so hmmm notice the picture below, is a negative angle, it lands on the 45° of the 3rd quadrant, where "x" and "y" are both negative, and also are both the same value
thus
[tex]\bf \begin{array}{llll}
4[cos(&-135^o)+i\ sin(&-135^o)]\\
\uparrow &\quad \uparrow &\quad \uparrow \\
r&\quad \theta&\quad \theta
\end{array}\qquad
\begin{cases}
x=rcos(\theta)\\
y=rsin(\theta)\\
----------\\
r=4\\
\theta=-135^o\\
cos(-135^o)=-\frac{\sqrt{2}}{2}\\
sin(-135^o)=-\frac{\sqrt{2}}{2}\\
\end{cases}
\\\\\\
x=4\left( -\frac{\sqrt{2}}{2} \right)\implies x=-2\sqrt{2}
\\\\\\
y=4\left( -\frac{\sqrt{2}}{2} \right)\implies y=-2\sqrt{2}\\\\
-----------------------------\\\\[/tex]
[tex]\bf \begin{array}{llll}
a&+&bi\\
x&&yi
\end{array}\implies -2\sqrt{2}-2\sqrt{2}\ i[/tex]
