Complex Numbers
Geometry of Complex Numbers
Grade 11
Question:
<p>Let \(z_1\) and \(z_2\) be two distinct complex numbers and let \(z = (1 - t)z_1 + tz_2\) for some real number \(t\) with \(0 < t < 1\). If \(\arg(w)\) denotes the principal argument of a nonzero complex number \(w\), then</p>
<p>(1) \(|z - z_1| + |z - z_2| = |z_1 - z_2|\)</p>
<p>(2) \((z - z_1) = (z - z_2)\)</p>
<p>(3) \(\begin{vmatrix} z - z_1 & \bar{z} - \bar{z}_1 \\ z_2 - z_1 & \bar{z}_2 - \bar{z}_1 \end{vmatrix} = 0\)</p>
<p>(4) \(\arg(z - z_1) = \arg(z_2 - z_1)\)</p>
Step-by-Step Solution
Key Concept: The locus of z = (1-t)z₁ + tz₂ for 0 < t < 1 represents the open line segment between z₁ and z₂ in the complex plane. Understanding that z lies strictly between these two points (excluding endpoints) is crucial for determining which statements are valid.
<p><strong>Step 1: Recognize the parametric form</strong> For 0 < t < 1, the expression z = (1-t)z₁ + tz₂ is a convex combination of z₁ and z₂, representing points on the open line segment between them.</p><p><strong>Step 2: Analyze endpoint behavior</strong> When t → 0⁺, z → z₁ (but never equals z₁). When t → 1⁻, z → z₂ (but never equals z₂). The endpoints are excluded.</p><p><strong>Step 3: Verify geometric properties</strong> Every point z satisfies: |z - z₁| + |z - z₂| = |z₂ - z₁| (triangle inequality becomes equality for collinear points). The points are strictly collinear with z₁ and z₂.</p><p><strong>Step 4: Check distance conditions</strong> For any 0 < t < 1: |z - z₁| = t|z₂ - z₁| and |z - z₂| = (1-t)|z₂ - z₁|, ensuring 0 < |z - z₁| < |z₂ - z₁| and 0 < |z - z₂| < |z₂ - z₁|.</p><p>∴ Answer: ACD</p>
Correct Answer: ACD