Let the curve z(1+i)+ z (1-i)=4, z ∈ C , divide the region |z-3| ≤ 1 into two parts of areas α and β . Then |α-β| equals :

Correct answer: A. 1+(π)/(2) aligned & Let z=x+i y & (x+i y)(1+i)+(x-i y)(1-i)=4 & x+i x+i y-y+x-i x-i y-y=4 & 2 x-2 y=4 & x-y=2 & |z-3| ≤ 1 & (x-3)^2+y^2 ≤ 1 aligned Area of shaded region = (π · 1^2)/(4)-(1)/(2) · 1 · 1= (π)/(4)-(1)/(2) Area of unshaded region inside the circle =(3)/(4) π · 1^2+(1)/(2) · 1 · 1=(3 π)/(4)+(1)/(2) difference of area =((3 π)/(4)+(1)/(2))-((π)/(4)-(1)/(2))=(π)/(2)+1

JEE Main 202522 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 22 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Complex Number

JEE Main 2025 · 22 January · Evening Shift · Mathematics · Complex Number

Let the curve z(1+i)+ z (1-i)=4, z ∈ C , divide the region |z-3| ≤...

Let the curve

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

, divide the region

|z-3| \leq 1

into two parts of areas

\alpha

and

\beta

. Then

|\alpha-\beta|

equals :

Let the curve z(1+i)+ z (1-i)=4, z ∈ C , divide the region |z-3| ≤ 1 into two parts of areas α and β . Then |α-β|

Options

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

Correct answer

A. $1+\frac{\pi}{2}$

Step-by-step solution

\begin{aligned} & \text{ Let } z=x+i y \\ & (x+i y)(1+i)+(x-i y)(1-i)=4 \\ & x+i x+i y-y+x-i x-i y-y=4 \\ & 2 x-2 y=4 \\ & x-y=2 \\ & |z-3| \leq 1 \\ & (x-3)^2+y^2 \leq 1 \end{aligned}

Area

of shaded region =

\frac{\pi \cdot 1^2}{4}-\frac{1}{2}

\cdot 1 \cdot 1=

\frac{\pi}{4}-\frac{1}{2} Area

of

unshaded region inside the circle =\frac{3}{4} \pi \cdot 1^2+\frac{1}{2} \cdot 1 \cdot 1=\frac{3 \pi}{4}+\frac{1}{2} \therefore$ difference of area =\left(\frac{3 \pi}{4}+\frac{1}{2}\right)-\left(\frac{\pi}{4}-\frac{1}{2}\right)=\frac{\pi}{2}+1

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