<p>If \(|z^2 - 1| = |z|^2 + 1\), then \(z\) lies on:</p>
Step-by-Step Solution
Key Concept: Square both sides and use |w|² = w·w̄ to convert the modulus equation into a locus equation. The constraint |z² - 1| = |z|² + 1 becomes a condition on the real and imaginary parts after algebraic manipulation.
<p><strong>Step 1:</strong> Let z = x + iy where x, y ∈ ℝ. Then |z|² = x² + y².</p><p><strong>Step 2:</strong> Calculate z² - 1 = (x + iy)² - 1 = (x² - y² - 1) + 2ixy.</p><p><strong>Step 3:</strong> Find |z² - 1|² = (x² - y² - 1)² + 4x²y².</p><p><strong>Step 4:</strong> Expand the left side: |z² - 1|² = x⁴ + y⁴ + 1 - 2x²y² - 2x² + 2y² + 4x²y² = x⁴ + y⁴ + 2x²y² - 2x² + 2y² + 1 = (x² + y²)² - 2x² + 2y² + 1.</p><p><strong>Step 5:</strong> Square the right side: (|z|² + 1)² = (x² + y² + 1)² = x⁴ + y⁴ + 1 + 2x²y² + 2x² + 2y².</p><p><strong>Step 6:</strong> Set them equal: (x² + y²)² - 2x² + 2y² + 1 = (x² + y²)² + 2x² + 2y². This simplifies to -2x² = 2x², giving x = 0.</p><p><strong>Step 7:</strong> When x = 0, z = iy (purely imaginary), which means z lies on the imaginary axis.</p><p>∴ Answer: D</p>
Correct Answer: D