Step-by-Step Solution
Key Concept: Separate the equation into real and imaginary parts by setting z = x + iy, then use the constraint |z| is always real to force imaginary parts to vanish independently of real parts.
<p><strong>Step 1:</strong> Let z = x + iy where x, y ∈ ℝ and |z| = √(x² + y²)</p><p><strong>Step 2:</strong> Substitute into z² + |z| = 0:</p><p>(x + iy)² + √(x² + y²) = 0</p><p>x² - y² + 2ixy + √(x² + y²) = 0</p><p><strong>Step 3:</strong> Equate real and imaginary parts:</p><p>Imaginary part: 2xy = 0 ⟹ x = 0 or y = 0</p><p>Real part: x² - y² + √(x² + y²) = 0</p><p><strong>Step 4:</strong> <strong>Case 1:</strong> If x = 0, then -y² + |y| = 0 ⟹ |y|(|y| - 1) = 0</p><p>So y = 0, 1, or -1</p><p>This gives z = 0, i, -i</p><p><strong>Step 5:</strong> <strong>Case 2:</strong> If y = 0, then x² + |x| = 0 ⟹ |x|(|x| + 1) = 0</p><p>Since |x| + 1 > 0, we need |x| = 0, so x = 0</p><p>This gives z = 0 (already found)</p><p><strong>Verification:</strong></p><p>• z = 0: 0 + 0 = 0 ✓</p><p>• z = i: (i)² + |i| = -1 + 1 = 0 ✓</p><p>• z = -i: (-i)² + |-i| = -1 + 1 = 0 ✓</p><p><strong>∴ Answer: z ∈ {0, i, -i}</strong></p>
Correct Answer: 0, i, -i