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