<p>Locus of the point <i>z</i> satisfying the equation <math>|iz - 1| + |z - 1| = 2</math>, is (where <i>i</i> = <math>\sqrt{-1}</math>)</p>
Step-by-Step Solution
Key Concept: Recognize that |iz - 1| and |z - 1| represent distances from z to specific points in the complex plane. The sum of distances from a point to two fixed points being constant defines an ellipse.
<p><strong>Step 1:</strong> Identify the fixed points. |iz - 1| = 0 when iz = 1, so z = 1/i = -i. Thus |iz - 1| = |z - (-i)| is the distance from z to point F₁ = -i.</p><p>Similarly, |z - 1| = 0 when z = 1, so |z - 1| is the distance from z to point F₂ = 1.</p><p><strong>Step 2:</strong> Rewrite the given equation as: |z - (-i)| + |z - 1| = 2</p><p>This is the sum of distances from point z to two fixed points F₁ = -i and F₂ = 1 equals 2.</p><p><strong>Step 3:</strong> Calculate the distance between the two fixed points: |F₁ - F₂| = |-i - 1| = |-(1 + i)| = √(1² + 1²) = √2</p><p><strong>Step 4:</strong> Check the condition for an ellipse. For an ellipse, we need: sum of distances = 2a > distance between foci = 2c</p><p>Here: 2 > √2 ✓ (since 2 ≈ 2 and √2 ≈ 1.414)</p><p>This confirms 2a = 2, so a = 1, and 2c = √2, so c = √2/2.</p><p><strong>Step 5:</strong> By the definition of an ellipse, the locus of all points whose sum of distances to two fixed points (foci) is constant is an ellipse.</p><p><strong>∴ Answer:</strong> C</p>
Correct Answer: C