<p>If \(k + |k + z^2| = |z|^2\) (\(k \in R^-\)), then possible argument of \(z\) is</p>
Step-by-Step Solution
Key Concept: Rearrange as |k + z²| = |z|² - k, then square both sides to eliminate the modulus. Since k < 0, we have |z|² - k > 0, which allows us to develop a constraint on z. Let z = x + iy and use the modulus condition to find the geometric relationship that determines the argument.
<p><strong>Step 1:</strong> Given k + |k + z²| = |z|² where k ∈ ℝ⁻</p><p>Rearrange: |k + z²| = |z|² - k</p><p><strong>Step 2:</strong> Since k < 0, let k = -a where a > 0. Then:</p><p>|-a + z²| = |z|² + a</p><p><strong>Step 3:</strong> Square both sides:</p><p>|-a + z²|² = (|z|² + a)²</p><p>|z²|² - 2a·Re(z²) + a² = |z|⁴ + 2a|z|² + a²</p><p>|z|⁴ - 2a·Re(z²) = |z|⁴ + 2a|z|²</p><p><strong>Step 4:</strong> Simplify:</p><p>-2a·Re(z²) = 2a|z|²</p><p>Re(z²) = -|z|²</p><p><strong>Step 5:</strong> Let z = re^(iθ). Then z² = r²e^(2iθ)</p><p>Re(z²) = r²cos(2θ) = -r²</p><p>cos(2θ) = -1</p><p>2θ = π + 2πn</p><p>θ = π/2 + πn</p><p><strong>Step 6:</strong> Therefore, argument of z is <strong>π/2 or 3π/2</strong> (or equivalently ±π/2)</p><p>∴ Answer: C</p>
Correct Answer: C