Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 10 - Exponents and Radicals - 10.8 The Complex Numbers - 10.8 Exercise Set - Page 686: 69

Answer

$\dfrac{3+5i}{17}$

Work Step by Step

Multiplying by the conjugate of the denominator, the given expression, $ \dfrac{2i}{5+3i} ,$ is equivalent to \begin{array}{l}\require{cancel} \dfrac{2i}{5+3i}\cdot\dfrac{5-3i}{5-3i} \\\\= \dfrac{2i(5)+2i(-3i)}{(5)^2-(3i)^2} \\\\= \dfrac{10i-6i^2}{25-9i^2} \\\\= \dfrac{10i-6(-1)}{25-9(-1)} \\\\= \dfrac{10i+6}{25+9} \\\\= \dfrac{6+10i}{34} \\\\= \dfrac{\cancel{2}(3+5i)}{\cancel{2}\cdot 17} \\\\= \dfrac{3+5i}{17} .\end{array}
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.