Chapter 8 - Triangle Trigonometry and Polar Coordinates - 8.3 Polar Coordinates - Exercises and Problems for Section 8.3 - Exercises and Problems - Page 347: 17

$$(2,\:-\frac{\pi }{6})$$

Converting the Cartesian coordinates to polar coordinates, we find: $$r=\sqrt{\left(-\sqrt{3}\right)^2+1^2} =2 \\ θ=\arctan \left(\frac{1}{-\sqrt{3}}\right) = -\frac{\pi }{6}$$ Thus, it follows: $$(2,\:-\frac{\pi }{6})$$