College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 1 - Equations and Inequalities - Exercise Set 1.4: 28

Answer

$\dfrac{3-4i}{4+3i}=-i$

Work Step by Step

$\dfrac{3-4i}{4+3i}$ Multiply the numerator and the denominator by the complex conjugate of the denominator: $\dfrac{3-4i}{4+3i}=\dfrac{3-4i}{4+3i}\cdot\dfrac{4-3i}{4-3i}=\dfrac{(3-4i)(4-3i)}{4^{2}-(3i)^{2}}=...$ $...=\dfrac{12-9i-16i+12i^{2}}{16-9i^{2}}=\dfrac{12-25i+12i^{2}}{16-9i^{2}}=...$ Substitute $i^{2}$ by $-1$ and simplify: $...=\dfrac{12-25i+12(-1)}{16-9(-1)}=\dfrac{12-12-25i}{16+9}=\dfrac{-25i}{25}=-i$
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.