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Question #40496

Difficulty: 8/10 | 🟢 Free

The Question

Let $\alpha$ and $\beta$ be the roots of the equation $z^2 - (\sqrt{3}-1 + i(\sqrt{3}+1))z - 2\sqrt{3} + 2i = 0$ . If $\text{Arg}(z)$ denotes the principal argument of a complex number $z$ such that $\text{Arg}(z) \in (-\pi, \pi]$ , then the value of $\text{Arg}(\alpha^{200}) + \text{Arg}(\beta^{200})$ is

Options

A.
$-\frac{4\pi}{3}$
B.
$\frac{2\pi}{3}$
C.
$-\frac{2\pi}{3}$
D.
$\frac{4\pi}{3}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Vieta’s Formulas in Complex Domain|Euler's Form and Rotations|Principal Value Branch Boundaries|Compound Angle Expansion

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to extract complex roots using algebraic manipulation and evaluate their arguments under exponentiation.

The 'Lethal Collision' occurs when students bypass finding individual roots and instead use Vieta's product formula to find $\text{Arg}((\alpha\beta)^{200})$ .

Because the principal argument function $\text{Arg}(z)$ is multi-valued and restricted to $(-\pi, \pi]$ , the identity $\text{Arg}(z_1) + \text{Arg}(z_2) = \text{Arg}(z_1 z_2)$ fails when the sum of the arguments crosses the branch cut.

Students who blindly apply this property will incorrectly obtain $2\pi/3$ , falling into the trap.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Complex Numbers and Quadratic Equations - Class 11
Concept	Vieta’s Formulas in Complex Domain
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free