Romanian Mathematical Olympiad

Applying the modulus to both members of the identity $(z^2 + zw + w^2) \cdot (z^2 - zw + w^2) = z^4 + z^2w^2 + w^4$ and, using the problem hypothesis, we obtain that $|z^2 - zw + w^2| = 3$.

Applying the modulus to both members of the identity $(z^4 + z^2w^2 + w^4) \cdot (z^4 - z^2w^2 + w^4) = z^8 + z^4w^4 + w^8$ and, using the problem hypothesis, we obtain that $|z^4 - z^2w^2 + w^4| = 13$.

Applying the modulus to both members of the identity $\frac{1}{2}(z^2 + zw + w^2)^2 + \frac{1}{2}(z^2 - zw + w^2)^2 + (z^4 - z^2w^2 + w^4) = 2(z^4 + z^2w^2 + w^4)$, using the modulus inequality and the above, we obtain that $\frac{1}{2} \cdot 49 + \frac{1}{2} \cdot 9 + 13 \ge 2 \cdot 21$.
We observe that equality holds, therefore there is a real number $t$ such that $(z^2 + zw + w^2)^2 = t^2(z^2 - zw + w^2)^2$. Moving to the modules, we find $t = \pm\frac{7}{3}$, so $3(z^2 + zw + w^2) = \pm 7(z^2 - zw + w^2)$.
For $t = \frac{7}{3}$ we obtain $2z^2 - 5zw + 2w^2 = 0 \Leftrightarrow (2z - w)(z - 2w) = 0$. It follows that $w = 2z$ or $z = 2w$. Substituting in $|z^2 + zw + w^2| = 7$, in the first case we obtain $|z| = 1$, and in the second $|w| = 1$.
For $t = -\frac{7}{3}$ we obtain $5z^2 - 2zw + 5w^2 = 0 \Leftrightarrow (\varphi z - w)(z - \varphi w) = 0$, where $\varphi = \frac{1+2\sqrt{6i}}{5}$, $5\varphi^2 - 2\varphi + 5 = 0$, $|\varphi| = 1$. Thus $w = \varphi z$ or $z = \varphi w$. Substituting in $|z^2 + zw + w^2| = 7$, in the first case we obtain $|z| = \sqrt{5}$, and in the second $|w| = \sqrt{5}$. None of these options are suitable, because they contradict equality $|z^4 - z^2w^2 + w^4| = 13$.

It is easy to verify that all pairs of the form $(z, 2z)$ and $(2z, z)$, where $z$ is a complex number with modulus equal to 1, have the property in the statement.