1. **State the problem:** Express $$\frac{\sqrt{3} + i}{1 + \sqrt{3}i}$$ in the form $$r(\cos \theta + i \sin \theta)$$ and then evaluate $$\left(\frac{\sqrt{3} + i}{1 + \sqrt{3}i}\right)^6$$. 2. **Convert to polar form:** Recall that a complex number $$z = x + yi$$ can be written as $$r(\cos \theta + i \sin \theta)$$ where $$r = \sqrt{x^2 + y^2}$$ and $$\theta = \tan^{-1}\left(\frac{y}{x}\right)$$. 3. **Simplify the fraction:** Multiply numerator and denominator by the conjugate of the denominator: $$\frac{\sqrt{3} + i}{1 + \sqrt{3}i} \times \frac{1 - \sqrt{3}i}{1 - \sqrt{3}i} = \frac{(\sqrt{3} + i)(1 - \sqrt{3}i)}{(1 + \sqrt{3}i)(1 - \sqrt{3}i)}$$ 4. **Calculate denominator:** $$ (1 + \sqrt{3}i)(1 - \sqrt{3}i) = 1 - (\sqrt{3}i)^2 = 1 - (-3) = 1 + 3 = 4 $$ 5. **Calculate numerator:** $$ (\sqrt{3} + i)(1 - \sqrt{3}i) = \sqrt{3} \times 1 - \sqrt{3} \times \sqrt{3}i + i \times 1 - i \times \sqrt{3}i $$ $$ = \sqrt{3} - 3i + i - \sqrt{3} i^2 $$ Since $$i^2 = -1$$, $$ = \sqrt{3} - 3i + i + \sqrt{3} = 2\sqrt{3} - 2i $$ 6. **Put numerator and denominator together:** $$ \frac{2\sqrt{3} - 2i}{4} = \frac{2\sqrt{3}}{4} - \frac{2i}{4} = \frac{\sqrt{3}}{2} - \frac{1}{2}i $$ 7. **Find modulus $$r$$:** $$ r = \sqrt{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(-\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4} + \frac{1}{4}} = \sqrt{1} = 1 $$ 8. **Find argument $$\theta$$:** $$ \theta = \tan^{-1}\left(\frac{-\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) = \tan^{-1}(-\frac{1}{\sqrt{3}}) = -\frac{\pi}{6} $$ Since the real part is positive and imaginary part negative, $$\theta = -\frac{\pi}{6}$$ is correct. 9. **Express in polar form:** $$ \frac{\sqrt{3} + i}{1 + \sqrt{3}i} = 1 \left( \cos\left(-\frac{\pi}{6}\right) + i \sin\left(-\frac{\pi}{6}\right) \right) $$ 10. **Evaluate the sixth power:** Using De Moivre's theorem: $$ \left( r(\cos \theta + i \sin \theta) \right)^6 = r^6 (\cos 6\theta + i \sin 6\theta) $$ Since $$r=1$$, $$ = \cos\left(6 \times -\frac{\pi}{6}\right) + i \sin\left(6 \times -\frac{\pi}{6}\right) = \cos(-\pi) + i \sin(-\pi) $$ $$ = -1 + 0i = -1 $$ **Final answers:** $$ \frac{\sqrt{3} + i}{1 + \sqrt{3}i} = \cos\left(-\frac{\pi}{6}\right) + i \sin\left(-\frac{\pi}{6}\right) $$ $$ \left(\frac{\sqrt{3} + i}{1 + \sqrt{3}i}\right)^6 = -1 $$