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: 68

Answer

$\dfrac{8+12i}{13}$

Work Step by Step

Multiplying by the conjugate of the denominator, the given expression, $ \dfrac{4}{2-3i} ,$ is equivalent to \begin{array}{l}\require{cancel} \dfrac{4}{2-3i}\cdot\dfrac{2+3i}{2+3i} \\\\= \dfrac{4(2)+4(3i)}{(2)^2-(3i)^2} \\\\= \dfrac{8+12i}{4-9i^2} \\\\= \dfrac{8+12i}{4-9(-1)} \\\\= \dfrac{8+12i}{4+9} \\\\= \dfrac{8+12i}{13} .\end{array}
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