Complex Numbers — JEE Main · 2022 · 27 Jun 2022 · Shift 1 Question with Solution

The area of the polygon, whose vertices are the non-real roots of the equation

\overset{z}{ˉ} = i z^{2}

is :

Answer: (A)

Solution

Let

z = x + i y

, where

x, y \in R

. Then

\overset{z}{ˉ} = x - i y

and

z^{2} = (x^{2} - y^{2}) + 2 x y i

.

Substituting into

\overset{z}{ˉ} = i z^{2}

x - i y = i (x^{2} - y^{2}) + i^{2} (2 x y) = i (x^{2} - y^{2}) - 2 x y

Equating real and imaginary parts:

Real: $x = - 2 x y \Rightarrow x (1 + 2 y) = 0$ , so $x = 0$ or $y = - \frac{1}{2}$
Imaginary: $- y = x^{2} - y^{2}$

Case 1:

x = 0 \Rightarrow - y = - y^{2} \Rightarrow y^{2} - y = 0 \Rightarrow y = 0

y = 1

. This gives

z = 0

(real) and

z = i

(non-real).

Case 2:

y = - \frac{1}{2} \Rightarrow \frac{1}{2} = x^{2} - \frac{1}{4} \Rightarrow x^{2} = \frac{3}{4} \Rightarrow x = \pm \frac{3}{2}

. This gives

z = \frac{3}{2} - \frac{i}{2}

and

z = - \frac{3}{2} - \frac{i}{2}

.

The three non-real roots form an equilateral triangle with vertices at

(0, 1)

(\frac{3}{2}, - \frac{1}{2})

, and

(- \frac{3}{2}, - \frac{1}{2})

. The base has length

3

and height

\frac{3}{2}

.

Area

= \frac{1}{2} \cdot 3 \cdot \frac{3}{2} = \frac{3 3}{4}

JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Single correct option