Elementary Differential Equations and Boundary Value Problems 9th Edition

Published by Wiley
ISBN 10: 0-47038-334-8
ISBN 13: 978-0-47038-334-6

Chapter 4 - Higher Order Linear Equations - 4.2 Homogenous Equations with Constant Coefficients - Problems - Page 232: 7

Answer

$Z_{0}=\;cos(0)+isin(0)=1\;, \;\;\;\;\; Z_{1}=\;cos(\frac{2\pi}{3})+isin(\frac{2\pi}{3})=-\frac{1}{2}+i\frac{\sqrt{3}}{2}\;,\;\;\;\;\; Z_{2}=\;cos(\frac{4\pi}{3})+isin(\frac{4\pi}{3})=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$

Work Step by Step

$1= cos(0)+isin(0)=e^{i(0)}\\\\$ $1=cos(0+2n\pi)+isin(0+2n\pi)=\;e^{i(0+2n\pi)}\\\\$ $1^{\frac{1}{3}}\;=\;e^{i\frac{2n\pi}{3}}\\\\$ $\;\;\;\;\;\;\;\;\;\;\Rightarrow 1^{\frac{1}{3}}\;=\;e^{i(\frac{2n\pi}{3})}=\;cos(\frac{2n\pi}{3})+isin(\frac{2n\pi}{3})$ At $\;\;\;\;n=0$ $Z_{0}=\;cos(0)+isin(0)=1\\\\$ At $\;\;\;n=1$ $Z_{1}=\;cos(\frac{2\pi}{3})+isin(\frac{2\pi}{3})=-\frac{1}{2}+i\frac{\sqrt{3}}{2}\\\\$ At $\;\;\;\;n=2$ $Z_{2}=\;cos(\frac{4\pi}{3})+isin(\frac{4\pi}{3})=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$
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