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: 59

Answer

$\dfrac{6i}{1-2i}=-\dfrac{12}{5}+\dfrac{6}{5}i$

Work Step by Step

$\dfrac{6i}{1-2i}$ Multiply the numerator and the denominator of this expression by the complex conjugate of the denominator: $\dfrac{6i}{1-2i}=\dfrac{6i}{1-2i}\cdot\dfrac{1+2i}{1+2i}=\dfrac{6i(1+2i)}{1^{2}-(2i)^{2}}=\dfrac{6i+12i^{2}}{1-4i^{2}}=...$ Substitute $i^{2}$ with $-1$ and simplify: $...=\dfrac{6i+12(-1)}{1-4(-1)}=\dfrac{-12+6i}{1+4}=\dfrac{-12+6i}{5}=-\dfrac{12}{5}+\dfrac{6}{5}i$
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