use polar notation to write $(-2+2i)^3$ in polar form

Question

I am supposed to go from eulers equation to polar. I am looking for solution verification:

$$r = \sqrt{-2^2+2^2} = \sqrt{8}$$

$$\theta = \arctan \Big(\frac{2}{-2} \Big) = \arctan(-1)= -45$$

since Graphing $-2+2i$ falls in the second quadrant I took $180 - (-45) = 225$

so we have $$\Big(\sqrt{8}e^{i225}\Big)^3 = \sqrt{8}^3[\cos((225)(3))+i\sin((225)(3))]$$

Answer 1

In polar we have $8^{3/2}e^{-\frac{3πi}4}$. This I did in radians. $π/4=45°$.