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: 77

Answer

$\dfrac{-23+43i}{58}$

Work Step by Step

Multiplying by the conjugate of the denominator, the given expression, $ \dfrac{4+5i}{3-7i} ,$ is equivalent to \begin{array}{l}\require{cancel} \dfrac{4+5i}{3-7i}\cdot\dfrac{3+7i}{3+7i} \\\\= \dfrac{4(3)+4(7i)+5i(3)+5i(7i)}{(3)^2-(7i)^2} \\\\= \dfrac{12+28i+15i+35i^2}{9-49i^2} \\\\= \dfrac{12+28i+15i+35(-1)}{9-49(-1)} \\\\= \dfrac{12+28i+15i-35}{9+49} \\\\= \dfrac{(12-35)+(28i+15i)}{9+49} \\\\= \dfrac{-23+43i}{58} .\end{array}
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