JEE Challenger
More from Complex Numbers and Quadratic Equations
MathHardAdv 2020 • Paper 1

Complex Numbers Satisfying Quadratic Absolute Value Equality

Let SS represent the set containing all complex numbers zz such that ∣z2+z+1∣=1|z^2 + z + 1| = 1. Which of the following statements is/are TRUE?

Options

A

∣z+12∣≤12\left|z + \frac{1}{2}\right| \le \frac{1}{2} for every z∈Sz \in S

B

∣z∣≤2|z| \le 2 for all z∈Sz \in S

Correct
C

∣z+12∣≥12\left|z + \frac{1}{2}\right| \ge \frac{1}{2} for every z∈Sz \in S

Correct
D

The set SS contains exactly four elements

Topics & Concepts

Complex Numbers and Quadratic EquationsComplex Numbers

Step-by-Step Solution

To determine which statements are true for the set S={z∈C:∣z2+z+1∣=1}S = \{z \in \mathbb{C} : |z^2 + z + 1| = 1\}, we rewrite the expression by completing the square: z2+z+1=(z+12)2+34z^2 + z + 1 = \left(z + \frac{1}{2}\right)^2 + \frac{3}{4}

Applying the triangle inequality to ∣(z+12)2+34∣=1\left|\left(z + \frac{1}{2}\right)^2 + \frac{3}{4}\right| = 1, we obtain: 1≤∣z+12∣2+34  ⟹  ∣z+12∣≥121 \le \left|z + \frac{1}{2}\right|^2 + \frac{3}{4} \implies \left|z + \frac{1}{2}\right| \ge \frac{1}{2} This confirms that statement (C) is TRUE and statement (A) is FALSE.

Using the reverse triangle inequality gives the upper bound: 1≥∣∣z+12∣2−34∣  ⟹  ∣z+12∣≤721 \ge \left| \left|z + \frac{1}{2}\right|^2 - \frac{3}{4} \right| \implies \left|z + \frac{1}{2}\right| \le \frac{\sqrt{7}}{2}

Bounding ∣z∣|z| using the triangle inequality: ∣z∣=∣(z+12)−12∣≤∣z+12∣+12≤7+12<2|z| = \left|\left(z + \frac{1}{2}\right) - \frac{1}{2}\right| \le \left|z + \frac{1}{2}\right| + \frac{1}{2} \le \frac{\sqrt{7} + 1}{2} < 2 Thus, statement (B) is TRUE.

Finally, ∣z2+z+1∣=1|z^2 + z + 1| = 1 defines a continuous one-dimensional curve in the complex plane, which contains uncountably infinitely many points. Hence, statement (D) is FALSE.

Correct Options: (B) and (C)

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