Solve the equation $z^3=z+\overline{z}$
Solution 1:
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, $z\in\{0,\sqrt2,-\sqrt2\}$.
Solution 2:
You have that $z^3$ is real, since it equals $z+\bar{z}$. So $z$ is on one of six regularly spaced rays pointing from the origin. In four cases out of six, $z^3$ and $z+\bar{z}$ would have opposing real parts and thus cannot be equal. This only leaves the two rays where $z$ itself is real. And your equation is reduced to $$z^3=2z\implies z(z-\sqrt{2})(z+\sqrt{2})=0$$
Solution 3:
You have $(a+bi)^3=2a$ Now expand the cube and equate real and imaginary parts
Solution 4:
Note that if $z=a+ib$ then $z+\overline{z}=2a$ and $$ z^3=2a $$ The cube roots of $2a$ are
\begin{align} \sqrt[3]{2a}\cdot\omega^{0} = & \sqrt[3]{2a} \\ \sqrt[3]{2a}\cdot\omega^{1} = & \sqrt[3]{2a}\big(\frac{1}{2}+i\frac{\sqrt{3}}{2}\big) \\ \sqrt[3]{2a}\cdot\omega^{2} = & \sqrt[3]{2a}\big(\frac{1}{2}+i\frac{\sqrt{3}}{2}\big)^{2}=\sqrt[3]{2a}\big(\frac{1}{2}-i\frac{\sqrt{3}}{2}\big) \end{align} where $\omega=\frac{1}{2}+i\frac{\sqrt[3]{3}}{2}$ is any cubic root of the unit.The possible values of $ a $ and $ b $ are obtained equaling $ z = a + ib$ to the roots of given above $2a$. \begin{align} \sqrt[3]{2a}\cdot\omega^{0} = & \sqrt[3]{2a} \\ \sqrt[3]{2a}\cdot\omega^{1} = & \sqrt[3]{2a}\big(\frac{1}{2}+i\frac{\sqrt{3}}{2}\big)^{1} \\ \sqrt[3]{2a}\cdot\omega^{2} = & \sqrt[3]{2a}\big(\frac{1}{2}+i\frac{\sqrt{3}}{2}\big)^{2}=\sqrt[3]{2a}\big(\frac{1}{2}-i\frac{\sqrt{3}}{2}\big) \end{align} More explicitly \begin{align} a+ib = & \sqrt[3]{2a} \\ a+ib = & \frac{1}{2}\sqrt[3]{2a}+i\frac{\sqrt{3}}{2}\sqrt[3]{2a} \\ a+ib = & \frac{1}{2}\sqrt[3]{2a}-i\frac{\sqrt{3}}{2}\sqrt[3]{2a} \end{align}
Solution 5:
$$z^3=z+\bar{z}=2\,{\rm Re}(z)\implies z^3\in{\bf R}\implies z=\omega r ~~(\omega^3=1,r\in{\bf R})$$
$$r^3=(\omega+\overline{\omega})r\implies\begin{cases} r=0 \\ r^2=2\,{\rm Re}(\omega) \implies\begin{cases} r=\pm\sqrt{2} & \omega=1 \\ r^2<0 & \omega\ne1\end{cases}\end{cases} $$
Thus the solutions are $0$ and $\pm\sqrt{2}$.