Question

If $z$ is a complex number and $\overline{z}$ is it's conjugate, then the number of solution(s) of $z^3+\overline{z}=0$ is

Answer 1

Let $z=x+iy,x,y\in R$, then
$z^3+\overline{z}=0\Rightarrow (x+iy{)}^3+(x-iy)=0\Rightarrow x^3+(iy{)}^3+3ixy(x+iy)+x-iy=0\Rightarrow (x^3+x-3x{y}^2)-i(y^3-3x^{2}y+y)=0\Rightarrow x^3+x-3x{y}^2=0,y^3-3x^{2}y+y=0$
So, solving
$x^3+x-3x{y}^2=0\cdots \left(1\right)\Rightarrow x=0\text{or}{x}^2-3y^2+1=0y^3-3x^{2}y+y=0\cdots \left(2\right)$

When $x=0$, using equation $\left(2\right)$, we get
$y(y^2+1)=0\Rightarrow y=0(\because y\in R)$

When $x^2-3y^2+1=0$ using equation $\left(2\right)$, we get
$y^3-3x^{2}y+y=0\Rightarrow y(y^2-3x^2+1)=0\Rightarrow y(y^2+1-3(3y^2-1\left)\right)=0\Rightarrow y(-8y^2+4)=0\Rightarrow y=0,y=\pm \frac{1}{\sqrt{2}}$
But when $y=0$
$x^2+1=0$, we don't get real solutions, so
$y=\pm \frac{1}{\sqrt{2}}\Rightarrow x=\pm \frac{1}{\sqrt{2}}$

Therefore,
$(x,y)=(0,0),(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}),(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}),(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}),(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$

Hence, the number of solution is $5.$