Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 10 - Exponents and Radicals - 10.8 The Complex Numbers - 10.8 Exercise Set: 81

Answer

$\dfrac{6-17i}{25}$

Work Step by Step

Multiplying by the conjugate of the denominator, the given expression, $ \dfrac{3-2i}{4+3i} ,$ is equivalent to \begin{array}{l}\require{cancel} \dfrac{3-2i}{4+3i}\cdot\dfrac{4-3i}{4-3i} \\\\= \dfrac{3(4)+3(-3i)-2i(4)-2i(-3i)}{(4)^2-(3i)^2} \\\\= \dfrac{12-9i-8i+6i^2}{16-9i^2} \\\\= \dfrac{12-9i-8i+6(-1)}{16-9(-1)} \\\\= \dfrac{12-9i-8i-6}{16+9} \\\\= \dfrac{(12-6)+(-9i-8i)}{16+9} \\\\= \dfrac{6-17i}{25} \end{array}
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