Question 226461
*[Tex \LARGE z=3\sqrt{2}-3\sqrt{2}i \Rightarrow x=3\sqrt{2}, y=-3\sqrt{2}]




*[Tex \LARGE r=\|z\|=\sqrt{x^2+y^2}=\sqrt{\left(3\sqrt{2}\right)^2+\left(-3\sqrt{2}\right)^2}=\sqrt{18+18}=\sqrt{36}=6]. *[Tex \LARGE r=6]



*[Tex \LARGE \theta=\tan^{-1}\left(\frac{y}{x}\right)=\tan^{-1}\left(\frac{-3\sqrt{2}}{3\sqrt{2}}\right)=\tan^{-1}(-1)=-\frac{\pi}{4}]


Polar Form: *[Tex \LARGE z=r\left(\cos\left(\theta\right)+i\sin\left(\theta\right)\right)=6\left(\cos\left(-\frac{\pi}{4}\right)+i\sin\left(-\frac{\pi}{4}\right)\right)=6\left(\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}\right)]



Answer: Polar form of *[Tex \LARGE z=3\sqrt{2}-3\sqrt{2}i] is *[Tex \LARGE  6\left(\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}\right)]