Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 1 - Review - Exercises - Page 134: 74

Answer

$a)$ $\dfrac{41}{25}-\dfrac{12}{25}i$ $b)$ $20+0i$

Work Step by Step

$a)$ $\dfrac{8+3i}{4+3i}$ Multiply the numerator and the denominator of this expression by the complex conjugate of the denominator: $\dfrac{8+3i}{4+3i}=\dfrac{8+3i}{4+3i}\cdot\dfrac{4-3i}{4-3i}=\dfrac{(8+3i)(4-3i)}{4^{2}-(3i)^{2}}=...$ $...=\dfrac{32-24i+12i-9i^{2}}{16-9i^{2}}=...$ Substitute $i^{2}$ by $-1$ and simplify: $...=\dfrac{32-24i+12i-9(-1)}{16-9(-1)}=\dfrac{32-12i+9}{16+9}=\dfrac{41-12i}{25}=...$ $...=\dfrac{41}{25}-\dfrac{12}{25}i$ $b)$ $\sqrt{-10}\cdot\sqrt{-40}$ Evaluate the product and simplify: $\sqrt{-10}\cdot\sqrt{-40}=\sqrt{(-10)(-40)}=\sqrt{400}=20+0i$
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