Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 8 - Complex Numbers, Polar Equations, and Parametric Equations - Section 8.4 De Moivre's Theorem: Powers and Roots of Complex Numbers - 8.4 Exercises - Page 376: 25

Answer

The two square roots are: $cos~0^{\circ}+i~sin~0^{\circ}$ $cos~180^{\circ}+i~sin~180^{\circ}$ We can graph the two square roots in the complex plane:
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Work Step by Step

$z = 1 + 0~i$ $z = cos~0^{\circ}+i~sin~0^{\circ}$ $r = 1$ and $\theta = 0^{\circ}$ We can use this equation to find the square roots: $z^{1/n} = r^{1/n}~[cos(\frac{\theta}{n}+\frac{360^{\circ}~k}{n})+i~sin(\frac{\theta}{n}+\frac{360^{\circ}~k}{n})]$, where $k \in \{0, 1, 2,...,n-1\}$ When k = 0: $z^{1/2} = 1^{1/2}~[cos(\frac{0^{\circ}}{2}+\frac{(360^{\circ})(0)}{2})+i~sin(\frac{0^{\circ}}{2}+\frac{(360^{\circ})(0)}{2})]$ $z^{1/2} = 1~[cos~0^{\circ}+i~sin~0^{\circ}]$ $z^{1/2} = cos~0^{\circ}+i~sin~0^{\circ}$ When k = 1: $z^{1/2} = 1^{1/2}~[cos(\frac{0^{\circ}}{2}+\frac{(360^{\circ})(1)}{2})+i~sin(\frac{0^{\circ}}{2}+\frac{(360^{\circ})(1)}{2})]$ $z^{1/2} = 1~[cos~180^{\circ}+i~sin~180^{\circ}]$ $z^{1/2} = cos~180^{\circ}+i~sin~180^{\circ}$ We can graph the two square roots in the complex plane:
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