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: 65

Answer

$\dfrac{16+15i}{-3i}=-5+\dfrac{16}{3}i$

Work Step by Step

$\dfrac{16+15i}{-3i}$ Multiply the numerator and the denominator of this expression by the complex conjugate of the denominator: $\dfrac{16+15i}{-3i}=\dfrac{16+15i}{-3i}\cdot\dfrac{3i}{3i}=\dfrac{3i(16+15i)}{-9i^{2}}=\dfrac{3(16i+15i^{2})}{-9i^{2}}=...$ Substitute $i^{2}$ with $-1$ and simplify: $...=\dfrac{3[16i+15(-1)]}{-9(-1)}=\dfrac{3(16i-15)}{9}=\dfrac{-15+16i}{3}=...$ $...=-\dfrac{15}{3}+\dfrac{16}{3}i=-5+\dfrac{16}{3}i$
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