Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 7 - Section 7.7 - Complex Numbers - Exercise Set: 81

Answer

$-\dfrac{5}{2}-4i$

Work Step by Step

Multiply both the numerator and the denominator by the conjugate of the denominator, which is $-2i$, to obtain: $=\dfrac{(8-5i)(-2i)}{2i(-2i)} \\=\dfrac{-16i+10i^2}{-4i^2}$ Use the fact that $i^2=-1$ to obtain: $=\dfrac{-16i+10(-1)}{-4(-1)} \\=\dfrac{-16i-10}{4} \\=-\dfrac{-10-16i}{4} \\=-\dfrac{10}{4} - \dfrac{16}{4}i \\=-\dfrac{5}{2}-4i$
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