Question

find all complex numbers $z$ which satisfy the following equation
$z^2=-\overline{z}$

Answer 1

$z^2=-\overline{z}$ or ${(x+iy)}^2=-(x-iy)$
or $x^2=y^{2}2ixy=-x+iy$.
Equating real and imaginary parts,
$x^2-y^2=-x$
and $2xy=y$ or $y(2x-1)=0$
From $\left(2\right),$ either $y=0$ or $x=\frac{1}{2}.$
When $y=0$, $\left(1\right)$ gives $x^2=-x$
or $x=-1$
Hence we get two sets of solution $x=0$, $y=0$, $x=-1$, $y=0$
When $x=\frac{1}{2},$ $\left(1\right)$ gives $\frac{1}{4}-y^2=-\frac{1}{2}$
or $y^2=\frac{3}{4}$ which gives $y=\pm \frac{\sqrt{3}}{2}$
Hence with obtain two more sets of solution:
$x=\frac{1}{2}$, $y=\frac{\sqrt{3}}{2}$; $x=\frac{1}{2}$, $y=-\frac{\sqrt{3}}{2}$
Thus in all we get the following four solution:
$z_1=0+i0=0$, $z_2=-1+i.0=-1$, $z_3=\frac{1}{2}+\frac{i\sqrt{3}}{2}$, $z_4=d\frac{1}{2}-\frac{i\sqrt{3}}{2}$