<p>Let \(z_1\) and \(z_2\) be complex numbers such that \(z_1 \neq z_2\) and \(|z_1| = |z_2|\). If \(z_1\) has positive real part and \(z_2\) has negative imaginary part, then \(\dfrac{z_1 + z_2}{z_1 - z_2}\) may be</p>
Step-by-Step Solution
Key Concept: When |z₁| = |z₂|, both points lie on a circle centered at origin. The expression (z₁ + z₂)/(z₁ - z₂) represents a ratio that depends on the geometric configuration of z₁ and z₂, which can be analyzed using the perpendicular bisector property of equal modulus points.
<p><strong>Step 1:</strong> Given |z₁| = |z₂| = r (say), z₁ ≠ z₂. Let z₁ = re^(iα) and z₂ = re^(iβ) where α, β ∈ [0, 2π).</p><p><strong>Step 2:</strong> z₁ has positive real part: Re(z₁) > 0, so cos(α) > 0 ⟹ α ∈ (-π/2, π/2)</p><p><strong>Step 3:</strong> z₂ has negative imaginary part: Im(z₂) < 0, so sin(β) < 0 ⟹ β ∈ (π, 2π) or equivalently β ∈ (-π, 0)</p><p><strong>Step 4:</strong> Calculate the ratio:<br/>z₁ + z₂ = r(e^(iα) + e^(iβ)) = r·e^(i(α+β)/2)·2cos((α-β)/2)<br/>z₁ - z₂ = r(e^(iα) - e^(iβ)) = r·e^(i(α+β)/2)·2i·sin((α-β)/2)</p><p><strong>Step 5:</strong> Therefore:<br/>(z₁ + z₂)/(z₁ - z₂) = cos((α-β)/2) / [i·sin((α-β)/2)] = -i·cot((α-β)/2)</p><p><strong>Step 6:</strong> Since α ∈ (-π/2, π/2) and β ∈ (-π, 0), we have (α - β) ∈ (π/2, 3π/2), so (α-β)/2 ∈ (π/4, 3π/4). In this range, cot((α-β)/2) can take any real value. Thus (z₁ + z₂)/(z₁ - z₂) = -i·cot((α-β)/2) is purely imaginary.</p><p><strong>Step 7:</strong> The result is a purely imaginary complex number of the form ki where k ∈ ℝ.</p><p>∴ Answer: D</p>
Correct Answer: D