Answer
$$\left( {2\sqrt 2 ,\frac{\pi }{4}} \right)$$ or $$\left( { - 2\sqrt 2 ,\frac{{5\pi }}{4}} \right)$$
Work Step by Step
$$\eqalign{
& \left( {2,2} \right) \Rightarrow x = 2{\text{ and }}y = 2 \cr
& {\text{Converting to cartesian coordinates }}\left( {r,\theta } \right).{\text{ Where}} \cr
& r = \sqrt {{x^2} + {y^2}} {\text{ and }}\theta = {\tan ^{ - 1}}\left( {\frac{y}{x}} \right) \cr
& {\text{We obtain}}{\text{,}} \cr
& r = \sqrt {{{\left( 2 \right)}^2} + {{\left( 2 \right)}^2}} = 2\sqrt 2 \cr
& {\text{Coordinate }}x{\text{ }}\left( {{\text{positive}}} \right){\text{, Coordinate }}y{\text{ }}\left( {{\text{positive}}} \right),{\text{ Quadrant I}} \cr
& \theta = {\tan ^{ - 1}}\left( {\frac{2}{2}} \right) = \frac{\pi }{4} \cr
& {\text{The polar coordinates are:}} \cr
& \left( {r,\theta } \right) = \left( {2\sqrt 2 ,\frac{\pi }{4}} \right) \cr
& or \cr
& \left( { - r,\theta + \pi } \right) = \left( { - 2\sqrt 2 ,\frac{{5\pi }}{4}} \right) \cr} $$
