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: 79

Answer

$\dfrac{19-4i}{29}$

Work Step by Step

Multiplying by the conjugate of the denominator, the given expression, $ \dfrac{2+3i}{2+5i} ,$ is equivalent to \begin{array}{l}\require{cancel} \dfrac{2+3i}{2+5i}\cdot\dfrac{2-5i}{2-5i} \\\\= \dfrac{2(2)+2(-5i)+3i(2)+3i(-5i)}{(2)^2-(5i)^2} \\\\= \dfrac{4-10i+6i-15i^2}{4-25i^2} \\\\= \dfrac{4-10i+6i-15(-1)}{4-25(-1)} \\\\= \dfrac{4-10i+6i+15}{4+25} \\\\= \dfrac{(4+15)+(-10i+6i)}{4+25} \\\\= \dfrac{19-4i}{29} \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.