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Question #40465

Difficulty: 8/10 | 🟢 Free

The Question

Let $z(\theta) = \frac{1 - \tan^2(\theta/2) + 2i\tan(\theta/2)}{1 + \tan^2(\theta/2)}$ for $\theta \in [0, 4\pi] \setminus \{\pi, 3\pi\}$ . Let $S$ be the set of all $\theta$ in the given domain satisfying the equation $\text{Re}(\sqrt{z(\theta)}) + \text{Im}(\sqrt{z(\theta)}) = -\frac{1}{\sqrt{2}}$ , where $\sqrt{w}$ denotes the principal square root of the complex number $w$ (with argument in $(-\pi, \pi]$ ). Find the sum of all elements in $S$ .

Options

A.
$\frac{13\pi}{3}$
B.
$5\pi$
C.
$10\pi$
D.
$\frac{14\pi}{3}$

Check Your Answer

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

Square Root of Complex Numbers|Tangent Half-Angle Substitution|Domain Constraint Intersections

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of tangent half-angle substitutions with the domain constraints of the principal complex square root.

Students commonly make the mistake of blindly applying the algebraic identity $\sqrt{e^{i\theta}} = e^{i\theta/2}$ across the entire domain.

However, the branch cut of the principal square root (defined for arguments in $(-\pi, \pi]$ ) creates a discontinuity at $\theta = \pi, 3\pi$ , mapping the output to $-e^{i\theta/2}$ in the interval $(\pi, 3\pi)$ .

Failing to account for this branch displacement leads to accepting extraneous roots (like $11\pi/6$ ) and missing valid ones, resulting in the trap answer $5\pi$ .

Quick Facts About This Question

Question Information

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