75th Romanian Mathematical Olympiad

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

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

Applying the modulus to both members of the identity $\frac{1}{2}(z^2+zw+w^2{)}^2+\frac{1}{2}(z^2-zw+w^2{)}^2+(z^4-z^2{w}^2+w^4)=2(z^4+z^2{w}^2+w^4)$, 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 $2z^2-5zw+2w^2=0\iff (2z-w)(z-2w)=0$. 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 $5z^2-2zw+5w^2=0\iff (\phi z-w)(z-\phi w)=0$, where $\phi =\frac{1+2\sqrt{6i}}{5}$, $5{\phi }^2-2\phi +5=0$, $|\phi |=1$. Thus $w=\phi z$ or $z=\phi w$. 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 $|z^4-z^2{w}^2+w^4|=13$.

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.