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: 78

Answer

$\dfrac{23+41i}{65}$

Work Step by Step

Multiplying by the conjugate of the denominator, the given expression is equivalent to \begin{array}{l}\require{cancel} \dfrac{5+3i}{7-4i} \\\\= \dfrac{5+3i}{7-4i}\cdot\dfrac{7+4i}{7+4i} \\\\= \dfrac{(5+3i)(7+4i)}{(7-4i)(7+4i)} .\end{array} Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to\begin{array}{l}\require{cancel} \dfrac{(5+3i)(7+4i)}{(7-4i)(7+4i)} \\\\= \dfrac{5(7)+5(4i)+3i(7)+3i(4i)}{(7-4i)(7+4i)} \\\\= \dfrac{35+20i+21i+12i^2}{(7-4i)(7+4i)} \\\\= \dfrac{35+41i+12i^2}{(7-4i)(7+4i)} .\end{array} Using the product of the sum and difference of like terms which is given by $(a+b)(a-b)=a^2-b^2,$ the expression above is equivalent \begin{array}{l}\require{cancel} \dfrac{35+41i+12i^2}{(7)^2-(4i)^2} \\\\= \dfrac{35+41i+12i^2}{49-16i^2} .\end{array} Using $i^2=-1$ and then combining like terms, the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{35+41i+12i^2}{49-16i^2} \\\\= \dfrac{35+41i+12(-1)}{49-16(-1)} \\\\= \dfrac{35+41i-12}{49+16} \\\\= \dfrac{23+41i}{65} .\end{array}
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