Thank you for reporting, we will resolve it shortly
Q.
The number of solutions of the equation $z^{2}+\bar{z}=0$, is
Complex Numbers and Quadratic Equations
Solution:
Let $z=x+i y$, so that $\bar{z}=x+i y$, therefore
$z^{2}+\bar{z}=0 \Leftrightarrow\left(x^{2}-y^{2}+x\right)+i(2 x y-y)=0$
equating real and imaginary parts, we get
$x^{2}-y^{2}+x=0$
and $2 x y-y=0 $
$\Rightarrow y=0$
or $x=\frac{1}{2}$
If $y=0$, then (1) gives $x^{2}+x=0$
$ \Rightarrow x=0$
or $x=-1$
If $x=\frac{1}{2}$,
Then $x^{2}-y^{2}+x=0$
$ \Rightarrow y^{2}=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}$
$ \Rightarrow y=\pm \frac{\sqrt{3}}{2}$
Hence, there are four solutions in all.