College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 1 - Equations and Inequalities - Exercise Set 1.4 - Page 142: 27



Work Step by Step

$\dfrac{2+3i}{2+i}$ Multiply the numerator and the denominator by the complex conjugate of the denominator: $\dfrac{2+3i}{2+i}=\dfrac{2+3i}{2+i}\cdot\dfrac{2-i}{2-i}=\dfrac{(2+3i)(2-i)}{2^{2}-i^{2}}=...$ $...=\dfrac{4-2i+6i-3i^{2}}{4-i^{2}}=\dfrac{4+4i-3i^{2}}{4-i^{2}}=...$ Substitute $i^{2}$ by $-1$ and simplify: $...=\dfrac{4+4i-3(-1)}{4-(-1)}=\dfrac{4+3+4i}{4+1}=\dfrac{7+4i}{5}=\dfrac{7}{5}+\dfrac{4}{5}i$
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