Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 10 - Section 10.7 - Complex Numbers - Exercise Set - Page 740: 69



Work Step by Step

$\dfrac{2-3i}{2+i}$ Multiply the numerator and the denominator of this expression by the complex conjugate of the denominator: $\dfrac{2-3i}{2+i}=\dfrac{2-3i}{2+i}\cdot\dfrac{2-i}{2-i}=\dfrac{(2-3i)(2-i)}{2^{2}-i^{2}}=...$ $...=\dfrac{(2-3i)(2-i)}{4-i^{2}}=\dfrac{4-2i-6i+3i^{2}}{4-i^{2}}=\dfrac{4-8i+3i^{2}}{4-i^{2}}=...$ Substitute $i^{2}$ with $-1$ and simplify: $...=\dfrac{4-8i+3(-1)}{4-(-1)}=\dfrac{4-8i-3}{4+1}=\dfrac{1-8i}{5}=\dfrac{1}{5}-\dfrac{8}{5}i$
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