<p>If \(\arg(z + a) = \pi/6\) and \(\arg(z - a) = 2\pi/3\) \((a \in R^+)\), then</p>
Step-by-Step Solution
Key Concept: Use the geometric interpretation that arg(z+a) and arg(z-a) represent angles from points -a and a on the real axis to point z. These two angle conditions uniquely determine z's position as the intersection of two rays.
<p><strong>Step 1:</strong> Interpret geometrically. arg(z+a) = π/6 means z lies on a ray from point (-a, 0) making angle π/6 with positive real axis. arg(z-a) = 2π/3 means z lies on a ray from point (a, 0) making angle 2π/3 with positive real axis.</p><p><strong>Step 2:</strong> Let z = x+iy. From arg(z+a) = π/6: the ray from (-a,0) has slope tan(π/6) = 1/√3, so y/(x+a) = 1/√3, giving y = (x+a)/√3.</p><p><strong>Step 3:</strong> From arg(z-a) = 2π/3: the ray from (a,0) has slope tan(2π/3) = -√3, so y/(x-a) = -√3, giving y = -√3(x-a).</p><p><strong>Step 4:</strong> Equate: (x+a)/√3 = -√3(x-a) ⟹ (x+a)/√3 = -√3x+√3a ⟹ x+a = -3x+3a ⟹ 4x = 2a ⟹ x = a/2.</p><p><strong>Step 5:</strong> Substitute back: y = (a/2+a)/√3 = 3a/(2√3) = a√3/2.</p><p><strong>Step 6:</strong> Therefore z = a/2 + i(a√3/2), so |z| = √(a²/4 + 3a²/4) = √(a²) = a, and arg(z) = arctan(√3) = π/3.</p><p>∴ Answer: |z| = a and arg(z) = π/3</p>
Correct Answer: BC