Let S 1 =\ z in C: z leq 5\ , S 2 = \ z in C: Im (frac z+1-√(3) i 1-√(3) i ) geq 0 \ and S 3 =\ z in C : Re ( z ) geq 0\ . Then

Mathematics · Complex Numbers

JEE Main 2024 — 5 April, Shift 2 — Question 3

Let $S_{1}=\{z \in C:|z| \leq 5\}$ , $S_{2}=\left\{z \in C: \operatorname{Im}\left(\frac{z+1-\sqrt{3} i}{1-\sqrt{3} i}\right) \geq 0\right\}$ and $\mathrm{S}_{3}=\{\mathrm{z} \in \mathrm{C}: \operatorname{Re}(\mathrm{z}) \geq 0\}$ .

Then

Option A:
$\frac{125 \pi}{6}$
Option B:
$\frac{125 \pi}{24}$
Option C:
$\frac{125 \pi}{4}$
Option D:
$\frac{125 \pi}{12}$
Correct

Answer: D

Step-by-step solution

$\mathrm{S}_{1}: \mathrm{x}^{2}+\mathrm{y}^{2} \leq 25$

$S_{2}: \operatorname{Im}$ of $\frac{z+(1-\sqrt{3} i)}{(1-\sqrt{3} i)} \geq 0$

$\operatorname{Im}$ of $\left(\frac{x+i y}{1-\sqrt{3} i}+1\right) \geq 0$

$\operatorname{Im}$ of $\left(\frac{(x+i y)(1+\sqrt{3} i)}{4}\right) \geq 0$

$\Rightarrow \sqrt{3} x+y \geq 0$

$\mathrm{S}_{3}: \mathrm{x} \geq 0$

Area $=\frac{5}{12}\left(\pi(5)^{2}\right)$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 5 April, Shift 2
Subject: Mathematics
Chapter: Complex Numbers
Topic: Properties of Complex Numbers