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Q.
The equation $z^2=\bar{z}$ has
Complex Numbers and Quadratic Equations
Solution:
$z^{2}=\bar{z}$
$\Rightarrow \left(x+iy\right)^{2}=x-iy$ [if $z = x + iy$]
$\Rightarrow x^{2}-y^{2}+2\,ixy=x-iy$
$\Rightarrow x^{2}-y^{2}=x$ and $2xy=-y$
$\Rightarrow x^{2}-y^{2}=x$ and $y\left(2x+1\right)=0$
Now $y \left(2x + 1\right) = 0$ gives $y = 0$
or $ x=-\frac{1}{2}$
when $y = 0$, we have $x^{2} - y^{2} - x = 0$
$\Rightarrow x^{2}-x=0$
$\Rightarrow x=0$ or $x=1$
when $x=-\frac{1}{2}, x^{2}-y^{2}-x=0$ gives
$\frac{1}{4}-y^{2}+\frac{1}{2}=0$
$\Rightarrow y^{2}=\frac{3}{4}$
$\Rightarrow y=\pm\frac{\sqrt{3}}{2}$.
Hence there are four solutions
$z_{1}=0+i\,0=0, z_{2}=1+i, 0=1$
$z_{3}=-\frac{1}{2}+\frac{\sqrt{3}}{2} i$ and $z_{4}=-\frac{1}{2}-\frac{\sqrt{3}}{2}i$