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Question #41114

Difficulty: 8/10 | 🟢 Free

The Question

Let $S = \{ z \in \mathbb{C} : z^{40} = 1 \}$ and $A = \{ z \in S : |z^2 - 1| \le \sqrt{2} \}$ . An equivalence relation $R$ is defined on $A$ by $z_1 R z_2 \iff z_1^4 = z_2^4$ . If $N$ is the number of elements in $A$ , and $M$ is the number of equivalence classes of $R$ on $A$ , find the value of $N + M$ .

Options

A.
32
B.
27
C.
30
D.
28

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

Equivalence Class Partitioning|De Moivre’s Theorem Applications|Geometry of Complex Inequalities

Why This Question Is Tricky

Common Mistakes

This question tests the geometry of complex inequalities and equivalence class partitioning on roots of unity.

The trap is classically JEE Advanced: students often treat boundary conditions (where $|z^2 - 1| = \sqrt{2}$ ) as exclusive due to over-reliance on open-set inequality intuition.

Missing the boundary points $m = \pm 5$ (where $\text{Re}(z^2) = 0$ ) incorrectly reduces the element count to $N=18$ .

Furthermore, because $m=5$ and $m=-5$ are the ONLY elements providing the equivalence class $m \equiv 5 \pmod{10}$ (which corresponds to $z^4 = -1$ ), excluding them drops the number of classes to $M=9$ , perfectly yielding the trap answer 27.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Relations and Functions - Class 12
Concept	Equivalence Class Partitioning
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free