Answer
$(3\sqrt 2, \frac{3\pi}{4})$, $(-3\sqrt 2, -\frac{\pi}{4})$
Work Step by Step
1. Given $(-3,3)$ (in quadrant II) in rectangular coordinates, we have $r=\sqrt {9+9}=3\sqrt 2$ and $\theta=tan^{-1}(\frac{3}{-3})=\frac{3\pi}{4}$, thus we have $(3\sqrt 2, \frac{3\pi}{4})$ in polar coordinates for $r\gt0$.
2. For $r\lt0$, we can choose $(-3\sqrt 2, -\frac{\pi}{4})$ for the same point.
