Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 8 - Radical Functions - Chapter Review Exercises - Page 671: 91



Work Step by Step

$\frac{5+7i}{3-4i}$ =$\frac{5+7i}{3-4i}\times\frac{3+4i}{3+4i}$ =$\frac{(5+7i)(3+4i)}{(3+4i)(3-4i)}$ =$\frac{5(3+4i)+7i(3+4i)}{3^{2}-(4i)^{2}}$ =$\frac{15+20i+21i+28i^{2}}{9-16i^{2}}$ =$\frac{15+41i+28(-1)}{9-16(-1)}$ =$\frac{-13+41i}{25}$ =$\frac{-13}{25}+\frac{41i}{25}$
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.