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Question #41118

Difficulty: 8/10 | 🟢 Free

The Question

Let $S = \{ z \in \mathbb{C} : |z| = 1, z \neq 1 \}$ . A function $f: S \to \mathbb{C}$ is defined as $f(z) = z \frac{z-1}{|z-1|}$ . Let $N(w)$ denote the number of distinct solutions $z \in S$ to the equation $f(z) = w$ . Which of the following statements is TRUE?

Options

A.
$N(w) = 1 \;\forall\; w \in f(S)$
B.
$N(w) = 2 \iff \text{Re}(w) < 0 \text{ and } w \in f(S)$
C.
$N(w) = 2 \iff \text{Im}(w) > 0 \text{ and } w \in f(S)$
D.
$N(w) = 2 \;\forall\; w \in f(S) \setminus \{1\}$

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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

Injective Mapping (One-One)|Locus Problems via Transformation|The Modulus Trap

Why This Question Is Tricky

Common Mistakes

This question tests the 'Modulus Trap' in complex transformations.

Students often blindly apply algebraic simplifications like $(z-1)/|z-1| = e^{i \arg(z-1)}$ without considering how the argument behaves across different branches of the domain.

Specifically, on the unit circle, $z-1$ crosses the origin's vertical alignment, causing the sign of its imaginary part to flip.

This forces a piecewise definition of the transformation.

Students who ignore the absolute value's piecewise nature will incorrectly assume the function is globally injective (Option A).

By carefully tracking the argument for $\text{Im}(z) > 0$ and $\text{Im}(z) < 0$ , the student discovers that the two halves of the unit circle map to overlapping arcs, breaking injectivity precisely in the left half-plane.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Relations and Functions - Class 12
Concept	Injective Mapping (One-One)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free