Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 7 - Section 7.7 - Complex Numbers - Exercise Set: 70

Answer

$-\dfrac{15}{13} + \dfrac{10}{13}i$

Work Step by Step

Multiply both the numerator and the denominator by the conjugate of the denominator, which is $2+3i$, to obtain: $=\dfrac{5i(2+3i)}{(2-3i)(2+3i)} \\=\dfrac{10i+15i^2}{(2-3i)(2+3i)}$ Simplify using the rule $(a-b)(a+b)=a^2-b^2$ to obtain: $=\dfrac{10i+15i^2}{2^2-(3i)^2} \\=\dfrac{10i+15i^2}{4-9i^2}$ Use the fact that $i^2=-1$ to obtain: $=\dfrac{10i+15(-1)}{4-9(-1)} \\=\dfrac{10i-15}{4+9} \\=\dfrac{-15+10i}{13} \\=-\dfrac{15}{13} + \dfrac{10}{13}i$
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.