Answer:
[tex]\large\boxed{C.\ (2,\ -45^o)}[/tex]
Step-by-step explanation:
Regular coordinates (x, y)
Polar coordinates (r, φ)
[tex]r=\sqrt{x^2+y^2}\\\\\psi=\arctan\frac{y}{x}[/tex]
We have the point
[tex](\sqrt2,\ -\sqrt2)[/tex]
Substitute:
[tex]r=\sqrt{(\sqrt2)^2+(-\sqrt2)^2}=\sqrt{2+2}=\sqrt4=2\\\\\psi=\arctan\left(\frac{-\sqrt2}{\sqrt2}\right)=\arctan(-1)=-45^o[/tex]
