## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 6 - Section 6.5 - Complex Numbers in Polar Form; DeMoivre's Theorem - Exercise Set - Page 768: 70

#### Answer

The complex fifth roots of the given complex numbers in rectangular form are ${{z}_{0}}=1+i\sqrt{3},\text{ }{{z}_{1}}=-1.3+1.5i,\ {{z}_{2}}=-1.8-0.8i,\text{ }{{z}_{3}}=0.2-2.0i,\text{ and }{{z}_{4}}=2.0-0.4i$.

#### Work Step by Step

Consider the given complex number $32\left( \cos \frac{5\pi }{3}+i\sin \frac{5\pi }{3} \right)$ For any complex number of the type $z=r\cos \theta +i\sin \theta$ If n is a positive integer, then ${{z}^{n}}$ is ${{z}_{k}}=\sqrt[n]{r}\left[ \cos \left( \frac{\theta +2\pi k}{n} \right)+i\sin \left( \frac{\theta +2\pi k}{n} \right) \right]$ ...... (1) Now, for the given complex number, use (1) to get \begin{align} & {{z}_{k}}=\sqrt[n]{r}\left[ \cos \left( \frac{\theta +2\pi k}{n} \right)+i\sin \left( \frac{\theta +2\pi k}{n} \right) \right] \\ & {{z}_{0}}=\sqrt[5]{32}\left[ \cos \frac{\frac{5\pi }{3}+2\pi \times 0}{5}+i\sin \frac{\frac{5\pi }{3}+2\pi \times 0}{5} \right] \\ & {{z}_{0}}=2\left[ \cos \frac{\pi }{3}+i\sin \frac{\pi }{3} \right] \\ & {{z}_{0}}=1+\sqrt{3}i \\ \end{align} Now, for the other roots, substitute the value of $k=1,2,3,4$. That is, \begin{align} & {{z}_{k}}=\sqrt[n]{r}\left[ \cos \left( \frac{\theta +2\pi k}{n} \right)+i\sin \left( \frac{\theta +2\pi k}{n} \right) \right] \\ & {{z}_{1}}=\sqrt[5]{32}\left[ \cos \frac{\frac{5\pi }{3}+2\pi \times 1}{5}+i\sin \frac{\frac{5\pi }{3}+2\pi \times 0}{5} \right] \\ & {{z}_{1}}=2\left[ \cos \frac{11\pi }{15}+i\sin \frac{11\pi }{15} \right] \\ & {{z}_{1}}=-1.3+1.5i \\ \end{align} Similarly, \begin{align} & {{z}_{k}}=\sqrt[n]{r}\left[ \cos \left( \frac{\theta +2\pi k}{n} \right)+i\sin \left( \frac{\theta +2\pi k}{n} \right) \right] \\ & {{z}_{2}}=\sqrt[5]{32}\left[ \cos \frac{\frac{5\pi }{3}+2\pi \times 2}{5}+i\sin \frac{\frac{5\pi }{3}+2\pi \times 2}{5} \right] \\ & {{z}_{2}}=2\left[ \cos \frac{17\pi }{15}+i\sin \frac{17\pi }{15} \right] \\ & {{z}_{2}}=-1.8-0.8i \\ \end{align} Also, \begin{align} & {{z}_{k}}=\sqrt[n]{r}\left[ \cos \left( \frac{\theta +2\pi k}{n} \right)+i\sin \left( \frac{\theta +2\pi k}{n} \right) \right] \\ & {{z}_{3}}=\sqrt[5]{32}\left[ \cos \frac{\frac{5\pi }{3}+2\pi \times 3}{5}+i\sin \frac{\frac{5\pi }{3}+2\pi \times 3}{5} \right] \\ & {{z}_{3}}=2\left[ \cos \frac{23\pi }{15}+i\sin \frac{23\pi }{15} \right] \\ & {{z}_{3}}=0.2-2.0i \\ \end{align} Also, \begin{align} & {{z}_{k}}=\sqrt[n]{r}\left[ \cos \left( \frac{\theta +2\pi k}{n} \right)+i\sin \left( \frac{\theta +2\pi k}{n} \right) \right] \\ & {{z}_{4}}=\sqrt[5]{32}\left[ \cos \frac{\frac{5\pi }{3}+2\pi \times 4}{5}+i\sin \frac{\frac{5\pi }{3}+2\pi \times 4}{5} \right] \\ & {{z}_{4}}=2\left[ \cos \frac{29\pi }{15}+i\sin \frac{29\pi }{15} \right] \\ & {{z}_{4}}=2.0-0.4i \\ \end{align} The fifth roots of the complex numbers in the polar form are ${{z}_{0}}=1+i\sqrt{3},\text{ }{{z}_{1}}=-1.3+1.5i,\ {{z}_{2}}=-1.8-0.8i,\text{ }{{z}_{3}}=0.2-2.0i,\text{ and }{{z}_{4}}=2.0-0.4i$.

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