Question

The area of the polygon, whose vertices are the non-real roots of the equation $\bar z = iz^2$ 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)

$\frac{3\sqrt 3 }{4}$

Answer 2

$\bar z = iz^2$

Let $z = x+ iy$

$x-iy=i(x^2-y^2+2xiy)$

$x-iy = i(x^2-y^2)-2xy$

$∴\; x = -2yx \;or \;x^2-y^2=-y$

$x = 0$ or

$y = - \frac{1}{2}$

Instance-A

$x=0$ and $-y^2 = -y$ , $y = 0,1$

Instance-B

$y=-\frac{1}{2}$

$⇒ x^2-\frac{1}{4}=\frac{1}{2}$

$⇒ x = \overset{+}{-}\frac{ \sqrt3}{2}$

$z = \\{0,i, \frac{√3}{2}-\frac{i}{2}, \frac{√3}{2}-\frac{i}{2}\\}$

Area of polygon

= $\frac{1}{2}\begin {vmatrix}0 &1& 1\\\ \frac{√3}{2} &\frac{-1}{2} &1 \\\\\frac{- √3}{2}& \frac{-1}{2} &1\end {vmatrix}$

= $\frac{1}{2}| - √3 - \frac{√3}{2}|$

= $\frac{3√3}{4}$