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: 9

Answer

$z_{0}=1\;\;\;\;\;\;,z_{1}=i\;\;\;\;\;\;\;z_{2}=-1\;\;\;\;\;\;\;\;z_{3}=-i$

Work Step by Step

$1= cos(0)+isin(0)=e^{i(0)}\\\\$ $1=cos(0+2n\pi)+isin(0+2n\pi)=\;e^{i(0+2n\pi)}\\\\$ $1^{\frac{1}{4}}\;=\;e^{i\frac{2n\pi}{4}}\\\\$ $\Rightarrow 1^{\frac{1}{4}}=\;e^{i(\frac{2n\pi}{4})}\;=\;cos(\frac{2n\pi}{4})+isin(\frac{2n\pi}{4})\\\\$ At: $\;\;\;n=0;$ $1^{\frac{1}{4}}=\;e^{0}\;=\;1\\\\$ At; $\;\;\;n=1;$ $1^{\frac{1}{4}}=\;e^{i(\frac{2\pi}{4})}\;=\;cos(\frac{2\pi}{4})+isin(\frac{2\pi}{4})=i\\\\$ At; $\;\;\;\;\;n=2;$ $1^{\frac{1}{4}}=\;e^{i(\frac{4\pi}{4})}\;=\;cos(\frac{4\pi}{4})+isin(\frac{4\pi}{4})=-1\\\\$ At; $\;\;\;\;n=3;$ $1^{\frac{1}{4}}=\;e^{i(\frac{6\pi}{4})}\;=\;cos(\frac{6\pi}{4})+isin(\frac{6\pi}{4})=-i\\\\$
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