Answer
$$z=6 \left ( \cos \frac{7\pi }{4} +i \sin \frac{7\pi }{4} \right )$$
Work Step by Step
We plot the complex number $z=3\sqrt{2}-3\sqrt{2}i$ the same way we plot $(3\sqrt{2}, -3\sqrt{2})$ in the rectangular coordinate system.
Please note that the complex number $z=a+bi$ is written in polar form as$$z=r(\cos \theta +i \sin \theta ),$$where $a=r\cos \theta$, $b=r\sin \theta$, $r=\sqrt{a^2+b^2}$, and $\tan \theta = \frac{b}{a}$, $0 \le \theta \lt 2\pi $.
So, for the complex number $z=3\sqrt{2}-3\sqrt{2}i$, we have$$r=\sqrt{(3\sqrt{2})^2+(-3\sqrt{2})^2}=\sqrt{36}=6, \\ \tan \theta =\frac{-3\sqrt{2}}{3\sqrt{2}}=-1.$$Since $\cos \theta =\frac{a}{r} = \frac{3\sqrt{2}}{\sqrt{36}} \gt 0$, the argument, $\theta$, must lie in quadrant $IV$. So$$\theta =\frac{3\pi }{4}+\pi = \frac{7\pi }{4}$$Thus, the complex number has the polar form$$z=6 \left ( \cos \frac{7\pi }{4} +i \sin \frac{7\pi }{4} \right ).$$
