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Question #41219

Difficulty: 8/10 | 🟢 Free

The Question

Let $S$ be the set of all complex numbers $z$ satisfying $|z| = 1$ such that the equation $\sin^{-1}\left( z^4 + \frac{1}{z^4} \right) + \cos^{-1}\left( i\left(z^2 - \frac{1}{z^2}\right) \right) = \frac{\pi}{2}$ holds true. The value of $\sum_{z \in S} \arg(z)$ , where $\arg(z) \in [0, 2\pi)$ , is

Options

A.
$5\pi$
B.
$\frac{13\pi}{2}$
C.
$3\pi$
D.
$4\pi$

Check Your Answer

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

Domain Constraint Intersections|Geometry of Complex Inequalities|De Moivre’s Theorem Applications

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of complex number geometry (De Moivre's Theorem) with the domain constraints of inverse trigonometric functions.

Students often solve the resulting trigonometric equation $2\sin^2(2\theta) - \sin(2\theta) - 1 = 0$ to find $\sin(2\theta) = 1$ or $-1/2$ , and blindly accept both.

The trap lies in treating the inverse trigonometric argument as a purely real variable without verifying the boundary conditions.

The De Moivre expansion forces the argument to be $-2\sin(2\theta)$ , which must lie in $[-1, 1]$ .

The root $\sin(2\theta) = 1$ yields an argument of $-2$ , falling outside the domain of $\cos^{-1}(x)$ , leading to extraneous solutions and the incorrect sum $\frac{13\pi}{2}$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Inverse Trigonometric Functions - Class 12
Concept	Domain Constraint Intersections
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free