Question:

<p><strong>Step 1:</strong> Simplify the complex fraction by multiplying numerator and denominator by the conjugate of the denominator.</p><p>$$\frac{1+i\sqrt{3}}{1-i\sqrt{3}} \cdot \frac{1+i\sqrt{3}}{1+i\sqrt{3}} = \frac{(1+i\sqrt{3})^2}{(1)^2 - (i\sqrt{3})^2}$$</p><p><strong>Step 2:</strong> Calculate the numerator: $(1+i\sqrt{3})^2 = 1 + 2i\sqrt{3} - 3 = -2 + 2i\sqrt{3}$</p><p><strong>Step 3:</strong> Calculate the denominator: $1 - (-3) = 4$</p><p>$$\frac{-2 + 2i\sqrt{3}}{4} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$$</p><p><strong>Step 4:</strong> Convert to polar form. Here $r = \sqrt{\frac{1}{4} + \frac{3}{4}} = 1$</p><p>For angle: $\cos\theta = -\frac{1}{2}$ and $\sin\theta = \frac{\sqrt{3}}{2}$ → $\theta = \frac{2\pi}{3}$ (second quadrant)</p><p><strong>Step 5:</strong> By De Moivre's theorem: $\left(e^{i\frac{2\pi}{3}}\right)^n = 1$</p><p>This requires: $n \cdot \frac{2\pi}{3} = 2\pi k$ where $k$ is a positive integer</p><p>$$n = 3k$$</p><p><strong>Step 6:</strong> The least positive integer value occurs when $k = 1$.</p><p>∴ <strong>Answer: n = 3</strong></p>