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: 80

Answer

$\dfrac{18-i}{25}$

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{18-i}{25} \end{array}
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