## Precalculus (6th Edition) Blitzer

The required polar form is $z=2\left( \cos \frac{5\pi }{6}+i\sin \frac{5\pi }{6} \right)$
Using $r=\sqrt{{{a}^{2}}+{{b}^{2}}}$ and, \begin{align} & \tan \theta =\frac{b}{a} \\ & =\frac{1}{-\sqrt{3}} \\ & \tan \theta =-\frac{1}{\sqrt{3}} \end{align} So, \begin{align} & r=\sqrt{{{a}^{2}}+{{b}^{2}}} \\ & =\sqrt{{{\left( -\sqrt{3} \right)}^{2}}+{{1}^{2}}} \\ & =\sqrt{3+1} \\ & =2 \end{align} Also, $\tan 30{}^\circ =\frac{1}{\sqrt{3}}$ and $\theta$ lies in the second quadrant. So, \begin{align} & \theta =180{}^\circ -30{}^\circ \\ & =150{}^\circ \end{align} Hence the polar form of $z=-\sqrt{3}+i$ is: \begin{align} & z=r\left( \cos \theta +i\sin \theta \right) \\ & =r\left( \cos 150{}^\circ +i\sin 150{}^\circ \right) \\ & z=2\left( \cos \frac{5\pi }{6}+i\sin \frac{5\pi }{6} \right) \end{align} So, $z=2\left( \cos \frac{5\pi }{6}+i\sin \frac{5\pi }{6} \right)$ Therefore, the required polar form is $z=2\left( \cos \frac{5\pi }{6}+i\sin \frac{5\pi }{6} \right)$.