College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 1 - Review Exercises: 26

Answer

$7-24i$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To simplify the given expression, $ (4-3i)^2 ,$ use the special product on squaring binomials. Then use the equivalence $i^2=-1.$ Finally, combine combine the real parts and the imaginary parts. $\bf{\text{Solution Details:}}$ Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to \begin{array}{l}\require{cancel} (4)^2-2(4)(3i)+(3i)^2 \\\\= 16-24i+9i^2 .\end{array} Since $i^2=-1,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 16-24i+9(-1) \\\\= 16-24i-9 .\end{array} Combining the real parts and the imaginary parts results to \begin{array}{l}\require{cancel} (16-9)-24i \\\\= 7-24i .\end{array}
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