[tex]\bf z=\stackrel{a}{-1}\stackrel{b}{+3}i\qquad
\begin{cases}
r=\sqrt{a^2+b^2}\\
\theta =tan^{-1}\left( \frac{b}{a} \right)
\end{cases}\qquad \qquad z=r[cos(\theta )i~sin(\theta )]\\\\
-------------------------------\\\\
r=\sqrt{(-1)^2+3^2}\implies r=\sqrt{10}
\\\\\\
\theta =tan^{-1}\left( \frac{3}{-1} \right)\implies \theta \approx -71.57\implies \theta \approx\stackrel{360-71.57}{288.43^o}\\\\
-------------------------------\\\\
z=\sqrt{10}\left[ cos(288.43^o)+i~sin(288.43^o) \right][/tex]
now, the calculation in the choices, show 289.5°, which I take it is due some early rounding up.
