Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 8 - Section 8.3 - Polar Form of Complex Numbers; De Moivre's Theorem - 8.3 Exercises - Page 610: 59

Answer

$z_{1}z_{2} = 4\displaystyle \sqrt{2}(\cos\frac{7\pi}{12}+i\sin\frac{7\pi}{12})$, $z_{1}/z_{2} =2\displaystyle \sqrt{2}(\cos\frac{13\pi}{12}+i\sin\frac{13\pi}{12})$ $1/z_{1}=\displaystyle \frac{1}{4}(\cos(-\frac{11\pi}{6})+i\sin(-\frac{11\pi}{6}))$

Work Step by Step

$z_{1}z_{2}=r_{1}r_{2}[\cos(\theta_{1}+\theta_{2})+i\sin(\theta_{1}+\theta_{2})]$ $\displaystyle \frac{z_{1}}{z_{2}}=\frac{r_{1}}{r_{2}}[\cos(\theta_{1}-\theta_{2})+i\sin(\theta_{1}-\theta_{2})],\quad z_{2}\neq 0$ ----------------- $z_{1}=2\sqrt{3}-2i$, ($\theta_{1}$ in quadrant IV). $r_{1}=\sqrt{(2\sqrt{3})^{2}+(-2)^{2}}=\sqrt{12+4}=4.$ $\displaystyle \tan\theta_{1}=\frac{-2}{2\sqrt{3}}=-\frac{1}{\sqrt{3}} \Rightarrow \displaystyle \theta_{1}=\frac{11\pi}{6}.$ $z_{2}=-1+i$, $(\theta_{2}$ in quadrant II $).$ $r_{2}=\sqrt{1+1}=\sqrt{2}$ $\tan\theta_{2}=-1 \Rightarrow \displaystyle \theta_{2}=\frac{3\pi}{4}.$ $z_{1}=4(\displaystyle \cos\frac{11\pi}{6}+i\sin\frac{11\pi}{6})$ and $z_{2}=\displaystyle \sqrt{2}(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4})$. $z_{1}z_{2} = 4\displaystyle \cdot\sqrt{2}[\cos(\frac{11\pi}{6}+\frac{3\pi}{4})+i\sin(\frac{11\pi}{6}+\frac{3\pi}{4})]$ $= 4\displaystyle \sqrt{2}(\cos\frac{7\pi}{12}+i\sin\frac{7\pi}{12})$, $z_{1}/z_{2} = \displaystyle \frac{4}{\sqrt{2}}[\cos(\frac{11\pi}{6}-\frac{3\pi}{4})+i\sin(\frac{11\pi}{6}-\frac{3\pi}{4})]$ $= 2\displaystyle \sqrt{2}(\cos\frac{13\pi}{12}+i\sin\frac{13\pi}{12})$ $1/z_{1}=\displaystyle \frac{1}{4}(\cos(-\frac{11\pi}{6})+i\sin(-\frac{11\pi}{6}))$.
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