Question

Find the complex number z if $z^2+\overline{z}=0$.

Answer 1

$z^2+\overline{z}=0$

Let $z=x+iy$

$\Rightarrow z^2+\overline{z}=0$

$\Rightarrow {(x+iy)}^2+(x-iy)=0$

$\Rightarrow x^2+i^2{y}^2+2xiy+x-iy=0$

$\Rightarrow x^2-y^2+x+i(2xy-y)=0$

$\Rightarrow x^2-y^2+x=0\text{and}y(2x-1)=0$

If $y=0\Rightarrow x(x+1)=0$

$\Rightarrow x=0,x=-1$

$\Rightarrow (0,0)\text{and}(-1,0)$

$\text{When}x=\frac{1}{2}$

$\Rightarrow y^2=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}$

$\Rightarrow y=\frac{\pm \sqrt{3}}{2}$

$\Rightarrow (\frac{1}{2},\frac{\sqrt{3}}{2})\text{and}(\frac{1}{2},\frac{-\sqrt{3}}{2})$

Hence, $z=0+0iorz=-1+0iorz=\frac{1}{2}+\frac{\sqrt{3}}{2}iorz=\frac{1}{2}-\frac{\sqrt{3}}{2}i$