Complex Numbers
Properties of complex numbers
Grade 11
Question:
<p><strong>For Problems 5–8</strong><br>Consider the complex numbers \(z_1\) and \(z_2\) satisfying the relation \(|z_1 + z_2|^2 = |z_1|^2 + |z_2|^2\).</p><p><strong>Problem 5.</strong> Complex number \(z_1\bar{z}_2\) is</p>
<p>(1) purely real</p>
<p>(2) purely imaginary</p>
<p>(3) zero</p>
<p>(4) none of these</p>
Step-by-Step Solution
Key Concept: The condition |z₁ + z₂|² = |z₁|² + |z₂|² expands to show that 2Re(z₁·z̄₂) = 0, meaning z₁·z̄₂ must be purely imaginary (or zero). This is the geometric condition for perpendicularity of complex numbers.
<p><strong>Step 1:</strong> Expand |z₁ + z₂|² using the property |w|² = w·w̄</p><p>|z₁ + z₂|² = (z₁ + z₂)(z̄₁ + z̄₂) = |z₁|² + |z₂|² + z₁z̄₂ + z̄₁z₂</p><p><strong>Step 2:</strong> Apply the given condition |z₁ + z₂|² = |z₁|² + |z₂|²</p><p>|z₁|² + |z₂|² + z₁z̄₂ + z̄₁z₂ = |z₁|² + |z₂|²</p><p><strong>Step 3:</strong> Simplify: z₁z̄₂ + z̄₁z₂ = 0</p><p>Note that z₁z̄₂ + z̄₁z₂ = 2Re(z₁z̄₂), so 2Re(z₁z̄₂) = 0</p><p><strong>Step 4:</strong> This means Re(z₁z̄₂) = 0, so z₁z̄₂ is purely imaginary (including the case where it equals 0)</p><p>∴ <strong>Answer: B</strong> (z₁z̄₂ is purely imaginary)</p>
Correct Answer: B