Answer
$$\left( {2\sqrt 2 , - 2\sqrt 2 } \right)$$
Work Step by Step
$$\eqalign{
& \left( { - 4,\frac{{3\pi }}{4}} \right) = \left( { - r,\theta } \right) \cr
& {\text{Where }}\left( { - r,\theta } \right) = \left( {r,\theta + \pi } \right) \cr
& \left( { - 4,\frac{{3\pi }}{4}} \right) = \left( {4,\frac{{3\pi }}{4} + \pi } \right) \cr
& \left( { - 4,\frac{{3\pi }}{4}} \right) = \left( {4,\frac{{7\pi }}{4}} \right) \cr
& \Rightarrow r = 4{\text{ and }}\theta = \frac{{7\pi }}{4} \cr
& {\text{Converting to cartesian coordinates }}\left( {x,y} \right).{\text{ Where}} \cr
& x = r\cos \theta {\text{ and }}y = r\sin \theta \cr
& {\text{We obtain}}{\text{,}} \cr
& x = 4\cos \left( {\frac{{7\pi }}{4}} \right) = 4\left( {\frac{{\sqrt 2 }}{2}} \right) = 2\sqrt 2 \cr
& y = 4\sin \left( {\frac{{7\pi }}{4}} \right) = 4\left( { - \frac{{\sqrt 2 }}{2}} \right) = - 2\sqrt 2 \cr
& {\text{The cartesian coordinates are:}} \cr
& \left( {x,y} \right) = \left( {2\sqrt 2 , - 2\sqrt 2 } \right) \cr} $$
