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: 77

Answer

$\dfrac{1-3i}{1+i}=-1-2i$

Work Step by Step

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