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Question #40391

Difficulty: 8/10 | 🟢 Free

The Question

Let $\arg(z)$ denote the principal argument of a complex number $z$ , where $-\pi < \arg(z) \le \pi$ . Consider the region $S$ in the complex plane defined by $S = \left\{ z \in \mathbb{C} : |z| \le 6, \ \arg(z^2) \ge \frac{\pi}{2}, \ \cos(\arg(z)) \le 0 \right\}$ Find the area of the region $S$ .

Options

A.
$\frac{9\pi}{2}$
B.
$9\pi$
C.
$\frac{27\pi}{4}$
D.
$18\pi$

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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

ASTC Quadrant Sign Convention|Argument and Principal Value|Geometry of Complex Inequalities

Why This Question Is Tricky

Common Mistakes

The problem requires finding the region of complex numbers $z$ satisfying an inequality involving $\arg(z^2)$ .

Students often fail to map the ASTC quadrants to the complex plane correctly, ignoring the branch cut of the principal argument.

The trap lies in incorrectly assuming $\arg(z^2) = 2\arg(z)$ universally.

This flawed logic leads to $2\arg(z) \ge \pi/2 \implies \arg(z) \ge \pi/4$ , which, when intersected with $\cos(\arg(z)) \le 0$ , yields the erroneous region $[\pi/2, \pi]$ (Area $9\pi$ ).

By properly accounting for the jump discontinuity at the negative real axis, the valid region is actually in the 3rd quadrant $[-3\pi/4, -\pi/2]$ , demonstrating how ignoring modulus-based angular boundaries excludes valid regions and includes invalid ones.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Trigonometric Functions - Class 11
Concept	ASTC Quadrant Sign Convention
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free