<s>[INST] Solve the equation $\lvert z\rvert^2z-3\overline z=0$ I was trying to solve the equation using the identities $z=x+iy$; $\overline z=x-iy$ and $\lvert z\rvert^2=x^2+y^2$ so as to get
$(x^2+y^2)(x+iy)-3(x-iy)=0$, that is $x^3+x^2iy+xy^2+iy^3-3x+3iy=0$
Now, a complex number is null $\iff Re(z)=Im(z)=0$, therefore
$x^3+xy^2-3x=0\implies x=0\lor x^2+y^2=3$, that is $\lvert z\rvert=\sqrt 3$
$x^2y+y^3+3y=0\implies y=0\lor x^2+y^2=-3$, that is $\lvert z\rvert=-\sqrt 3$
Is it correct to say that the solutions are $z=0$, $z=\sqrt 3$, and $-\sqrt 3$?
 [/INST] HINT:
As $|z|^2=z\cdot\bar z,$
we have $\bar z(z^2-3)=0$
</s>