Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 1 - Review - Exercises: 73

Answer

$a)$ $\dfrac{6}{5}+\dfrac{8}{5}i$ $b)$ $2+0i$

Work Step by Step

$a)$ $\dfrac{4+2i}{2-i}$ Multiply the numerator and the denominator of this expression by the complex conjugate of the denominator: $\dfrac{4+2i}{2-i}=\dfrac{4+2i}{2-i}\cdot\dfrac{2+i}{2+i}=\dfrac{(4+2i)(2+i)}{2^{2}-i^{2}}=...$ $...=\dfrac{8+4i+4i+2i^{2}}{4-i^{2}}=...$ Substitute $i^{2}$ by $-1$ and simplify: $...=\dfrac{8+4i+4i+2(-1)}{4-(-1)}=\dfrac{8+8i-2}{4+1}=\dfrac{6+8i}{5}=...$ $...=\dfrac{6}{5}+\dfrac{8}{5}i$ $b)$ $(1-\sqrt{-1})(1+\sqrt{-1})$ Evaluate the product: $(1-\sqrt{-1})(1+\sqrt{-1})=1^{2}-(\sqrt{-1})^{2}=1-(-1)=...$ $...=1+1=2+0i$
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