Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

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 384: 53

Answer

The fifth roots of 1 are: 1 $0.309+0.951~i$ $-0.809+0.588~i$ $-0.809-0.588~i$ $0.309-0.951~i$

Work Step by Step

We can find the fifth roots of 1: $\theta = 0^{\circ}$: $1^{1/5} = cos~\theta^{\circ}+i~sin~\theta^{\circ}$ $1^{1/5} = cos~0^{\circ}+i~sin~0^{\circ}$ $1^{1/5} = 1+0i$ $\theta = 72^{\circ}$: $1^{1/5} = cos~\theta^{\circ}+i~sin~\theta^{\circ}$ $1^{1/5} = cos~72^{\circ}+i~sin~72^{\circ}$ $1^{1/5} = 0.309+0.951~i$ $\theta = 144^{\circ}$: $1^{1/5} = cos~\theta^{\circ}+i~sin~\theta^{\circ}$ $1^{1/5} = cos~144^{\circ}+i~sin~144^{\circ}$ $1^{1/5} = -0.809+0.588~i$ $\theta = 216^{\circ}$: $1^{1/5} = cos~\theta^{\circ}+i~sin~\theta^{\circ}$ $1^{1/5} = cos~216^{\circ}+i~sin~216^{\circ}$ $1^{1/5} = -0.809-0.588~i$ $\theta = 288^{\circ}$: $1^{1/5} = cos~\theta^{\circ}+i~sin~\theta^{\circ}$ $1^{1/5} = cos~288^{\circ}+i~sin~288^{\circ}$ $1^{1/5} = 0.309-0.951~i$
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