Complex Numbers
Modulus of complex numbers
Grade 11
Question:
<p><b>For Problems 17–19:</b> Suppose \(z\) and \(\omega\) are two complex numbers such that \(|z| \leq 1\), \(|\omega| \leq 1\), and \(|z + i\omega| = |z - i\bar{\omega}| = 2\).</p><p>Which of the following is true about \(|z|\) and \(|\omega|\)?</p>
<p>(1) \(|z| = |\omega| = \dfrac{1}{2}\)</p>
<p>(2) \(|z| = \dfrac{1}{2}, |\omega| = \dfrac{3}{4}\)</p>
<p>(3) \(|z| = |\omega| = \dfrac{3}{4}\)</p>
<p>(4) \(|z| = |\omega| = 1\)</p>
Step-by-Step Solution
Key Concept: Use the equality of two modulus expressions to derive constraints on z and ω by squaring and expanding, then apply the given inequality constraints |z| ≤ 1 and |ω| ≤ 1 to find exact values.
<p><strong>Step 1:</strong> Square both modulus equations.</p><p>From |z + iω| = 2: (z + iω)(z̄ - iω̄) = 4</p><p>This gives: |z|² + |ω|² + i(zω̄ - z̄ω) = 4</p><p><strong>Step 2:</strong> Square the second equation |z - iω̄| = 2.</p><p>(z - iω̄)(z̄ + iω) = 4</p><p>This gives: |z|² + |ω|² - i(zω - z̄ω̄) = 4</p><p><strong>Step 3:</strong> From both equations, the imaginary parts must sum to zero (comparing conditions). This yields |z|² + |ω|² = 4.</p><p><strong>Step 4:</strong> Apply the constraints |z| ≤ 1 and |ω| ≤ 1.</p><p>If |z|² + |ω|² = 4 and both |z|² ≤ 1 and |ω|² ≤ 1, then we must have |z|² = 1 and |ω|² = 1 (equality in both constraints).</p><p><strong>Step 5:</strong> Therefore |z| = 1 and |ω| = 1.</p><p>∴ Answer: D</p>
Correct Answer: D