Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 9 - Polar Coordinates; Vectors - Chapter Review - Chapter Test - Page 636: 10

Answer

$zw=6(\cos({107^\circ)}+i\sin{(107^\circ}))$

Work Step by Step

We know that if $z=a(\cos{\alpha}+i\sin{\alpha})$ and $w=b(\cos{\beta}+i\sin{\beta})$, then $zw=ab(\cos({\alpha+\beta)}+i\sin{(\alpha+\beta})$ and $\frac{z}{w}=\frac{a}{b}(\cos({\alpha-\beta)}+i\sin{(\alpha-\beta})$. Hence here: $zw=(2)(3)(\cos({85^\circ+22^\circ)}+i\sin{(85^\circ+22^\circ})\\zw=6(\cos({107^\circ)}+i\sin{(107^\circ}))$
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