1. **State the problem:** Convert the complex number $-\sqrt{3} + i$ into polar form, expressing the angle $\theta$ in radians in terms of $\pi$ over the interval $0 \leq \theta < 2\pi$. 2. **Recall the polar form formula:** A complex number $z = x + yi$ can be written in polar form as $$z = r(\cos \theta + i \sin \theta)$$ where $$r = \sqrt{x^2 + y^2}$$ and $$\theta = \tan^{-1}\left(\frac{y}{x}\right)$$ with adjustments to $\theta$ depending on the quadrant. 3. **Identify $x$ and $y$:** Here, $x = -\sqrt{3}$ and $y = 1$. 4. **Calculate the magnitude $r$:** $$r = \sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2$$ 5. **Calculate the angle $\theta$:** $$\theta = \tan^{-1}\left(\frac{1}{-\sqrt{3}}\right) = \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right)$$ 6. **Evaluate $\tan^{-1}(-1/\sqrt{3})$:** We know $\tan(\pi/6) = 1/\sqrt{3}$, so $$\tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6}$$ 7. **Adjust $\theta$ for the correct quadrant:** Since $x < 0$ and $y > 0$, the point lies in the second quadrant. The angle in the second quadrant is $$\theta = \pi - \left| -\frac{\pi}{6} \right| = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$$ 8. **Write the polar form:** $$z = 2 \left( \cos \frac{5\pi}{6} + i \sin \frac{5\pi}{6} \right)$$ **Final answer:** The polar form of $-\sqrt{3} + i$ is $$2 \left( \cos \frac{5\pi}{6} + i \sin \frac{5\pi}{6} \right)$$