Complex Numbers
Geometry in Complex Plane
Grade 11
Question:
<p>If <em>P</em> and <em>Q</em> are represented by the complex numbers \(z_1\) and \(z_2\), such that \(|1/z_2 + 1/z_1| = |1/z_2 - 1/z_1|\), then</p>
<p>(1) △<em>OPQ</em> (where <em>O</em> is the origin) is equilateral</p>
<p>(2) △<em>OPQ</em> is right angled</p>
<p>(3) the circumcenter of △<em>OPQ</em> is \(\dfrac{1}{2}(z_1 + z_2)\)</p>
<p>(4) the circumcenter of △<em>OPQ</em> is \(\dfrac{1}{3}(z_1 + z_2)\)</p>
Step-by-Step Solution
Key Concept: The condition |1/z₂ + 1/z₁| = |1/z₂ - 1/z₁| means the two complex numbers 1/z₂ + 1/z₁ and 1/z₂ - 1/z₁ have equal moduli. This occurs when their real parts are equal and imaginary parts are opposite, which happens when 1/z₁ is purely imaginary.
<p><strong>Step 1:</strong> Let |1/z₂ + 1/z₁| = |1/z₂ - 1/z₁|</p><p><strong>Step 2:</strong> Square both sides: |1/z₂ + 1/z₁|² = |1/z₂ - 1/z₁|²</p><p><strong>Step 3:</strong> Expand using |w|² = w·w̄:</p><p>(1/z₂ + 1/z₁)(1/z̄₂ + 1/z̄₁) = (1/z₂ - 1/z₁)(1/z̄₂ - 1/z̄₁)</p><p><strong>Step 4:</strong> Simplifying: 1/|z₂|² + 1/(z₂z̄₁) + 1/(z̄₂z₁) + 1/|z₁|² = 1/|z₂|² - 1/(z₂z̄₁) - 1/(z̄₂z₁) + 1/|z₁|²</p><p><strong>Step 5:</strong> This gives: 2[1/(z₂z̄₁) + 1/(z̄₂z₁)] = 0</p><p><strong>Step 6:</strong> Therefore: 1/(z₂z̄₁) + 1/(z̄₂z₁) = 0, which means 1/z₁ is purely imaginary</p><p><strong>Step 7:</strong> This occurs when z₁ is purely imaginary (z₁ = bi where b ∈ ℝ)</p><p><strong>Step 8:</strong> Thus OP ⊥ OQ or the points P and Q lie such that ∠POQ = 90°</p><p>∴ Answer: The locus is a circle with PQ as diameter, or OP ⊥ OQ</p>
Correct Answer: 2