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Question #40472

Difficulty: 8/10 | 🟢 Free

The Question

Let $S$ be the set of all complex numbers $z$ satisfying the following conditions:

$|z - \sqrt{3}| + |z + \sqrt{3}| = 4$
$\big| |z|^2 - 2 \big| + \big| |z|^2 - 3 \big| = 1$
$\text{Arg}(z^3) = 3\text{Arg}(z)$
$\text{Im}(z) \ge 0$

Let $A$ be the area of the region in the complex plane bounded by the arc $S$ and the line segments joining the origin to the endpoints of $S$ . Find the value of $A$ .

Options

A.
$\sin^{-1}\left(\frac{1}{3}\right)$
B.
$2\sin^{-1}\left(\frac{1}{3}\right)$
C.
$\cos^{-1}\left(\frac{1}{3}\right)$
D.
$\frac{1}{2}\sin^{-1}\left(\frac{1}{3}\right)$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Argument and Principal Value|Ellipse Auxiliary Circle|Modulus Function Breakdown

Why This Question Is Tricky

Common Mistakes

The question integrates complex numbers with ellipse geometry via the auxiliary circle parameter $\theta$ .

The core trap lies in the Modulus Function Breakdown ( $\big| |z|^2 - 2 \big| + \big| |z|^2 - 3 \big| = 1$ ) which yields two symmetric arcs on the ellipse.

Students often ignore the principal value constraint of the argument function ( $\text{Arg}(z^3) = 3\text{Arg}(z) \implies \text{Arg}(z) \in (-\pi/3, \pi/3]$ ) and calculate the area for both arcs, leading to the trap answer $2\sin^{-1}(1/3)$ .

The problem forces a rigorous check of the argument's domain to reject the second-quadrant solution.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Complex Numbers and Quadratic Equations - Class 11
Concept	Argument and Principal Value
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free