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 - Page 104: 84

Answer

$I=8+3i$

Work Step by Step

Plug in the given values to the formula $E=IZ$. $57+67i=I(9+5i)$ Divide both sides by $(9+5i)$. $\frac{57+67i}{9+5i}=I$ Multiply by top and bottom by the complex conjugate of the denominator. $(\frac{9-5i}{9-5i})\frac{57+67i}{9+5i}=I$ Expand. $\frac{513+603i-285i-335i^2}{81+45i-45i-25i^2}=I$ Combine like terms using the fact that $i^2=-1$. $\frac{848+318i}{106}=8+3i=I$ $I=8+3i$
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