Complex Numbers
Modulus and triangle conditions
Grade 11
Question:
<p>Let \(z_1, z_2\) and \(z_3\) be three complex numbers satisfying \(|z| = 1\) and \(4z_3 = 3(z_1 + z_2)\), then \(|z_1 - z_2|\) is equal to</p>
<p>\(\frac{2}{3}\)</p>
<p>\(\frac{\sqrt{5}}{3}\)</p>
<p>\(\frac{3}{2}\)</p>
<p>\(\frac{2\sqrt{5}}{3}\)</p>
Step-by-Step Solution
Key Concept: Since |z| = 1 for all three numbers, they lie on the unit circle. Use the constraint 4z₃ = 3(z₁ + z₂) combined with |z₃| = 1 to establish that z₁ + z₂ has a specific modulus, then apply the identity |z₁ - z₂|² = |z₁|² + |z₂|² - 2Re(z₁·z̄₂).
<p><strong>Step 1:</strong> From 4z₃ = 3(z₁ + z₂), we have |4z₃| = |3(z₁ + z₂)|, so 4|z₃| = 3|z₁ + z₂|.</p><p><strong>Step 2:</strong> Since |z₃| = 1, we get 4 = 3|z₁ + z₂|, therefore |z₁ + z₂| = 4/3.</p><p><strong>Step 3:</strong> Let z₁ = e^(iα) and z₂ = e^(iβ). Then |z₁ + z₂|² = |e^(iα) + e^(iβ)|² = 2 + 2cos(α - β) = (4/3)² = 16/9.</p><p><strong>Step 4:</strong> Solving: 2 + 2cos(α - β) = 16/9 gives 2cos(α - β) = -2/9, so cos(α - β) = -1/9.</p><p><strong>Step 5:</strong> Now |z₁ - z₂|² = 2 - 2cos(α - β) = 2 - 2(-1/9) = 2 + 2/9 = 20/9.</p><p><strong>Step 6:</strong> Therefore |z₁ - z₂| = √(20/9) = 2√5/3.</p><p>∴ Answer: D</p>
Correct Answer: D