Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 6 - Section 6.5 - Complex Numbers in Polar Form; DeMoivre's Theorem - Exercise Set - Page 768: 76


$\approx 1+0.5i$ ; $\approx -0.2+1.1i$ ; $\approx -1.1+0.2i$ ; $\approx -0.5-1.0i$ ; $0.8-0.8i$

Work Step by Step

$\approx 1+0.5i$ ; $\approx -0.2+1.1i$ ; $\approx -1.1+0.2i$ ; $\approx -0.5-1.0i$ ; $0.8-0.8i$ Roots of a complex number represented in polar form can be found by the application of De Moivre’s theorem. DeMoivre’s theorem for finding complex roots states that every complex number has two distinct complex square roots, three distinct complex cube roots, four distinct complex fourth roots, and so on. Each root has the same modulus. $\begin{align} & r=z_{k}^{n} \\ & {{z}_{k}}=\sqrt[n]{r}\left[ \cos \left( \frac{\theta +2\pi k}{n} \right)+i\sin \left( \frac{\theta +2\pi k}{n} \right) \right] \\ \end{align}$ where $k$ is the number of distinct $n\text{th}$ root and $\theta $ is in radians. There are exactly five fifth roots of $-1+i$ as per De Moivre’stheorem. Convert the above expression into polar form as follows: $\begin{align} & -1+i=r\left( \cos \,\theta +i\,\sin \,\theta \right) \\ & =\sqrt{2}\left( \cos \left( \frac{3\pi }{4} \right)+i\,\sin \left( \frac{3\pi }{4} \right) \right) \end{align}$ Apply DeMoivre’stheorem to find all fifth roots of $-1+i$. $\begin{align} & {{z}_{k}}=\sqrt[5]{\sqrt{2}}\left[ \cos \left( \frac{\frac{3\pi }{4}+2\pi k}{5} \right)+i\,\sin \left( \frac{\frac{3\pi }{4}+2\pi k}{5} \right) \right] \\ & =\sqrt[10]{2}\left[ \cos \left( \frac{\frac{3\pi }{4}+2\pi k}{5} \right)+i\,\sin \left( \frac{\frac{3\pi }{4}+2\pi k}{5} \right) \right] \end{align}$ Here, $k=0,1,2,3,4$. Insert the above values of $k$ to find five distinct fifth roots of the provided number. For $k=0$, $\begin{align} & {{z}_{0}}=\sqrt[5]{\sqrt{2}}\left[ \cos \left( \frac{\frac{3\pi }{4}+2\pi \left( 0 \right)}{5} \right)+i\,\sin \left( \frac{\frac{3\pi }{4}+2\pi \left( 0 \right)}{5} \right) \right] \\ & =\sqrt[10]{2}\left[ \cos \left( \frac{\frac{3\pi }{4}+2\pi \left( 0 \right)}{5} \right)+i\,\sin \left( \frac{\frac{3\pi }{4}+2\pi \left( 0 \right)}{5} \right) \right] \\ & =\sqrt[10]{2}\left[ \cos \left( \frac{3\pi }{20} \right)+i\,\sin \left( \frac{3\pi }{20} \right) \right] \end{align}$ Substitute the values of $\cos \left( \frac{3\pi }{20} \right)$ and $\sin \left( \frac{3\pi }{20} \right)$: $\begin{align} & {{z}_{0}}=\sqrt[10]{2}\left[ 0.8910+i0.4539 \right] \\ & =0.9549+i0.4866 \\ & \approx 1+0.5i \end{align}$ For $k=1$, $\begin{align} & {{z}_{1}}=\sqrt[5]{\sqrt{2}}\left[ \cos \left( \frac{\frac{3\pi }{4}+2\pi \left( 1 \right)}{5} \right)+i\,\sin \left( \frac{\frac{3\pi }{4}+2\pi \left( 1 \right)}{5} \right) \right] \\ & =\sqrt[10]{2}\left[ \cos \left( \frac{11\pi }{20} \right)+i\,\sin \left( \frac{11\pi }{20} \right) \right] \end{align}$ Substitute the values of $\cos \left( \frac{11\pi }{20} \right)$ and $\sin \left( \frac{11\pi }{20} \right)$: For $k=2$, $\begin{align} & {{z}_{2}}=\sqrt[5]{\sqrt{2}}\left[ \cos \left( \frac{\frac{3\pi }{4}+2\pi \left( 2 \right)}{5} \right)+i\,\sin \left( \frac{\frac{3\pi }{4}+2\pi \left( 2 \right)}{5} \right) \right] \\ & =\sqrt[10]{2}\left[ \cos \left( \frac{19\pi }{20} \right)+i\,\sin \left( \frac{19\pi }{20} \right) \right] \end{align}$ Substitute the values of $\cos \left( \frac{19\pi }{20} \right)$ and $\sin \left( \frac{19\pi }{20} \right)$: $\begin{align} & {{z}_{2}}=\sqrt[10]{2}\left[ -0.9877+i\left( 0.1564 \right) \right] \\ & =-1.0586+0.1677i \\ & \approx -1.1+0.2i \end{align}$ For $k=3$, $\begin{align} & {{z}_{3}}=\sqrt[5]{\sqrt{2}}\left[ \cos \left( \frac{\frac{3\pi }{4}+2\pi \left( 3 \right)}{5} \right)+i\,\sin \left( \frac{\frac{3\pi }{4}+2\pi \left( 3 \right)}{5} \right) \right] \\ & =\sqrt[10]{2}\left[ \cos \left( \frac{27\pi }{20} \right)+i\,\sin \left( \frac{27\pi }{20} \right) \right] \end{align}$ Substitute the values of $\cos \left( \frac{27\pi }{20} \right)$ and $\sin \left( \frac{27\pi }{20} \right)$: $\begin{align} & {{z}_{3}}=\sqrt[10]{2}\left[ -0.4539+i\left( -0.8910 \right) \right] \\ & =-0.4865-i0.9549 \\ & \approx -0.5-1.0i \end{align}$ For $k=4$, $\begin{align} & {{z}_{4}}=\sqrt[5]{\sqrt{2}}\left[ \cos \left( \frac{\frac{3\pi }{4}+2\pi \left( 4 \right)}{5} \right)+i\,\sin \left( \frac{\frac{3\pi }{4}+2\pi \left( 4 \right)}{5} \right) \right] \\ & =\sqrt[10]{2}\left[ \cos \left( \frac{35\pi }{20} \right)+i\,\sin \left( \frac{35\pi }{20} \right) \right] \\ & =\sqrt[10]{2}\left[ \cos \left( \frac{7\pi }{4} \right)+i\,\sin \left( \frac{7\pi }{4} \right) \right] \end{align}$ Substitute the values of $\cos \left( \frac{7\pi }{4} \right)$ and $\sin \left( \frac{7\pi }{4} \right)$: $\begin{align} & {{z}_{4}}=\sqrt[10]{2}\left[ 0.7071+i\left( -0.7071 \right) \right] \\ & =0.7579-i0.7579 \\ & \approx 0.8-0.8i \end{align}$
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