## Precalculus: Mathematics for Calculus, 7th Edition

$z_{1}z_{2}= 40(\displaystyle \cos\frac{7\pi}{6}+i\sin\frac{7\pi}{6})$, $z_{1}/z_{2}= 10 (\displaystyle \cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6})$, $1/z_{1}=\displaystyle \frac{1}{20}[\cos(-\pi)+i\sin(-\pi)]$
$z_{1}z_{2}=r_{1}r_{2}[\cos(\theta_{1}+\theta_{2})+i\sin(\theta_{1}+\theta_{2})]$ $\displaystyle \frac{z_{1}}{z_{2}}=\frac{r_{1}}{r_{2}}[\cos(\theta_{1}-\theta_{2})+i\sin(\theta_{1}-\theta_{2})],\quad z_{2}\neq 0$ ----------------- $z_{1}=-20$, $\theta_{1}=\pi$ and $r_{1}=20$. $z_{1}=20(\cos\pi+i\sin\pi)$ $z_{2}=\sqrt{3}+i$, $\theta_{2}$ terminates in quadrant I, $\displaystyle \tan\theta_{2}=\frac{1}{\sqrt{3}}$ $\Rightarrow \displaystyle \theta_{2}=\frac{\pi}{6}$, $r_{2}=\sqrt{3+1}=2$. $z_{2}=2(\displaystyle \cos\frac{\pi}{6}+i\sin\frac{\pi}{6})$. $z_{1}z_{2} =20\displaystyle \cdot 2[\cos(\pi+\frac{\pi}{6})+i\sin(\pi+\frac{\pi}{6})]$ $= 40(\displaystyle \cos\frac{7\pi}{6}+i\sin\frac{7\pi}{6})$, $z_{1}/z_{2} = \displaystyle \frac{20}{2}[\cos(\pi-\frac{\pi}{6})+i\sin(\pi-\frac{\pi}{6})]$ $= 10 (\displaystyle \cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6})$, $1/z_{1}=\displaystyle \frac{1}{20}[\cos(-\pi)+i\sin(-\pi)]$