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: 71


$2$ ; $0.6+1.9i$ ; $-1.6+1.2i$ ; $-1.6-1.2i$ ; $0.6-1.9i$

Work Step by Step

Roots of a complex number represented in polar form can be found by the application of DeMoivre’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 $32$ as per DeMoivre’s theorem. First write the provided number in rectangular form as follows: $32=32+0i$ Convert the above expression into polar form as follows: $\begin{align} & 32=r\left( \cos \,\theta +i\,\sin \,\theta \right) \\ & =32\left( \cos \,0+i\,\sin \,0 \right) \end{align}$ Apply De Moivre’s theorem to find all fifth roots of $32$. ${{z}_{k}}=\sqrt[5]{32}\left[ \cos \left( \frac{0+2\pi k}{5} \right)+i\,\sin \left( \frac{0+2\pi k}{5} \right) \right]$ Here, $k=0,1,2,3,4$. Insert the above values of $k$ to find five distinct fifth roots of the provided number. For $k=0$, $\begin{align} & {{z}_{0}}=\sqrt[5]{32}\left[ \cos \left( \frac{0+2\pi \left( 0 \right)}{5} \right)+i\,\sin \left( \frac{0+2\pi \left( 0 \right)}{5} \right) \right] \\ & =2\left[ \cos \,0+i\,\sin \,0 \right] \\ & =2 \end{align}$ For $k=1$, $\begin{align} & {{z}_{1}}=\sqrt[5]{32}\left[ \cos \left( \frac{0+2\pi \left( 1 \right)}{5} \right)+i\,\sin \left( \frac{0+2\pi \left( 1 \right)}{5} \right) \right] \\ & =2\left[ \cos \left( \frac{2\pi }{5} \right)+i\,\sin \left( \frac{2\pi }{5} \right) \right] \end{align}$ Substitute the values of $\cos \left( \frac{2\pi }{5} \right)$ and $\sin \left( \frac{2\pi }{5} \right)$: $\begin{align} & {{z}_{1}}=2\left[ 0.3090+i0.9511 \right] \\ & =0.6180+i1.9021 \\ & \approx 0.6+1.9i \end{align}$ For $k=2$, $\begin{align} & {{z}_{2}}=\sqrt[5]{32}\left[ \cos \left( \frac{0+2\pi \left( 2 \right)}{5} \right)+i\,\sin \left( \frac{0+2\pi \left( 2 \right)}{5} \right) \right] \\ & =2\left[ \cos \left( \frac{4\pi }{5} \right)+i\,\sin \left( \frac{4\pi }{5} \right) \right] \end{align}$ Substitute the values of $\cos \left( \frac{4\pi }{5} \right)$ and $\sin \left( \frac{4\pi }{5} \right)$: $\begin{align} & {{z}_{2}}=2\left[ -0.8090+i0.5877 \right] \\ & =-1.618+i1.1755 \\ & \approx -1.6+1.2i \end{align}$ For $k=3$, $\begin{align} & {{z}_{3}}=\sqrt[5]{32}\left[ \cos \left( \frac{0+2\pi \left( 3 \right)}{5} \right)+i\,\sin \left( \frac{0+2\pi \left( 3 \right)}{5} \right) \right] \\ & =2\left[ \cos \left( \frac{6\pi }{5} \right)+i\,\sin \left( \frac{6\pi }{5} \right) \right] \end{align}$ Substitute the values of $\cos \left( \frac{6\pi }{5} \right)$ and $\sin \left( \frac{6\pi }{5} \right)$: $\begin{align} & {{z}_{3}}=2\left[ -0.8090+i\left( -0.5878 \right) \right] \\ & =-1.6180-i1.1756 \\ & \approx -1.6-1.2i \end{align}$ For $k=4$, $\begin{align} & {{z}_{4}}=\sqrt[5]{32}\left[ \cos \left( \frac{0+2\pi \left( 4 \right)}{5} \right)+i\,\sin \left( \frac{0+2\pi \left( 4 \right)}{5} \right) \right] \\ & =2\left[ \cos \left( \frac{8\pi }{5} \right)+i\,\sin \left( \frac{8\pi }{5} \right) \right] \end{align}$ Substitute the values of $\cos \left( \frac{8\pi }{5} \right)$ and $\sin \left( \frac{8\pi }{5} \right)$: $\begin{align} & {{z}_{4}}=2\left[ 0.3090+i\left( -0.9511 \right) \right] \\ & =0.6180-i1.9021 \\ & \approx 0.6-1.9i \end{align}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.