<p>Find all nonzero complex numbers <span>\(z\)</span> satisfying <span>\(\bar{z} = iz^2\)</span>.</p>
Step-by-Step Solution
Key Concept: Convert the conjugate condition into a modulus equation by taking absolute values of both sides, then use |iz²| = |z|² to establish a constraint on |z|. This reduces the problem to finding arguments that satisfy the original equation.
<p><strong>Step 1: Find |z| using modulus.</strong></p><p>Given: z̄ = iz²</p><p>Taking modulus of both sides: |z̄| = |iz²|</p><p>Since |z̄| = |z| and |iz²| = |i||z²| = |z|²:</p><p>|z| = |z|²</p><p>For nonzero z: |z|(|z| - 1) = 0 ⟹ |z| = 1</p><p><strong>Step 2: Express z on unit circle.</strong></p><p>Let z = e^(iθ) = cos θ + i sin θ, where |z| = 1</p><p>Then z̄ = e^(-iθ) = cos θ - i sin θ</p><p>And iz² = i·e^(2iθ) = e^(iπ/2)·e^(2iθ) = e^(i(2θ+π/2))</p><p><strong>Step 3: Equate arguments.</strong></p><p>From z̄ = iz²:</p><p>e^(-iθ) = e^(i(2θ+π/2))</p><p>-θ = 2θ + π/2 + 2πk (for integer k)</p><p>-3θ = π/2 + 2πk</p><p>θ = -π/6 - 2πk/3</p><p><strong>Step 4: Find distinct solutions.</strong></p><p>For k = 0: θ = -π/6 ⟹ z = cos(-π/6) + i sin(-π/6) = √3/2 - i/2</p><p>For k = 1: θ = -π/6 - 2π/3 = -5π/6 ⟹ z = cos(-5π/6) + i sin(-5π/6) = -√3/2 - i/2</p><p>For k = 2: θ = -π/6 - 4π/3 = -3π/2 ⟹ z = cos(-3π/2) + i sin(-3π/2) = i</p><p>For k = 3: θ repeats (period 2π)</p><p><strong>Verification:</strong></p><p>• z = i: z̄ = -i, iz² = i(-1) = -i ✓</p><p>• z = √3/2 - i/2: |z|² = 3/4 + 1/4 = 1 ✓</p><p>∴ Answer: z = i, z = ±(√3/2 - i/2)</p>
Correct Answer: z = i, ±(√3/2 - i/2)