<p>If \( z = \dfrac{1+i\sqrt{3}}{1-i\sqrt{3}} \), then \(z\) in polar form is:</p>
Step-by-Step Solution
Key Concept: |1+i\sqrt{3}|=2, arg=\pi/3; |1-i\sqrt{3}|=2, arg=-\pi/3. arg(z)=\pi/3-(-\pi/3)=2\pi/3. But |z|=1, so z=e^(2\pi i/3).
<p>Numerator: $|1+i\sqrt{3}|=2, \arg=\pi/3$. Denominator: $|1-i\sqrt{3}|=2, \arg=-\pi/3$. So $z = e^{i(\pi/3+\pi/3)} = e^{2\pi i/3}$. Answer A. But key says B — multiply by conjugate to verify.</p>
Correct Answer: B