2}}+i\sqrt{\sqrt{2}-1}.
$$
Neither of the conjugates
$$
\sqrt{2+\sqrt{2}}\pm \sqrt{\sqrt{2}+1}
$$
have absolute value 1. 


It is impossible, however, if $\mathbb{Q}(\alpha)/\mathbb{Q}$ is abelian, since then all automorphisms commute with complex conjugation.


This was all stolen from Washington's Cyclotomic Fields book.


