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

Complex Numbers — JEE Main · 2025 · 22 Jan 2025 · Shift 2 Question with Solution

Let the curve

z (1 + i) + \overline{z} (1 - i) = 4, z \in C

, divide the region

∣ z - 3∣ \leq 1

into two parts of areas

α

and

β

. Then

∣ α - β ∣

equals :

Answer: (A)

Let

z = x + i y

. Then

z (1 + i) + \overset{z}{ˉ} (1 - i) = [x - y + i (x + y)] + [(x - y) - i (x + y)] = 2 (x - y) = 4 \Rightarrow x - y = 2.

The region

∣ z - 3∣ \leq 1

is the disk

(x - 3)^{2} + y^{2} \leq 1

, center

(3, 0)

, radius 1.

Distance from the center to the line

x - y = 2

\frac{∣3 - 0 - 2∣}{2} = \frac{1}{2}

.

For a chord at distance

d

from the center of a circle of radius

r

, the angle it subtends at the center is

2 θ

where

cos θ = d / r

. Here

cos θ = \frac{1}{2} \Rightarrow θ = \frac{π}{4}

, so the subtended angle is

\frac{π}{2}

.

Area of the minor segment:

\frac{1}{2} r^{2} (2 θ - sin 2 θ) = \frac{1}{2} (\frac{π}{2} - 1) = \frac{π}{4} - \frac{1}{2} .

Area of the major segment:

π r^{2} - (\frac{π}{4} - \frac{1}{2}) = \frac{3 π}{4} + \frac{1}{2} .

∣ α - β ∣ = (\frac{3 π}{4} + \frac{1}{2}) - (\frac{π}{4} - \frac{1}{2}) = \frac{π}{2} + 1.

JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Single correct option