Romanian Mathematical Olympiad

Put $z^{3n}+{\bar{z}}^{3n}=f_n(z)$. As $f_n(z)=\overline{f_n(z)}$, we get $f_n(z)\in \mathbb{R}$, for all $z\in \mathbb{C}$. We should determine the set $M=\{z\in \mathbb{C}/f_n(z)\ge 0\}$. If $z\in M$, then $f_0(z)=z+\bar{z}\ge 0$, so $\text{Re}(z)\ge 0$.

For non-negative $z\in \mathbb{R}$, $f_n(z)=2z^{3n}\ge 0$, for all non-negative integers $n$, implying $[0,\infty )\subset M$.

If $z=bi\in \mathbb{C}$, $b\in \mathbb{R}$, then $\bar{z}=-z$, so $f_n(z)=0$, for $n\in \mathbb{N}$, so $\{bi/b\in \mathbb{R}\}\subset M$. As $f_n(z)=f_n(\bar{z})$, we have $z\in M\iff \bar{z}\in M$.

It remains to find $z\in M$ such that $\text{Re}(z)>0$ and $\text{Im}(z)>0$. Let $z=r(\cos t+i\sin t)$ with $r>0$, $t\in (0,\frac{\pi }{2})$. Then $f_n(z)=2r^{3n}\cos (3^{n}t)$, so $z\in M\iff \cos (3^{n}t)\ge 0$, for all $n\in \mathbb{N}$.

As $t\in (0,\frac{\pi }{2})$, there is $k\in \mathbb{N}$ such that $3^{k}t\le \frac{\pi }{2}<3^{k+1}t$ (in fact $k=\lfloor {\log }_3\frac{\pi }{2t}\rfloor$). If $3^{k}t=\frac{\pi }{2}$, then $\cos (3^{n}t)>0$, for all $n\in \{0,1,2,\dots ,k-1\}$ and $\cos (3^{n}t)=0$, for all $n\ge k$, that is $z\in M$. If $3^{k}t<\frac{\pi }{2}<3^{k+1}t$, then $\frac{\pi }{2}<3^{k+1}t<\frac{3\pi }{2}$, implying $\cos (3^{k+1}t)<0$, that is $z\notin M$.

In conclusion $$M=[0,\infty )\cup \left\{r\left(\cos \frac{\pi }{2\cdot 3^k}\pm i\sin \frac{\pi }{2\cdot 3^k}\right)/r>0,k\in \mathbb{N}\right\}.$$