[tex]\bf \begin{array}{lclll}
\underline{2}&\underline{-2} i\\
a&b
\end{array}\qquad
\begin{cases}
r=\sqrt{a^2+b^2}\\\\
\theta=tan^{-1}\left( \frac{b}{a} \right)
\end{cases}\\\\
-----------------------------\\\\
r=\sqrt{2^2+(-2)^2}\implies r=\sqrt{8}\implies r=2\sqrt{2}
\\\\\\
\theta=tan^{-1}\left( \cfrac{-2}{2} \right)\implies \theta=tan^{-1}(-1)\qquad \theta=
\begin{cases}
\frac{3\pi }{4}\\\\
\frac{7\pi }{4}
\end{cases}[/tex]
so, either of those angles, have a tangent of -1, check your Unit Circle
however, notice the rectangular is 2 and -2i, or 2,-2i, that's on the 4th quadrant, graph it and check, so, the angle for that rectangular will the be [tex]\bf \frac{7\pi }{4}[/tex]
now, the polar form of any rectangular is [tex]\bf r[cos(\theta)-i\ sin(\theta)][/tex]
thus [tex]\bf 2,-2i\implies 2\sqrt{2}\left[ cos\left( \frac{7\pi }{4} \right) +i\ sin\left( \frac{7\pi }{4} \right)\right]
[/tex]
