Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 8 - Section 8.1 - Complex Numbers - 8.1 Problem Set - Page 426: 54



Work Step by Step

Knowing that $i^{2}=-1$, we also know that $i^{4}=(i^{2})^{2}=(-1)^{2}=1$. We can use the fact that $i^{4}=1$ to evaluate higher powers of $i$. $i^{35}=i^{32}\times i^{2}\times i=(i^{4})^{8}\times i^{2}\times i=(1)^{8}\times -1\times i=1\times-1\times i=-i$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.