Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 1 - Equations and Inequalities - 1.3 Complex Numbers - 1.3 Exercises: 76

Answer

$\dfrac{4-3i}{4+3i}=\dfrac{7}{25}-\dfrac{24}{25}i$

Work Step by Step

$\dfrac{4-3i}{4+3i}$ Begin the evaluation of the quotient by multiplying the numerator and the denominator by the complex conjugate of the denominator: $\dfrac{4-3i}{4+3i}\cdot\dfrac{4-3i}{4-3i}=\dfrac{(4-3i)^{2}}{4^{2}-(3i)^{2}}=...$ Evaluate the operations indicated in the numerator and in the denominator: $...=\dfrac{16-2(4)(3i)+(3i)^{2}}{16-9i^{2}}=\dfrac{16-24i+9i^{2}}{16-9i^{2}}=...$ Substitute $i^{2}$ by $−1$ and simplify: $...=\dfrac{16-24i+9(-1)}{16-9(-1)}=\dfrac{16-24i-9}{16+9}=\dfrac{7-24i}{25}=...$ $...=\dfrac{7}{25}-\dfrac{24}{25}i$
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