Solution. From the given condition, we get $z^{2}-a z+1=0$. This means that $z$ is a solution to the quadratic equation $x^{2}-a x+1=0$. The latter equation has real solutions, and then $z$ is a real number, or the solutions are the pair of conjugate complex numbers $z$ and $\bar{z}$. However, according to Vieta's formulas, we get $z \bar{z}=1$, which means $|z|=1$.