Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 10 - Section 10.7 - Complex Numbers - Practice - Page 738: 20



Work Step by Step

$i^{50}=i^{48}\times i^{2}=(i^{4})^{12}\times i^{2}$ As $i^{4}=1$ and $i^{2}=-1$, $(i^{4})^{12}\times i^{2}=(1)^{12}\times (-1)=1\times-1=-1$. Therefore, $i^{50}=-1$.
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