Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 8 - Section 8.3 - Products and Quotients in Trigonometric Form - 8.3 Problem Set - Page 440: 46

Answer

$\frac{z_{1}}{z_{2}}=2\,cis\,0=2$

Work Step by Step

Dividing in standard form, we have $\frac{z_{1}}{z_{2}}=\frac{2-2i}{1-i}=\frac{2-2i}{1-i}\cdot\frac{1+i}{1+i}$ $=\frac{2+2i-2i-2i^{2}}{1+1}=\frac{4}{2}=2$ $Put z_{1}$ and $z_{2}$ in trigonometric form: $z_{1}=2-2i=2\sqrt {2}(\cos \frac{7\pi}{4}+i\sin\frac{7\pi}{4})$ $z_{2}=1-i=\sqrt {2}(\cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4})$ Dividing again, we obtain $\frac{z_{1}}{z_{2}}=\frac{2\sqrt {2}\,cis\,\frac{7\pi}{4}}{\sqrt {2}\,cis\,\frac{7\pi}{4}}=\frac{2\sqrt {2}}{\sqrt {2}}\,cis\,(\frac{7\pi}{4}-\frac{7\pi}{4})=2\,cis\,0$ This gives: $=2(1+i\cdot0)=2$
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