<p>Given Im \ w ≠ 0. If \ w - \(\bar{w}\)z = k(1-z), then which of the following is true about z?</p><p>More specifically: Let \ w - \bar{w}z = k(1-z) where k is real. Then \(|z|\) equals:</p>
Step-by-Step Solution
Key Concept: Since w - w̄z = k(1-z) with k real, separate w into real and imaginary parts (w = a + bi where b ≠ 0) and equate real/imaginary components on both sides to establish a constraint that forces |z| = 1.
<p><strong>Step 1:</strong> Write w = a + bi where a, b ∈ ℝ and b ≠ 0 (given Im(w) ≠ 0).</p><p><strong>Step 2:</strong> Let z = x + yi. The equation becomes: (a + bi) - (a - bi)(x + yi) = k(1 - x - yi)</p><p><strong>Step 3:</strong> Expand LHS: a + bi - (ax + ayi - bxi + by) = a - ax + by + i(b - ay - bx)</p><p><strong>Step 4:</strong> RHS: k(1-x) - kyi</p><p><strong>Step 5:</strong> Equate real parts: a - ax + by = k(1-x) ... (1)</p><p><strong>Step 6:</strong> Equate imaginary parts: b - ay - bx = -ky ... (2)</p><p><strong>Step 7:</strong> From equation (2): b(1-x) - ay = -ky, so b(1-x) = a(y) - ky = y(a-k)</p><p><strong>Step 8:</strong> From equation (1): a(1-x) + by = k(1-x), so a(1-x) = k(1-x) - by</p><p><strong>Step 9:</strong> If 1-x ≠ 0, from (1): a - bky/(1-x) + by = k, which combined with (2) yields x² + y² = 1</p><p><strong>Step 10:</strong> If 1-x = 0, then x = 1, and (2) gives -ay = -ky, so y = 0 (since a ≠ k generally), making z = 1 with |z| = 1</p><p>∴ Answer: <strong>|z| = 1</strong></p>
Correct Answer: A