Complex Numbers
Geometry of Complex Numbers
Grade 11
Question:
<p>If \(z_0,z_1\) represents points P, Q on the locus \(|z-1|=1\) and the line segment PQ subtends an angle \(\dfrac{\pi}{2}\) at the point \(z=1\), then</p>
<p>\(z_1 = 1+i(z_0-1)\)</p>
<p>\(-i = \dfrac{z_1-1}{z_0-1}\)</p>
<p>\(z_1 = 1-i(z_0-1)\)</p>
<p>\(z_1 = -i(z_0-1)\)</p>
Step-by-Step Solution
Key Concept: Points on the circle |z-1|=1 that subtend a right angle at the center z=1 must satisfy the condition that the vectors from center to P and center to Q are perpendicular, meaning (z₀-1)·(z₁-1) interpreted as complex numbers gives a purely imaginary product.
<p><strong>Step 1:</strong> The locus |z-1|=1 is a circle with center C=1 and radius r=1. Points P and Q lie on this circle, so z₀=1+e^(iα) and z₁=1+e^(iβ) for some angles α, β.</p><p><strong>Step 2:</strong> The line segment PQ subtends angle π/2 at center z=1 means the vectors CP and CQ are perpendicular. Thus (z₀-1)⊥(z₁-1).</p><p><strong>Step 3:</strong> For perpendicular directions: e^(iα)⊥e^(iβ) implies e^(i(β-α))=±i, so β-α=±π/2.</p><p><strong>Step 4:</strong> This gives z₁-1=±i(z₀-1), which means z₁=1±i(z₀-1).</p><p><strong>Step 5:</strong> Computing: (z₀-1)(z₁-1)̄ = (z₀-1)(∓i(z₀-1)̄) = ∓i|z₀-1|²=∓i (since |z₀-1|=1).</p><p><strong>Step 6:</strong> Therefore Re[(z₀-1)(z₁-1)̄]=0 and |z₀-1|²+|z₁-1|²=1+1=2, while |z₀-z₁|²=|z₀-1-(z₁-1)|²=|z₀-1|²+|z₁-1|²∓2Im[(z₀-1)(z₁-1)̄]=2.</p><p>∴ The relationship is (z₀-1)·(z₁-1)̄ is purely imaginary, and |z₀-z₁|=√2</p>
Correct Answer: A