Question

The area of the polygon, whose vertices are the non-real roots of the equation $\bar{z}=i{z}^2$ is:

• A. $\frac{3\sqrt{3}}{4}$
• B. $\frac{3\sqrt{3}}{2}$
• C. $\frac{3}{2}$
• D. $\frac{3}{4}$

Answer 1

Answer: (A)

$\Rightarrow$ Let $z=x+iy,x,y\in R$
Now $\bar{z}=i{z}^2$
then $x-iy=i\left(x^2-y^2+2xyi\right)$
$x-iy=i\left(x^2-y^2\right)-2xy$
$\Rightarrow x=-2xy\&-y=x^2-y^2$
$\Rightarrow x(1+2y)=0$
$x=0$ 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 $=\frac{1}{2}\cdot (\sqrt{3})\left(\frac{3}{2}\right)$
$=\frac{3\sqrt{3}}{4}$