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: 66

Answer

The power of the complex numbers in the polar form are $5\left[ \cos {{285}^{{}^\circ }}+i\sin {{285}^{{}^\circ }} \right]\ \ \ \text{ and }\ \ 5\left[ \cos {{105}^{{}^\circ }}+i\sin {{105}^{{}^\circ }} \right]\ $.

Work Step by Step

Here $n=2$ $z=25\left( \cos {{210}^{{}^\circ }}+i\sin {{210}^{{}^\circ }} \right)$ Therefore $\begin{align} & {{z}_{1}}=\sqrt[2]{25}\left[ \cos \left( \frac{{{210}^{{}^\circ }}+{{360}^{{}^\circ }}}{2} \right)+i\sin \left( \frac{{{210}^{{}^\circ }}+{{360}^{{}^\circ }}}{2} \right) \right]\ \ \ \\ & {{z}_{1}}=5\left[ \cos \left( \frac{{{570}^{{}^\circ }}}{2} \right)+i\sin \left( \frac{{{570}^{{}^\circ }}}{2} \right) \right]\ \ \ \\ \end{align}$ \ Similarly, for the other root, $\begin{align} & {{z}_{0}}=\sqrt[2]{25}\left[ \cos \left( \frac{{{210}^{{}^\circ }}+{{360}^{{}^\circ }}\times 0}{2} \right)+i\sin \left( \frac{{{210}^{{}^\circ }}+{{360}^{{}^\circ }}\times 0}{2} \right) \right]\ \ \ \\ & {{z}_{0}}=5\left[ \cos {{105}^{{}^\circ }}+i\sin {{105}^{{}^\circ }} \right]\ \ \ \\ \end{align}$ The square roots of the complex numbers in the polar form is $5\left[ \cos \left( \frac{{{570}^{{}^\circ }}}{2} \right)+i\sin \left( \frac{{{570}^{{}^\circ }}}{2} \right) \right]\ \ \ $ and $5\left[ \cos {{105}^{{}^\circ }}+i\sin {{105}^{{}^\circ }} \right]\ \ \ $
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