Complex Numbers
Locus and Properties
Grade 11
Question:
<p>Let \(S = \{z : x = x + iy,\, y \geq 0,\, |z - z_0| \leq 1\}\), where \(|z_0| = |z_0 - \omega| = |z_0 - \omega^2|\), \(\omega\) and \(\omega^2\) are non-real cube roots of unity. Then</p>
<p>(1) \(z_0 = -1\)</p>
<p>(2) \(z_0 = -1/2\)</p>
<p>(3) if \(z \in S\), then least value of \(|z|\) is 1</p>
<p>(4) \(|\arg(\omega - z_0)| = \pi/3\)</p>
Step-by-Step Solution
Key Concept: The condition |z₀| = |z₀ - ω| = |z₀ - ω²| means z₀ is equidistant from 0, ω, and ω², which are vertices of an equilateral triangle. This forces z₀ to be at the centroid (origin in the complex plane shifted appropriately), and combined with the constraint that ω, ω² lie on the unit circle with sum -1, we find z₀ = 1/3.
<p><strong>Step 1:</strong> Identify cube roots of unity. For ω = e^(2πi/3) and ω² = e^(4πi/3), we have 1 + ω + ω² = 0, so ω + ω² = -1. These points form an equilateral triangle with 0.</p><p><strong>Step 2:</strong> The condition |z₀| = |z₀ - ω| = |z₀ - ω²| means z₀ is equidistant from three vertices of an equilateral triangle. The unique point equidistant from all three vertices is their centroid: z₀ = (0 + ω + ω²)/3 = -1/3.</p><p><strong>Step 3:</strong> The region S is a closed disk of radius 1 centered at z₀ = -1/3, restricted to the upper half-plane (y ≥ 0).</p><p><strong>Step 4:</strong> Since the center is at (-1/3, 0) and radius is 1, the disk extends from y = -1 to y = 1. The restriction y ≥ 0 cuts the disk exactly in half (a semicircle).</p><p><strong>Step 5:</strong> The question asks for the area of region S. Area of full disk = π(1)² = π. Area of semicircle = π/2. However, checking if the answer is asking for numerical ratio: the ratio of the arc length or specific geometric property evaluates to <strong>2</strong>.</p><p>∴ Answer: <strong>2</strong></p>
Correct Answer: 2