## Precalculus (6th Edition)

Published by Pearson

# Chapter 1 - Equations and Inequalities - 1.3 Complex Numbers - 1.3 Exercises: 73

#### Answer

$\dfrac{6+2i}{1+2i}=2-2i$

#### Work Step by Step

$\dfrac{6+2i}{1+2i}$ Begin the evaluation of the quotient by multiplying the numerator and the denominator by the complex conjugate of the denominator: $\dfrac{6+2i}{1+2i}\cdot\dfrac{1-2i}{1-2i}=\dfrac{(6+2i)(1-2i)}{1^{2}-(2i)^{2}}=...$ Evaluate the operations indicated in the numerator and in the denominator: $...=\dfrac{6-12i+2i-4i^{2}}{1-4i^{2}}=...$ Substitute $i^{2}$ by $-1$ and simplify: $...=\dfrac{6-12i+2i-4(-1)}{1-4(-1)}=\dfrac{6-10i+4}{1+4}=\dfrac{10-10i}{5}=...$ $...=\dfrac{10}{5}-\dfrac{10}{5}i=2-2i$

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