Complex Numbers
Modulus of complex numbers
Grade 11
Question:
<p><strong>261.</strong> Let \(z\) be the complex number satisfying \(|z + 16| = 4|z + 1|\), then:</p>
<p>(a) \(|z| = 4\)</p>
<p>(b) \(|z| = 5\)</p>
<p>(c) \(|z| = 6\)</p>
<p>(d) \(4 < |z| < 64\)</p>
Step-by-Step Solution
Key Concept: The equation |z + 16| = 4|z + 1| represents the locus of points whose distance from -16 is 4 times the distance from -1, which is an Apollonius circle. Rewrite as |z - (-16)| = 4|z - (-1)| and complete the square to find the circle's center and radius.
<p><strong>Step 1:</strong> Start with |z + 16| = 4|z + 1|. Square both sides:</p><p>|z + 16|² = 16|z + 1|²</p><p><strong>Step 2:</strong> Let z = x + iy. Then:</p><p>(x + 16)² + y² = 16[(x + 1)² + y²]</p><p><strong>Step 3:</strong> Expand both sides:</p><p>x² + 32x + 256 + y² = 16[x² + 2x + 1 + y²]</p><p>x² + 32x + 256 + y² = 16x² + 32x + 16 + 16y²</p><p><strong>Step 4:</strong> Simplify by canceling 32x:</p><p>x² + 256 + y² = 16x² + 16 + 16y²</p><p>0 = 15x² + 15y² - 240</p><p>x² + y² = 16</p><p><strong>Step 5:</strong> This is a circle with center at the origin (0, 0) and radius 4.</p><p>∴ z lies on a circle with |z| = 4</p>
Correct Answer: B