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Q.
The area of the polygon, whose vertices are the non-real roots of the equation $\bar{z}=i z^{2}$ is:
JEE MainJEE Main 2022Complex Numbers and Quadratic Equations
Solution:
$\Rightarrow$ Let $z=x+i y, x, y \in R$
Now $\bar{z}=i z^{2}$
then $x-i y=i\left(x^{2}-y^{2}+2 x y i\right)$
$x-i y=i\left(x^{2}-y^{2}\right)-2 x y$
$\Rightarrow x=-2 x y \&-y=x^{2}-y^{2}$
$\Rightarrow x (1+2 y )=0$
$x=0$ or $y=-\frac{1}{2}$
Put $x=0$ in $-y=x^{2}-y^{2}$
We get $y=y^{2}$
$\Rightarrow y =0,1$
Similarly
Put $y=-\frac{1}{2}$ in $-y=x^{2}-y^{2}$
$\Rightarrow \frac{1}{2}=x^{2}-\frac{1}{4}$
$\Rightarrow x^{2}=\frac{3}{4}$
$x=\pm \frac{\sqrt{3}}{2}$
$z=\left(0, i, \frac{\sqrt{3}}{2}-\frac{1}{2} i,-\frac{\sqrt{3}}{2}-\frac{1}{2} i\right)$
Area $=\frac{1}{2} \cdot(\sqrt{3})\left(\frac{3}{2}\right)$
$=\frac{3 \sqrt{3}}{4}$
