Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Appendix G - Complex Numbers - G Exercises - Page A56: 33

Answer

$-1024$

Work Step by Step

We are given: $z=1+i=1+1*i$ To find $r$ of a complex number $a+bi$, we use: $\sqrt{a^2+b^2}$: $r=\sqrt{1^2+1^2}=\sqrt{2}$ To find $\theta$, we use $\tan{\theta}=\frac{b}{a}$: $\displaystyle \tan\theta=\frac{1}{1}=1 $ And since (1,1) is in the 1st quadrant: $\displaystyle \theta=\frac{\pi}{4} $ To put the number in polar form, we use $r(\cos{\theta}+i\sin{\theta})$: $z=\displaystyle \sqrt{2}(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4})$ We apply the exponent to the polar form of the function and use De Moivre's Theorem: $(1+i)^{20}=[\sqrt{2}(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4})]^{20}=(2^{1/2})^{20}(\cos\frac{20\pi}{4}+i\sin\frac{20\pi}{4})=2^{10}(\cos 5\pi+i\sin 5\pi) =2^{10}(-1+i*0)=-2^{10}=-1024$
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