<p>Find the complex number \(z\) satisfying \(\text{Re}(z^2) = 0\), \(|z| = \sqrt{3}\).</p>
Step-by-Step Solution
Key Concept: If Re(z²) = 0, then z² is purely imaginary, meaning z² = ki for some real k. Combined with |z| = √3, we get |z²| = 3, so k² = 3, giving z² = ±3i. Then solve for z by taking square roots of purely imaginary numbers.
<p><strong>Step 1:</strong> Let z = a + bi where a, b ∈ ℝ.</p><p><strong>Step 2:</strong> Calculate z² = (a + bi)² = a² - b² + 2abi.</p><p><strong>Step 3:</strong> From Re(z²) = 0: a² - b² = 0, so a² = b², giving a = ±b.</p><p><strong>Step 4:</strong> From |z| = √3: a² + b² = 3.</p><p><strong>Step 5:</strong> Substitute a² = b²: 2a² = 3, so a² = 3/2, giving a = ±√(3/2).</p><p><strong>Step 6:</strong> Since a = ±b, we have b = ±√(3/2) (with b = a or b = -a).</p><p><strong>Step 7:</strong> This yields four solutions:</p><p>z = √(3/2) + i√(3/2), z = √(3/2) - i√(3/2), z = -√(3/2) + i√(3/2), z = -√(3/2) - i√(3/2)</p><p><strong>Verification:</strong> For z = √(3/2) + i√(3/2): z² = 3/2 - 3/2 + 2i(3/2) = 3i (Re = 0 ✓); |z|² = 3/2 + 3/2 = 3 ✓</p><p>∴ Answer: z = ±√(3/2) ± i√(3/2)</p>
Correct Answer: z = ±√(3/2) ± i√(3/2)