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: 78

Answer

$\dfrac{-3+4i}{2-i}=-2+i$

Work Step by Step

$\dfrac{-3+4i}{2-i}$ Begin the evaluation of the quotient by multiplying the numerator and the denominator by the complex conjugate of the denominator: $\dfrac{-3+4i}{2-i}\cdot\dfrac{2+i}{2+i}=\dfrac{(-3+4i)(2+i)}{2^{2}-i^{2}}=...$ Evaluate the operations indicated in the numerator and in the denominator: $...=\dfrac{-6-3i+8i+4i^{2}}{4-i^{2}}=...$ Substitute $i^{2}$ by $−1$ and simplify: $...=\dfrac{-6-3i+8i+4(-1)}{4-(-1)}=\dfrac{-6+5i-4}{4+1}=\dfrac{-10+5i}{5}=...$ $...=-\dfrac{10}{5}+\dfrac{5}{5}i=-2+i$
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