## Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole

# Chapter 8 - Section 8.3 - Polar Form of Complex Numbers; De Moivre's Theorem - 8.3 Exercises - Page 610: 58

#### Answer

$z_{1}z_{2}=2\displaystyle \sqrt{2}(\cos\frac{3\pi}{2}+i\sin\frac{3\pi}{2})$ $z_{1}/z_{2} = \sqrt{2}(\cos 0+i\sin 0)$ $1/z_{1}=\displaystyle \frac{1}{2}[\cos(-\frac{7\pi}{4})+i\sin(-\frac{7\pi}{4})]$

#### Work Step by Step

$z_{1}z_{2}=r_{1}r_{2}[\cos(\theta_{1}+\theta_{2})+i\sin(\theta_{1}+\theta_{2})]$ $\displaystyle \frac{z_{1}}{z_{2}}=\frac{r_{1}}{r_{2}}[\cos(\theta_{1}-\theta_{2})+i\sin(\theta_{1}-\theta_{2})],\quad z_{2}\neq 0$ ----------------- $z_{1}=\sqrt{2}-\sqrt{2}i$, ($\theta_{1}$ in quadrant IV). $r_{1}=\sqrt{2+2}=2$ $\tan\theta_{1}=-1 \Rightarrow \displaystyle \theta_{1}=\frac{7\pi}{4}$. . $z_{2}=1-i$, ($\theta_{2}$ in quadrant IV). $r_{2}=\sqrt{1+1}=\sqrt{2}$. $\tan\theta_{2}=-1 \Rightarrow \displaystyle \theta_{2}=\frac{7\pi}{4}$. $z_{1}=2(\displaystyle \cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4})$ and $z_{2}=\displaystyle \sqrt{2}(\cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4})$. $z_{1}z_{2}=2\displaystyle \cdot\sqrt{2}[\cos(\frac{7\pi}{4}+\frac{7\pi}{4})+i\sin(\frac{7\pi}{4}+\frac{7\pi}{4})]$ $=2\displaystyle \sqrt{2}(\cos\frac{7\pi}{2}+i\sin\frac{7\pi}{2})\qquad$ ... $\displaystyle \frac{7\pi}{2}=\frac{3\pi}{2}+2\pi$ $=2\displaystyle \sqrt{2}(\cos\frac{3\pi}{2}+i\sin\frac{3\pi}{2})$ $z_{1}/z_{2} = \displaystyle \frac{2}{\sqrt{2}}[\cos(\frac{7\pi}{4}-\frac{7\pi}{4})+i\sin(\frac{7\pi}{4}-\frac{7\pi}{4})]$ $= \sqrt{2}(\cos 0+i\sin 0)$ $1/z_{1}=\displaystyle \frac{1}{2}[\cos(-\frac{7\pi}{4})+i\sin(-\frac{7\pi}{4})]$

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