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: 34

Answer

$16+16\sqrt{3}i$

Work Step by Step

We are given; $z=1-\sqrt{3}i$ To find $r$ of a complex number $a+bi$, we use: $\sqrt{a^2+b^2}$: $r=\sqrt{1^{2}+(-\sqrt{3})^{2}}=2$ To find $\theta$, we use $\tan{\theta}=\frac{b}{a}$: $\displaystyle \tan\theta=\frac{-\sqrt{3}}{1}=-\sqrt{3}$ And since $(1,-\sqrt{3})$ is in the 4th quadrant: $\displaystyle \theta=\frac{5\pi}{3} $ To put the number in polar form, we use $r(\cos{\theta}+i\sin{\theta})$: $z=2(\displaystyle \cos\frac{5\pi}{3}+i\sin\frac{5\pi}{3})$ . We apply the exponent to the polar form of the function and use De Moivre's Theorem: $(1-\sqrt{3}i)^{5}=[2(\cos\frac{5\pi}{3}+i\sin\frac{5\pi}{3})]^{5}=2^{5}(\cos\frac{5*5\pi}{3}+i\sin\frac{5*5\pi}{3})=2^{5}(\cos{(8\pi+\frac{\pi}{3})}+i\sin{(8\pi+\frac{\pi}{3})})=2^{5}(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}) =32(\frac{1}{2}+\frac{\sqrt{3}}{2}i)=16+16\sqrt{3}i$
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