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 - Exercise Set: 58



Work Step by Step

$\dfrac{9}{1-2i}$ Multiply the numerator and the denominator of this expression by the complex conjugate of the denominator: $\dfrac{9}{1-2i}=\dfrac{9}{1-2i}\cdot\dfrac{1+2i}{1+2i}=\dfrac{9(1+2i)}{1^{2}-(2i)^2}=\dfrac{9(1+2i)}{1-4i^{2}}=...$ Substitute $i^{2}$ with $-1$ and simplify: $...=\dfrac{9+18i}{1-4(-1)}=\dfrac{9+18i}{1+4}=\dfrac{9+18i}{5}=\dfrac{9}{5}+\dfrac{18}{5}i$
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.