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: 61

Answer

$z_{1}z_{2}=20\displaystyle \sqrt{2}(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}),$ $z_{1}/z_{2}=\displaystyle \frac{5\sqrt{2}}{4}(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4})$, $1/z_{1}=\displaystyle \frac{1}{5\sqrt{2}}(\cos(-\frac{\pi}{4})+i\sin(-\frac{\pi}{4}))$

Work Step by Step

$z_{1}z_{2}=r_{1}r_{2}[\cos(\theta_{1}+\theta_{2})+i\sin(\theta_{1}+\theta_{2})]$ $\displaystyle \frac{z_{1}}{z_{2}}=\frac{r_{1}}{r_{2}}[\cos(\theta_{1}-\theta_{2})+i\sin(\theta_{1}-\theta_{2})],\quad z_{2}\neq 0$ ----------------- $z_{1}=5+5i$, $\theta_{1}$ terminates in quadrant I $\displaystyle \tan\theta_{1}=\frac{5}{5}=\mathrm{l} \Rightarrow$ $\displaystyle \theta_{1}=\frac{\pi}{4}$, $r_{1}=\sqrt{25+25}=5\sqrt{2}$ $z_{1}=5\displaystyle \sqrt{2}(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4})$ $z_{2}=4$, $\theta_{2}=0$, $r_{2}=4$. $z_{2}=4(\cos 0+i\sin 0)$. $z_{1}z_{2}=5\displaystyle \sqrt{2}\cdot 4[\cos(\frac{\pi}{4}+0)+i\sin(\frac{\pi}{4}+0)]$ $=20\displaystyle \sqrt{2}(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}),$ $z_{1}/z_{2}=\displaystyle \frac{5\sqrt{2}}{4}(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4})$, $1/z_{1}=\displaystyle \frac{1}{5\sqrt{2}}(\cos(-\frac{\pi}{4})+i\sin(-\frac{\pi}{4}))$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.