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Question #40399

Difficulty: 8/10 | 🟢 Free

The Question

Let $z = x + iy$ be a complex number and $\vec{v} = x\hat{i} + y\hat{j} + 0\hat{k}$ be its corresponding position vector in $\mathbb{R}^3$ . Let $S$ be the set of all complex numbers $z$ satisfying the following system of conditions:

$|z(2\sqrt{2}-i) - 1|^2 - |z - (2\sqrt{2}+i)|^2 \ge 24$
$|\vec{v} \times \vec{u}|^2 + (\vec{v} \cdot \vec{u})^2 \le 400$ , where $\vec{u} = 3\hat{i} + 4\hat{j} + 0\hat{k}$
$\operatorname{Im}(z^4) \ge 0$

Find the area of the region $S$ in the complex plane.

Options

A.
$3\pi$
B.
$6\pi$
C.
$8\pi$
D.
$12\pi$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Factor-Product Conversion|Lagrange Identity|Geometry of Complex Inequalities

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to bypass tedious algebraic expansion by recognizing and applying factor-product conversions and the Lagrange identity in both complex and vector forms.

The primary trap lies in the first condition: students who attempt to expand $z = x+iy$ algebraically will be consumed by massive cross-terms, whereas recognizing the structure $|z\bar{w}-1|^2 - |z-w|^2 = (|z|^2-1)(|w|^2-1)$ yields an immediate locus.

The second condition reinforces this by using the vector Lagrange identity to define the outer boundary.

Finally, the phase-angle constraint $\operatorname{Im}(z^4) \ge 0$ acts as a secondary trap, testing if students correctly identify that the valid argument intervals sum to exactly half the plane ( $\pi$ radians) rather than mistakenly assuming it restricts to a single quadrant.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Trigonometric Functions - Class 11
Concept	Factor-Product Conversion
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free