Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

APPENDIX H - Complex Numbers - H Exercises - Page A 64: 31

Answer

a) $4\sqrt 2 [\cos (31\pi/12)+i \sin (31\pi/12)]$ b) $2 \sqrt 2 [\cos (13\pi/12)+i \sin (13\pi/12)]$ c) $\dfrac{1}{4}[\cos (-11\pi/6)+i \sin (-11\pi/6)]$ Others angles can be considered.

Work Step by Step

a) Here, we have $z=4 [\cos (11\pi/6)+i \sin (11\pi/6)]; w=\sqrt 2 [\cos (3\pi/4)+i \sin (3\pi/4)]$ $zw=4 [\cos (11\pi/6)+i \sin (11\pi/6)] \times \sqrt 2 [\cos (3\pi/4)+i \sin (3\pi/4)]$ or, $=4\sqrt 2 [\cos (31\pi/12)+i \sin (31\pi/12)]$ b) Here, we have $z=4 [\cos (11\pi/6)+i \sin (11\pi/6)]; w=\sqrt 2 [\cos (3\pi/4)+i \sin (3\pi/4)]$ $\dfrac{z}{w}=\dfrac{4 [\cos (11\pi/6)+i \sin (11\pi/6)]}{\sqrt 2 [\cos (3\pi/4)+i \sin (3\pi/4)]}$ or, $=2 \sqrt 2 [\cos (13\pi/12)+i \sin (13\pi/12)]$ c) Here, we have $z=4 [\cos (11\pi/6)+i \sin (11\pi/6)]; w=\sqrt 2 [\cos (3\pi/4)+i \sin (3\pi/4)]$ $\dfrac{1}{z}=\dfrac{1}{4 [\cos (11\pi/6)+i \sin (11\pi/6)]}$ or, $=\dfrac{1}{4}[\cos (-11\pi/6)+i \sin (-11\pi/6)]$
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