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: 17

Answer

$128+128i$

Work Step by Step

Since, $(-2-2i)=2\sqrt 2(\cos 225^{\circ}+i \sin 225^{\circ})$ and $(-2-2i)^{5}=2\sqrt 2(\cos 225^{\circ}+i \sin 225^{\circ})^{5}$ 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 $[ r cis\theta]^{n}=[ r^{n} (cis \theta)]$ $2\sqrt 2(\cos 225^{\circ}+i \sin 225^{\circ})^{5}=(2\sqrt 2)^{5}(\cos 5\times 225^{\circ}+i \sin 5\times 225^{\circ})$ $=32\sqrt {32}(\cos 1125^{\circ}+i \sin 1125^{\circ})$ $=128\sqrt 2(\cos 45^{\circ}+i \sin 45^{\circ})$ $=128\sqrt 2(\frac{1}{\sqrt 2}+i.\frac{ 1}{\sqrt 2})$ $=128+128i$
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