Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 8 - Complex Numbers, Polar Equations, and Parametric Equations - Section 8.4 De Moivre's Theorem: Powers and Roots of Complex Numbers - 8.4 Exercises - Page 376: 12

Answer

$-8-8i$

Work Step by Step

Since, $(-1+i)=\sqrt 2(\cos 135^{\circ}+i \sin 135^{\circ})$ and $(-1+i)^{7}=\sqrt 2(\cos 135^{\circ}+i \sin 135^{\circ})^{7}$ 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 $[ r cis\theta]^{n}=[ r^{n} (cis \theta)]$ $\sqrt 2(\cos 135^{\circ}+i \sin 135^{\circ})^{7}=(\sqrt 2)^{7}(\cos 7\times 135^{\circ}+i \sin 7\times 135^{\circ})$ $=8\sqrt 2(\cos 945^{\circ}+i \sin 945^{\circ})$ $=8\sqrt 2(\cos 225^{\circ}+i \sin 225^{\circ})$ $=8\sqrt 2(\frac{-1}{\sqrt 2}-i.\frac{ 1}{\sqrt 2})$ $=-8-8i$
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