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: 39

Answer

$\dfrac{2-3i}{1-2i}=\dfrac{8}{5}+\dfrac{1}{5}i$

Work Step by Step

$\dfrac{2-3i}{1-2i}$ Multiply the fraction by $\dfrac{1+2i}{1+2i}$: $\Big(\dfrac{2-3i}{1-2i}\Big)\Big(\dfrac{1+2i}{1+2i}\Big)=\dfrac{2+4i-3i-6i^{2}}{1-(2i)^{2}}=\dfrac{2+i-6i^{2}}{1-4i^{2}}=...$ Substitute $i^{2}$ with $-1$: $...=\dfrac{2+i-6(-1)}{1-4(-1)}=\dfrac{8+i}{5}=\dfrac{8}{5}+\dfrac{1}{5}i$
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