## Precalculus (6th Edition) Blitzer

Step 1. see figure. Step 2. $4cos(\frac{5\pi}{6})+4isin(\frac{5\pi}{6})$
Step 1. Given the complex number $-2\sqrt 3+2i$, we can identify $a=-2\sqrt 3, b=2$ $(-2\sqrt 3,2)$ in complex coordinates as shown in the figure. Step 2. Use the above results, we can get the modulus as $r=\sqrt {a^2+b^2}=\sqrt {(-2\sqrt 3)^2+(2)^2}=4$. The polar angle can be found as $tan\theta=\frac{b}{a}=\frac{2}{-2\sqrt 3}=-\frac{\sqrt 3}{3}$ which gives $\theta=\frac{5\pi}{6}$ (in quadrant II). Thus, we can write the complex number in polar form as $-2\sqrt 3+2i=4cos(\frac{5\pi}{6})+4isin(\frac{5\pi}{6})$