Solve a complex equation with conjugate and square This seems like a rather simple equation:
$$z^2 = \bar{z}$$
I tried $z=x+iy$ and then
$$x^2-y^2 + 2xyi=x-iy,$$
but this is an equation in two variables. Does anyone have a clever hint without giving the entire solution away? I want to solve it myself.
 A: Note that$$z^2=\overline z\implies|z|^2=\bigl|\overline z\bigr|=|z.|$$Therefore, $|z|=0(\iff z=0)$  or $|z|=1$. It turns out that $0$ is indeed a solution. If $|z|=1$, then $\overline z=\frac1z$. So, solve the equation $z^2=\frac1z(\iff z^3=1)$.
A: $$z^2=\overline z\to z^3=|z|^2\to r^3\text{ cis }3\theta=r^2.$$
Then 

$$z=0$$ or $$r=1,\\3\theta=2k\pi$$

and $z$ is a cubic root of unit.

The Cartesian equations say
$$\begin{cases}x^2-y^2=x,\\2xy=-y\end{cases}.$$
We can put aside $y=0$ which implies $x=0$ or $x=1$, and the second equation says $2x=-1$, then by the first $y=\pm\sqrt{\frac34}$.

 $$(0,0),(1,0),(-\frac12,\frac{\sqrt3}2),(-\frac12,-\frac{\sqrt3}2).$$

This solution is quite manageable.
A: Let $z=re^{it},r\ge 0$ and $t$ is real
$\implies r^2e^{2it}=re^{-it}\iff r(re^{3it}-1)=0$
If $r\ne0,$
$r=e^{-3it}\implies|r|=|e^{-3it}|=1\implies r=1$
$\implies e^{-3it}=1\implies-3t=2m\pi$ where $m$ is any integer
$\implies t=-\dfrac{2m\pi}3$ where $m\equiv0,1,2\pmod3$
A: We have that
$$z^2=\bar z \iff z^2z=\bar zz\iff z^3=|z|^2$$
therefore $z^3$ is real that is
$$z^3=r^3 \implies r^3=r^2 \implies r(r^2-1)=0 \implies r=0, 1$$
and then the solutions are


*

*$z=0$

*$z=1,e^{2i\pi/3},e^{-2i\pi/3}$

