Complex Numbers
Modulus and argument
Grade 11
Question:
<p>If a complex number \(z\) satisfies \(|z| = 1\) and \(\arg(z - 1) = \dfrac{2\pi}{3}\), (\(\omega\) is complex cube root of unity), then</p>
<p>\(z^2 + z\) is purely imaginary number</p>
<p>\(z = -\omega^2\)</p>
<p>\(z = -\omega\)</p>
<p>\(|z - 1| = 1\)</p>
Step-by-Step Solution
Key Concept: A complex number z with |z| = 1 lies on the unit circle. The condition arg(z - 1) = 2π/3 means the vector from point 1 to z makes angle 2π/3 with positive real axis. The intersection of these two loci determines z uniquely.
<p><strong>Step 1:</strong> Since |z| = 1, point z lies on the unit circle centered at origin.</p><p><strong>Step 2:</strong> The condition arg(z - 1) = 2π/3 means the vector from (1, 0) to z makes angle 2π/3 with positive real axis. This defines a ray: z = 1 + r·e^(i·2π/3) where r > 0.</p><p><strong>Step 3:</strong> Substituting into |z| = 1: |1 + r·e^(i·2π/3)| = 1</p><p><strong>Step 4:</strong> Let e^(i·2π/3) = -1/2 + i√3/2. Then z = 1 + r(-1/2 + i√3/2) = (1 - r/2) + ir√3/2</p><p><strong>Step 5:</strong> |z|² = (1 - r/2)² + (r√3/2)² = 1</p><p><strong>Step 6:</strong> 1 - r + r²/4 + 3r²/4 = 1 → r² - r = 0 → r = 1 (since r > 0)</p><p><strong>Step 7:</strong> Therefore z = 1/2 + i√3/2 = e^(iπ/3)</p><p><strong>Step 8:</strong> This equals ω² where ω = e^(i·2π/3), confirming z is related to cube roots of unity.</p><p>∴ Answer: ACD</p>
Correct Answer: ACD