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: 55


$z_{1}z_{2}=100(\cos 350^{o}+i\sin 350^{o})$ $\displaystyle \frac{z_{1}}{z_{2}}=\frac{4}{25}(\cos 50^{o}+i\sin 50^{o})$

Work Step by Step

See p. 605, If the two complex numbers $z_{1}$ and $z_{2}$ have the polar forms If $z_{1}=r_{1}(\cos\theta_{1}+i\sin\theta_{1})$, $z_{2}=r_{2}(\cos\theta_{2}+i\sin\theta_{2})$, then $z_{1}z_{2}=r_{1}r_{2}[\cos(\theta_{1}+\theta_{2})+i\sin(\theta_{1}+\theta_{2})]$ $\displaystyle \frac{z_{1}}{z_{2}}=\frac{r_{1}}{r_{2}}[\cos(\theta_{1}-\theta_{2})+i\sin(\theta_{1}-\theta_{2})],\quad z_{2}\neq 0$ ---------- $r_{1}r_{2}=4\cdot 25=100,$ $200^{o}+150^{o}=350^{o}$ $ \displaystyle \frac{r_{1}}{r_{2}}=\frac{4}{25}$ $200^{o}-150^{o}=50^{o}$ $z_{1}z_{2}=100(\cos 350^{o}+i\sin 350^{o})$ $\displaystyle \frac{z_{1}}{z_{2}}=\frac{4}{25}(\cos 50^{o}+i\sin 50^{o})$
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