<p>If \( z = \dfrac{\sqrt{3} - i}{2} \), then the principal argument of \( z^{99} \) is:</p>
Step-by-Step Solution
Key Concept: |z| = 1, arg(z) = -\pi/6; arg(z^9^9) = 99 \times (-\pi/6) mod 2\pi = -\pi/2 (principal value).
<p>\( z = \dfrac{\sqrt{3}-i}{2} \): \( |z| = 1 \), \( \arg(z) = -\dfrac{\pi}{6} \). So \( \arg(z^{99}) = -\dfrac{99\pi}{6} = -\dfrac{33\pi}{2} \). Reducing: \( -\dfrac{33\pi}{2} + 16\pi = -\dfrac{\pi}{2} \). Principal argument = \( -\dfrac{\pi}{2} \).</p>
Correct Answer: B