[tex]\bf -2\implies -2+0i\implies \stackrel{a}{-2}+\stackrel{b}{0}i\quad
\begin{cases}
r=\sqrt{a^2+b^2}\\
r=\sqrt{(-2)^2+0^2}\\
r=2\\
----------\\
\theta =tan^{-1}\left( \frac{b}{a} \right)\\
\theta =tan^{-1}\left( \frac{0}{-2} \right)\\
\theta =tan^{-1}(0)\\
\measuredangle \theta =0~,~\pi
\end{cases}[/tex]
now, there are two possible angles for that tangent... however, let's look at our a,b components.
"a" is negative whilst "b" is positive, so is not 0 then, thus
[tex]\bf -2+0i\implies r[cos(\theta )+i~sin(\theta )]\implies 2[cos(\pi )+i~sin(\pi )][/tex]
