Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 8 - Application of Trigonometry - 8.6 De Moivre's Theorem; Powers and Roots of Complex Numbers - 8.6 Exercises - Page 808: 14

Answer

$-128+128\sqrt 3i$

Work Step by Step

Since, $(2-2i\sqrt 3)=4(\cos 300^{\circ}+i \sin 300^{\circ})$ and $(2-2i\sqrt 3)^{4}=4(\cos 300^{\circ}+i \sin 300^{\circ})^{4}$ De Moivre’s Theorem states that when $ r (\cos\theta+i \sin\theta)$ is a complex number, and if $n$ is any real number, then the following relationship holds. $[ r (\cos\theta+i \sin\theta)]^{n}=[ r^{n} (\cos n\theta+i \sin n\theta)]$ In compact form, this is written $[ r cis\theta]^{n}=[ r^{n} (cis \theta)]$ $4(\cos 300^{\circ}+i \sin 300^{\circ})^{4}=4^{4}(\cos 4\times300^{\circ}+i \sin 4\times300^{\circ})$ $=256(\cos 1200^{\circ}+i \sin 1200^{\circ})$ $=[256(\frac{-1}{2}+i.\frac{ \sqrt 3}{2})]$ $=-128+128\sqrt 3i$
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