College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 1 - Section 1.3 - Complex Numbers - 1.3 Exercises: 83

Answer

$Z=12+8i$

Work Step by Step

Plug in the given values to the formula $E=IZ$. $88+128i=(10+4i)Z$ Divide both sides by $(10+4i)$. $\frac{88+128i}{10+4i}=Z$ Reduce. $\frac{44+64i}{5+2i}=Z$ Multiply by top and bottom by the complex conjugate of the denominator. $(\frac{5-2i}{5-2i})\frac{44+64i}{5+2i}=Z$ Expand. $\frac{220+320i-88i-128i^2}{25+10i-10i-4i^2}=Z$ Combine like terms using the fact that $i^2=-1$. $\frac{348+232i}{29}=12+8i=Z$ $Z=12+8i$
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.