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 769: 100

Answer

See the explanation below.

Work Step by Step

The power of a complex number provided in polar form can be found by the use of DeMoivre’s Theorem. According to this theorem, the ${{n}^{th}}$ power of a complex number $z$ can be found by providing the power $n$ to the modulus and multiplying the argument by $n$. Let $z=r\left( \cos \theta +i\sin \theta \right)$ be a complex number in polar form. If $n$ is a positive integer, then $z$ to the ${{n}^{th}}$ power, ${{z}^{n}}$ is, ${{z}^{n}}={{r}^{n}}\left[ \cos \left( n\theta \right)+i\sin \left( n\theta \right) \right]$. Example: The above explanation can be justified with the help of an example. Let $z=5\left( \cos 30{}^\circ +i\sin 30{}^\circ \right)$ be a complex number in polar form. To find the ${{3}^{rd}}$ power of this complex number, raise the modulus to the power $3$ and multiply the argument with $3$ in the provided complex number. $\begin{align} & {{z}^{3}}={{5}^{3}}\left( \cos \left( 30{}^\circ \times 3 \right)+i\sin \left( 30{}^\circ \times 3 \right) \right) \\ & =125\left( \cos 90{}^\circ +i\sin 90{}^\circ \right) \end{align}$ Above is the ${{3}^{rd}}$ power of the provided complex number which is found with the help of DeMoivre’s Theorem.
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