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Question #40488

Difficulty: 8/10 | 🟢 Free

The Question

Let $C$ be the curve in the complex plane obtained by applying the transformation $w = \frac{3}{2}z + \frac{1}{2}\bar{z}$ to the unit circle $|z| = 1$ . Let $S$ be the set of all complex numbers $w$ satisfying $\text{Re}(w^2) = 3$ from which exactly two mutually perpendicular tangents can be drawn to $C$ . Find the value of $\sum_{w \in S} |w|^2$ .

Options

A.
0
B.
16
C.
20
D.
24

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

Locus Problems via Transformation|Condition of Tangency|Tangent/Normal Slope Interplay

Why This Question Is Tricky

Common Mistakes

This question tests the locus of perpendicular tangents (director circle) and the algebraic limitations of slope-based tangency conditions.

A common trap occurs when students derive the director circle using the quadratic slope equation $m^2(u^2-a^2) - 2uvm + (v^2-b^2) = 0$ and set the product of roots to $-1$ .

This algebraic method strictly requires the leading coefficient $u^2-a^2 \neq 0$ .

If students blindly follow this algebraic constraint, they will incorrectly exclude the points where $u = \pm a$ (the vertices), assuming tangents cannot be perpendicular there because the slope of a vertical tangent is undefined.

In this problem, the intersection of the director circle and the given hyperbola lands exactly on these 'undefined' vertex points $(\pm 2, \pm 1)$ .

Students falling for the trap will conclude the set $S$ is empty and incorrectly choose 0.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Complex Numbers and Quadratic Equations - Class 11
Concept	Locus Problems via Transformation
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free