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 - Page 64: 46



Work Step by Step

$\dfrac{(1+2i)(3-i)}{2+i}$ Evaluate the product in the numerator: $\dfrac{(1+2i)(3-i)}{2+i}=\dfrac{3-i+6i-2i^{2}}{2+i}=\dfrac{3+5i-2i^{2}}{2+i}=...$ Substitute $i^{2}$ with $-1$: $...=\dfrac{3+5i-2(-1)}{2+i}=\dfrac{5+5i}{2+i}=...$ Multiply the fraction by $\dfrac{2-i}{2-i}$: $...=\Big(\dfrac{5+5i}{2+i}\Big)\Big(\dfrac{2-i}{2-i}\Big)=\dfrac{10-5i+10i-5i^{2}}{4-i^{2}}=\dfrac{10+5i-5i^{2}}{4-i^{2}}$ Once again, substitute $i^{2}$ with $-1$: $...=\dfrac{10+5i-5(-1)}{4-(-1)}=\dfrac{15+5i}{5}=3+i$
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