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Mathematics · Complex Numbers

JEE Main 2024 — 4 April, Shift 2 — Question 10

The area (in sq. units) of the region S={z∈C;∣z−1∣≤2;(z+z‾)+i(z−z‾)≤2,lm⁡(z)≥0}\mathrm{S}=\{\mathrm{z} \in \mathbb{C} ;|\mathrm{z}-1| \leq 2 ;(\mathrm{z}+\overline{\mathrm{z}})+\mathrm{i}(\mathrm{z}-\overline{\mathrm{z}}) \leq 2, \operatorname{lm}(z) \geq 0\} is

  1. Option A:

    7π3\frac{7 \pi}{3}

  2. Option B:

    3π2\frac{3 \pi}{2}

    Correct
  3. Option C:

    17π8\frac{17 \pi}{8}

  4. Option D:

    7π4\frac{7 \pi}{4}

Answer: B

Step-by-step solution

Put z=x+\mathrm{z}=\mathrm{x}+iy

 ∣z−1∣≤2⇒(x−1)2+y2≤4#(1) (z+z‾ )+i(z−z‾ )≤2⇒2x+i(2iy)≤2#(1) ⇒x−y≤1#(2)\begin{matrix}~\left| z-1 \right|\le 2\Rightarrow {{(x-1)}^{2}}+{{y}^{2}}\le 4\#\left( 1 \right) \\~\left( z+\overline{{\mathrm{z}}}\, \right)+i\left( z-\overline{{\mathrm{z}}}\, \right)\le 2\Rightarrow 2x+i\left( 2iy \right)\le 2\#\left( 1 \right) \\~\Rightarrow x-y\le 1\#\left( 2 \right) \\\end{matrix}

Required area== Area of semi-circle - area of sector A

Required area=12π(2)2−π2=\frac{1}{2} \pi(2)^{2}-\frac{\pi}{2}

Required area=3π2=\frac{3 \pi}{2}

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
4 April, Shift 2
Subject
Mathematics
Chapter
Complex Numbers
Topic
Geometry of Complex Numbers