Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 7 - Section 7.7 - Complex Numbers - 7.7 Exercises - Page 490: 12

Answer

-1

Work Step by Step

By the definition of $i$ shown on Page 485, we know that $\sqrt -1=i$ because $i^{2}=-1$. Therefore, $(-i)^{2}=(-i\times-i)=(i\times i\times-1\times-1)=(i^{2}\times-1\times-1)=(-1\times-1\times-1)=-1$.
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