Question

Solve the equation, $z^2=\overline{z},$ where z is a complex number.

Answer 1

Let z=x+iy. Then,
$z^2=\overline{z}\Rightarrow (x+iy{)}^2=x-iy$
$\Rightarrow (x^2-y^2)+2ixy=x-iy$. ...(i)
Equating real parts and imaginary parts on both sides of (i) separately, we get
$x^2-y^2=x$ ...(ii) and $2xy=-y$ ...(iii)
From (iii), we get
$2xy+y=0\Rightarrow y(2x+1)=0\Rightarrow y=0orx=\frac{-1}{2}$.

Case I When y=0
Putting y=0 in (ii), we get
$x^2-x=0\Rightarrow x(x-1)=0\Rightarrow x=0orx=1$.
$\therefore (x=0,y=0)or(x=1,y=0)$
Thus, z=(0+i0)~or~z=(1+i0)

Case II When $x=\frac{-1}{2}$
Putting $x+\frac{-1}{2}$ in (ii), we get
${\left(\frac{-1}{2}\right)}^2-y^2=\left(\frac{-1}{2}\right)\Rightarrow y^2=(\frac{1}{4}+\frac{1}{2})=\frac{3}{4}\Rightarrow y=\pm \frac{\sqrt{3}}{2}$.

$\therefore (x=\frac{-1}{2},y=\frac{\sqrt{3}}{2})or(x=\frac{-1}{2},y=\frac{-\sqrt{3}}{2})$.
Thus, $z=(\frac{-1}{2}+\frac{\sqrt{3}}{2}i)$ or $z=(\frac{-1}{2}-\frac{\sqrt{3}}{2}i)$.
Hence, z=0, 1, $(\frac{-1}{2}+\frac{\sqrt{3}}{2}i)$ and $(\frac{-1}{2}-\frac{\sqrt{3}}{2}i)$ are the required roots of the given equation.