Question

find all complex numbers $z$ which satisfy the following equation
$z^3=\overline{z}$

Answer 1

$z^3=\overline{z}$ or $(x+iy{)}^3=x-iy$
or $x^3+3x^2.iy+3x{i}^2{y}^2=x-iy$
or $(x^3-3{xy}^2+i(3x^{2}y-y^3)=x-iy$
Equating real and imaginary parts,
$x^3-3x{y}^2=x\therefore x(x^2-3y^2-1)=0$
$3x^{2}y-y^3=-y\therefore y(3x^2-y^2+1)=0$
Putting $x=0$ from $\left(1\right)$ in $\left(2\right)$, we get $-y^3=-y$
or $y(y^2-1)=0\therefore y=0,1,-1$
$\therefore z=0+0i,0+i,0-i$
i.e. $z=0,i,-i$
We may say put $y=0$ from $\left(2\right)$ in $\left(1\right)$, then
$x(x^2-1)=0$
$\therefore x=0,1,-1$ but these will give real numbers not complex.
Again cancelling $x$ and $y$ from $\left(1\right)$ and $\left(2\right)$, we get $x^2-3y^2=1$ and $3x^2-y^2=-1$. These when solved give $x^2=y^2=-\frac{1}{2}$ which is not possible as $x$ and $y$ are real. Hence the value of $z$ are given by $\left(A\right).$