## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# APPENDIX H - Complex Numbers - H Exercises: 35

#### Answer

$-512\sqrt{3}+512i$

#### Work Step by Step

We are given: $z=2\sqrt{3}+2i$ To find $r$ of a complex number $a+bi$, we use: $\sqrt{a^2+b^2}$: $r=\sqrt{(2\sqrt{3})^{2}+2^{2}}=\sqrt{16}=4$ To find $\theta$, we use $\tan{\theta}=\frac{b}{a}$: $\displaystyle \tan\theta=\frac{2}{2\sqrt{3}}=\frac{1}{\sqrt{3}}$ And since $z$ is in the 1st quadrant (by looking at the signs of $a$ and $b$), we have: $\displaystyle \theta=\frac{\pi}{6}$ To put the number in polar form, we use $r(\cos{\theta}+i\sin{\theta})$: $z=4(\displaystyle \cos\frac{\pi}{6}+i\sin\frac{\pi}{6})$ . We apply the exponent to the polar form of the function and use De Moivre's Theorem: $(2\displaystyle \sqrt{3}+2i)^{5}=[4(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6})]^{5}=4^{5}(\cos\frac{5*\pi}{6}+i\sin\frac{5*\pi}{6})=1024(-\frac{\sqrt{3}}{2}+\frac{1}{2}i)=-512\sqrt{3}+512i$

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