Answer
$\frac{8}{1+2i}=\frac{8}{5}-\frac{16}{5}i$
Work Step by Step
$\frac{8}{1+2i}=\frac{8}{1+2i}\frac{1-2i}{1-2i}=\frac{8(1-2i)}{(1+2i)(1-2i)}=\frac{8-16i}{1^2-(2i)^2}=\frac{8-16i}{1-4i^2}=\frac{8-16i}{5}=\frac{8}{5}-\frac{16}{5}i$
Algebra and Trigonometry 10th Edition
by Larson, Ron
Published by
Cengage Learning
ISBN 10:
9781337271172
ISBN 13:
978-1-33727-117-2
Chapter 1 - Chapter Test - Page 156: 18
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