Precalculus (6th Edition) Blitzer

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

Chapter 6 - Section 6.7 - The Dot Product - Exercise Set - Page 799: 76


$7\left( \cos 25{}^\circ +i\sin 25{}^\circ \right),7\left( \cos 205{}^\circ +i\sin 205{}^\circ \right)$

Work Step by Step

Method for finding complex roots: To find $n$ distinct complex roots for any complex number $z=r\left( \cos \theta +i\sin \theta \right)$, if $z\ne 0$, in radians we use the formula given below: ${{z}_{k}}=\sqrt[n]{r}\left[ \cos \left( \frac{\theta +360{}^\circ \cdot k}{n} \right)+i\sin \left( \frac{\theta +360{}^\circ \cdot k}{n} \right) \right]$ Where $k=0,1,2,3,....,n-1$. So, the square roots of $49\left( \cos 50{}^\circ +i\sin 50{}^\circ \right)$ are ${{z}_{k}}=\sqrt[2]{49}\left[ \cos \left( \frac{50{}^\circ +360{}^\circ \cdot k}{2} \right)+i\sin \left( \frac{50{}^\circ +360{}^\circ \cdot k}{2} \right) \right],k=0,1$ Therefore, we can find the two complex square roots in the following manner: $\begin{align} & {{z}_{0}}=\sqrt[2]{49}\left[ \cos \left( \frac{50{}^\circ +360{}^\circ \cdot 0}{2} \right)+i\sin \left( \frac{50{}^\circ +360{}^\circ \cdot 0}{2} \right) \right] \\ & =7\left( \cos \frac{50{}^\circ }{2}+i\sin \frac{50{}^\circ }{2} \right) \\ & =7\left( \cos 25{}^\circ +i\sin 25{}^\circ \right) \end{align}$ $\begin{align} & {{z}_{1}}=\sqrt[2]{49}\left[ \cos \left( \frac{50{}^\circ +360{}^\circ \cdot 1}{2} \right)+i\sin \left( \frac{50{}^\circ +360{}^\circ \cdot 1}{2} \right) \right] \\ & =7\left( \cos \frac{410{}^\circ }{2}+i\sin \frac{410{}^\circ }{2} \right) \\ & =7\left( \cos 205{}^\circ +i\sin 205{}^\circ \right) \end{align}$ So, the square roots are $7\left( \cos 25{}^\circ +i\sin 25{}^\circ \right)\text{ and }7\left( \cos 205{}^\circ +i\sin 205{}^\circ \right)$.
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