Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 10 - Section 10.7 - Complex Numbers - Exercise Set: 61

Answer

$\dfrac{3+5i}{1+i}=4+i$

Work Step by Step

$\dfrac{3+5i}{1+i}$ Multiply the numerator and the denominator of this expression by the complex conjugate of the denominator: $\dfrac{3+5i}{1+i}=\dfrac{3+5i}{1+i}\cdot\dfrac{1-i}{1-i}=\dfrac{(3+5i)(1-i)}{1^{2}-i^{2}}=...$ $...=\dfrac{3-3i+5i-5i^{2}}{1-i^{2}}=...$ Substitute $i^{2}$ by $-1$ and simplify: $...=\dfrac{3-3i+5i-5(-1)}{1-(-1)}=\dfrac{3+2i+5}{1+1}=\dfrac{8+2i}{2}=4+i$
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.