TITLE: Solve the equation $z^3=z+\overline{z}$
QUESTION [11 upvotes]: I have been trying to solve an equation $z^3=z+\overline{z}$, where $\overline{z}=a-bi$ if $z=a+bi$. But I cant find any clues on how to move forward on that one. Please help.

REPLY [15 votes]: Writing out real and imaginary parts of $z$ and separating real and imaginary parts yields
$$
x^3+3ix^2y-3xy^2-iy^3=2x
$$
therefore,
$$
x^3-3xy^2=2x\implies x=0\quad\text{or}\quad x^2-3y^2=2
$$
and
$$
3x^2y-y^3=0\implies y=0\quad\text{or}\quad3x^2=y^2
$$
If $y=0$, then $x^3=2x\implies x\in\{0,\sqrt2,-\sqrt2\}$.
If $x=0$, then $y=0$.
If $x^2-3y^2=2$ and $3x^2=y^2$, then $-8x^2=2$.
Thus, $(x,y)\in\{(0,0),(\sqrt2,0),(-\sqrt2,0)\}$; that is, $z\in\{0,\sqrt2,-\sqrt2\}$.

REPLY [4 votes]: $$z^3=z+\bar{z}=2\,{\rm Re}(z)\implies z^3\in{\bf R}\implies z=\omega r ~~(\omega^3=1,r\in{\bf R})$$
$$r^3=(\omega+\overline{\omega})r\implies\begin{cases} r=0 \\ r^2=2\,{\rm Re}(\omega) \implies\begin{cases} r=\pm\sqrt{2} & \omega=1 \\ r^2<0 & \omega\ne1\end{cases}\end{cases} $$
Thus the solutions are $0$ and $\pm\sqrt{2}$.