Answer
$(4 \sqrt 2, 135^{\circ})$
Work Step by Step
Here, $r=\sqrt {x^2+y^2}=\sqrt{(-4)^2+(4)^2}=4 \sqrt 2$
$\tan \theta =\dfrac{y}{x}$
and $\theta=arctan[\dfrac{4}{-4}]=arctan {-1}$
Thus, $\theta =135^{\circ}$
Hence, $(4 \sqrt 2, 135^{\circ})$
