Question

The equation $z^2=\bar{z}$ has

• A. no solution
• B. two solutions
• C. four solutions
• D. an infinite number of solutions

Answer 1

Answer: (C)

$z^2=\bar{z}$
$\Rightarrow {\left(x+iy\right)}^2=x-iy$ [if $z=x+iy$]
$\Rightarrow x^2-y^2+2ixy=x-iy$
$\Rightarrow x^2-y^2=x$ 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
$z_1=0+i0=0,z_2=1+i,0=1$
$z_3=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$ and $z_4=-\frac{1}{2}-\frac{\sqrt{3}}{2}i$