[tex]\bf \stackrel{a}{-14}+\stackrel{b}{8}i\quad
\begin{cases}
r[cos(\theta )+i~sin(\theta )]\\
----------\\
r=\sqrt{a^2+b^2}\\
\theta =tan^{-1}\left( \frac{b}{a} \right)
\end{cases}
\\\\\\
r=\sqrt{(-14)^2+8^2}\implies r=\sqrt{260}\implies \boxed{r=2\sqrt{65}}
\\\\\\
\theta =tan^{-1}\left( \frac{8}{-14} \right)\implies \theta \approx -29.74^o[/tex]
now.... that angle is correct for that tangent value, however, let's take a peek at our a,b elements, "a" is negative and "b" is positive, that means the x-coordinate is negative and the y-coordinate is positive, that means, the II quadrant, NOT the IV quadrant.
so, let's get the twin of -29.74°, but adding 180° to it, and we'll land at 150.26° or thereabouts, which is in the II quadrant, where our a,b point is at.
[tex]\bf \begin{cases}
r=2\sqrt{65}\\
\theta =150.26^o
\end{cases}\implies 2\sqrt{65}[cos(150.26^o)+i~sin(150.26^o)][/tex]
