Calculus 8th Edition

Published by Cengage

Appendix G - Complex Numbers - G Exercises - Page A56: 32

Answer

a) $24\sqrt 2 [\cos (17\pi/12)+i \sin (17\pi/12)]$ b) $\dfrac{4 \sqrt 2}{3} [\cos (-13\pi/12)+i \sin (-13\pi/12)]$ c) $\dfrac{1}{8}[\cos (-\pi/6)+i \sin (-\pi/6)]$

Work Step by Step

a) Here, we have $z=8[\cos (\dfrac{\pi}{6})+i \sin (\dfrac{\pi}{6})]$ and $w=3 \sqrt 2 [\cos (5\pi/4)+i \sin (5\pi/4)]$ $zw=8[\cos (\pi/6)+i \sin (\dfrac{\pi}{6})] \times 3 \sqrt 2 [\cos (5\pi/4)+i \sin (5\pi/4)]=24\sqrt 2 [\cos (17\pi/12)+i \sin (17\pi/12)]$ b) Here, we have $z=8[\cos (\dfrac{\pi}{6})+i \sin (\dfrac{\pi}{6})]$ and $w=3 \sqrt 2 [\cos (5\pi/4)+i \sin (5\pi/4)]$ $\dfrac{z}{w}=\dfrac{8[\cos (\dfrac{\pi}{6})+i \sin (\dfrac{\pi}{6})]}{3 \sqrt 2 [\cos (5\pi/4)+i \sin (5\pi/4)]}=\dfrac{4 \sqrt 2}{3} [\cos (-13\pi/12)+i \sin (-13\pi/12)]$ c) Here, we have $z=8[\cos (\dfrac{\pi}{6})+i \sin (\dfrac{\pi}{6})]$ and $w=3 \sqrt 2 [\cos (5\pi/4)+i \sin (5\pi/4)]$ $\dfrac{1}{z}=\dfrac{1}{8}[\cos (-\pi/6)+i \sin (-\pi/6)]$

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