<p>If \(|z^2 - 1| = |z|^2 + 1\), then \(z\) lies on</p>
Step-by-Step Solution
Key Concept: Square both sides and separate into real and imaginary parts by writing z = x + iy, then use the constraint that both sides must be equal to derive a relationship between x and y that represents a geometric locus.
<p><strong>Step 1:</strong> Let z = x + iy where x, y ∈ ℝ. Then z² = x² - y² + 2ixy.</p><p><strong>Step 2:</strong> z² - 1 = (x² - y² - 1) + 2ixy, so |z² - 1|² = (x² - y² - 1)² + 4x²y²</p><p><strong>Step 3:</strong> Also |z|² + 1 = x² + y² + 1, so |z|² + 1)² = (x² + y² + 1)²</p><p><strong>Step 4:</strong> Squaring the original equation: (x² - y² - 1)² + 4x²y² = (x² + y² + 1)²</p><p><strong>Step 5:</strong> Expanding LHS: x⁴ + y⁴ + 1 - 2x²y² - 2x² + 2y² + 4x²y² = x⁴ + y⁴ + 1 + 2x²y² - 2x² + 2y²</p><p><strong>Step 6:</strong> Expanding RHS: x⁴ + y⁴ + 1 + 2x²y² + 2x² + 2y²</p><p><strong>Step 7:</strong> Equating: x⁴ + y⁴ + 1 + 2x²y² - 2x² + 2y² = x⁴ + y⁴ + 1 + 2x²y² + 2x² + 2y²</p><p><strong>Step 8:</strong> This simplifies to -2x² = 2x², giving 4x² = 0, so x = 0</p><p><strong>Step 9:</strong> Therefore z lies on the <strong>imaginary axis</strong> (the line x = 0).</p><p>∴ Answer: D</p>
Correct Answer: D