Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 1 - Equations and Inequalities - 1.3 Complex Numbers - 1.3 Exercises - Page 112: 75



Work Step by Step

$\dfrac{2-i}{2+i}$ Begin the evaluation of the quotient by multiplying the numerator and the denominator by the complex conjugate of the denominator: $\dfrac{2-i}{2+i}\cdot\dfrac{2-i}{2-i}=\dfrac{(2-i)^{2}}{2^{2}-i^{2}}=...$ Evaluate the operations indicated in the numerator and in the denominator: $...=\dfrac{4-(2)(2)i+i^{2}}{4-i^{2}}=\dfrac{4-4i+i^{2}}{4-i^{2}}=...$ Substitute $i^{2}$ by $-1$ and simplify: $...=\dfrac{4-4i-1}{4-(-1)}=\dfrac{3-4i}{4+1}=\dfrac{3-4i}{5}=\dfrac{3}{5}-\dfrac{4}{5}i$
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