Answer
$$z=\sqrt{8} \left ( \cos \frac{\pi }{4} +i \sin \frac{\pi }{4} \right )$$
Work Step by Step
We plot the complex number $z=2+2i$ the same way we plot $(2,2)$ 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 $z=2+2i$, we have$$r=\sqrt{2^2+2^2}=\sqrt{8}, \\ \tan \theta =\frac{2}{2}=1.$$Since $\cos \theta =\frac{a}{r} = \frac{2}{\sqrt{4}} \gt 0$, the argument, $\theta$, must lie in quadrant $I$. So$$\theta = \frac{\pi }{4}.$$Thus, the complex number has the polar form$$z=\sqrt{8} \left ( \cos \frac{\pi }{4} +i \sin \frac{\pi }{4} \right ).$$
