Mathematics · Complex Numbers

JEE Main 2025 — 22 January, Evening Shift — Question 19

Let the curve $z(1+i) + \bar{z}(1-i) = 4, z \in C,$ divide the region $|z-3| \le 1$ into two parts of areas $\alpha$ and $\beta.$ Then $|\alpha - \beta|$ equals:

Option A:
$1 + \frac{\pi}{2}$
Correct
Option B:
$1 + \frac{\pi}{3}$
Option C:
$1 + \frac{\pi}{4}$
Option D:
$1 + \frac{\pi}{6}$

Answer: A

Step-by-step solution

\text{Let\; } z = x + iy

(x+iy)(1+i) + (x-iy)(1-i) = 4

x + ix + iy - y + x - ix - iy - y = 4

2x - 2y = 4

x - y = 2

|z - 3| \le 1

(x-3)^2 + y^2 \le 1

\text{Area\; of\; shaded\; region\; } = \frac{\pi . 1^2}{4} - \frac{1}{2} . 1 . 1 = \frac{\pi}{4} - \frac{1}{2}

\text{Area\; of\; unshaded\; region\; inside\; the\; circle\; } = \frac{3}{4} \pi . 1^2 + \frac{1}{2} . 1 . 1 = \frac{3\pi}{4} + \frac{1}{2}

\therefore \text{ difference\; of\; area\; } = \left( \frac{3\pi}{4} + \frac{1}{2} \right) - \left( \frac{\pi}{4} - \frac{1}{2} \right)

$\therefore \text{ difference\; of\; area\; }$$1 + \frac{\pi}{2}$

Answer key and solution verified before publishing.

Practise Complex Numbers

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2025
Paper: 22 January, Evening Shift
Subject: Mathematics
Chapter: Complex Numbers
Topic: Geometry of Complex Numbers