JEE Main 2022 · 27 Jun · Shift 1 · Mathematics Q1

MathematicsSingle correct+4 / -1Question 1 of 90

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

\dfrac{3\sqrt3}4

\dfrac{3\sqrt3}2

\dfrac32

\dfrac34

Answer and worked solution

Official answer: A

Polar form. With $z=re^{i\theta}$ : $re^{-i\theta}=r^2e^{i(2\theta+\pi/2)}$ , so $r=1$ for non-zero roots and $3\theta=-\frac\pi2+2k\pi$ .

Roots. $\theta=\frac\pi2,-\frac\pi6,-\frac{5\pi}6$ : all non-real, three points on the unit circle $\frac{2\pi}3$ apart.

Area of the equilateral triangle in a unit circle: $\dfrac{3\sqrt3}4$ .

(A)

Topics:Polar and Euler form Geometry in the complex plane

JEE Main 2022 · 27 Jun · Shift 1, Mathematics question 1