Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 7 - Section 7.7 - Complex Numbers - Exercise Set: 39



Work Step by Step

We are given the expression $(4-2i)^{2}$. Found on page 285, we know that the square of a binomial $(a-b)^{2}$ simplifies to $a^{2}-2ab+b^{2}$. Therefore, $(4-2i)^{2}=(4)^{2}-2(4)(2i)+(2i)^{2}=16-16i+4i^{2}=16-16i+(4\times-1)=16-16i-4=12-16i$ We know that $i^{2}=-1$.
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