Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.6 - Complex Numbers - 1.6 Exercises: 41

Answer

$\dfrac{10i}{1-2i}=-4+2i$

Work Step by Step

$\dfrac{10i}{1-2i}$ Multiply the fraction by $\dfrac{1+2i}{1+2i}$: $\Big(\dfrac{10i}{1-2i}\Big)\Big(\dfrac{1+2i}{1+2i}\Big)=\dfrac{10i+20i^{2}}{1-(2i)^{2}}=\dfrac{10i+20i^{2}}{1-4i^{2}}=...$ Substitute $i^{2}$ with $-1$: $...=\dfrac{10i+20(-1)}{1-4(-1)}=\dfrac{10i-20}{5}=-4+2i$
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