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: 40



Work Step by Step

We are given the expression $(6-3i)^{2}$. Found on page 285, we know that the square of a binomial $(a-b)^{2}$ simplifies to $a^{2}-2ab+b^{2}$. Therefore, $(6-3i)^{2}=(6)^{2}-2(6)(3i)+(3i)^{2}=36-36i+9i^{2}=36-36i+(9\times-1)=36-36i-9=27-36i$ We know that $i^{2}=-1$.
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