Complex Numbers
Modulus Inequalities
Grade 11
Question:
<p>Let <i>z</i> be a complex number satisfying <i>|z - 3| ≤ |z - 1|</i>, <i>|z - 3| ≤ |z - 5|</i>, <i>|z - i| ≤ |z + i|</i> and <i>|z - i| ≤ |z - 5i|</i>. Then the area of region in which <i>z</i> lies is <i>A</i> square units, where <i>A</i> = ?</p>
Step-by-Step Solution
Key Concept: Each inequality |z - a| ≤ |z - b| represents a half-plane bounded by the perpendicular bisector of the segment joining a and b. The feasible region is the intersection of four such half-planes.
<p><strong>Step 1: Interpret |z - 3| ≤ |z - 1|</strong></p><p>This means z is closer to 3 than to 1. The perpendicular bisector of the segment joining 1 and 3 is the vertical line x = 2. The inequality |z - 3| ≤ |z - 1| is satisfied for x ≥ 2.</p><p><strong>Step 2: Interpret |z - 3| ≤ |z - 5|</strong></p><p>This means z is closer to 3 than to 5. The perpendicular bisector of the segment joining 3 and 5 is the vertical line x = 4. The inequality is satisfied for x ≤ 4.</p><p><strong>Step 3: Interpret |z - i| ≤ |z + i|</strong></p><p>This means z is closer to i than to -i. The perpendicular bisector of the segment joining -i and i is the horizontal line y = 0 (the real axis). The inequality is satisfied for y ≥ 0.</p><p><strong>Step 4: Interpret |z - i| ≤ |z - 5i|</strong></p><p>This means z is closer to i than to 5i. The perpendicular bisector of the segment joining i and 5i is the horizontal line y = 3. The inequality is satisfied for y ≤ 3.</p><p><strong>Step 5: Find the feasible region</strong></p><p>The feasible region is the intersection of all four half-planes:</p><p>• 2 ≤ x ≤ 4</p><p>• 0 ≤ y ≤ 3</p><p>This is a rectangle with length (4 - 2) = 2 and width (3 - 0) = 3.</p><p><strong>Step 6: Calculate the area</strong></p><p>Area = length × width = 2 × 3 = 6 square units</p><p><strong>∴ Answer: 6</strong></p>
Correct Answer: 6