College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 1 - Equations and Inequalities - Exercise Set 1.4: 22

Answer

$\dfrac{3}{4+i}=\dfrac{12}{17}-\dfrac{3}{17}i$

Work Step by Step

$\dfrac{3}{4+i}$ Multiply the numerator and the denominator of the fraction by the complex conjugate of the denominator: $\dfrac{3}{4+i}=\dfrac{3}{4+i}\cdot\dfrac{4-i}{4-i}=\dfrac{3(4-i)}{4^{2}-i^{2}}=\dfrac{12-3i}{16-i^{2}}=...$ Substitute $i^{2}$ by $-1$ and simplify: $...=\dfrac{12-3i}{16-(-1)}=\dfrac{12-3i}{16+1}=\dfrac{12-3i}{17}=\dfrac{12}{17}-\dfrac{3}{17}i$
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