Complex Numbers
Modulus Inequalities
Grade 11
Question:
<p>If \(|z - (1/z)| = 1\), then</p>
<p>(1) \(|z|_{\max} = \dfrac{1 + \sqrt{5}}{2}\)</p>
<p>(2) \(|z|_{\min} = \dfrac{\sqrt{5} - 1}{2}\)</p>
<p>(3) \(|z|_{\max} = \dfrac{\sqrt{5} - 2}{2}\)</p>
<p>(4) \(|z|_{\min} = \dfrac{\sqrt{5} - 1}{\sqrt{2}}\)</p>
Step-by-Step Solution
Key Concept: For complex numbers on the unit circle (|z| = 1), the condition |z - 1/z| = 1 becomes |z - z̄| = 1, which translates to |2i·Im(z)| = 1. This severely restricts z to lie on specific points where Im(z) = ±1/2.
<p><strong>Step 1:</strong> Let |z - 1/z| = 1. Multiply through by |z|: |z² - 1| = |z|</p><p><strong>Step 2:</strong> Assume |z| = r > 0. Then |z² - 1| = r. Squaring: |z² - 1|² = r²</p><p><strong>Step 3:</strong> For z = x + iy: (x² - y² - 1)² + (2xy)² = r² = x² + y²</p><p><strong>Step 4:</strong> Expanding the left side: (x² - y²)² - 2(x² - y²) + 1 + 4x²y² = x² + y²</p><p><strong>Step 5:</strong> Simplifying using (x² + y²)² - 2(x² - y²) + 1 = x² + y², which yields: r⁴ - r² - 2(x² - y²) + 1 = 0</p><p><strong>Step 6:</strong> Testing r = 1: 1 - 1 - 2(x² - y²) + 1 = 0 ⟹ x² - y² = 1/2 ✓ (consistent for points on unit circle)</p><p><strong>Step 7:</strong> For r ≠ 1, the equation leads to contradictions. Therefore |z| = 1 is necessary.</p><p>∴ <strong>The answer is |z| = 1</strong></p>
Correct Answer: 1