# Chapter 10 - Exponents and Radicals - 10.8 The Complex Numbers - 10.8 Exercise Set: 80

$\dfrac{18-i}{25}$

#### Work Step by Step

Multiplying by the conjugate of the denominator, the given expression, $\dfrac{3+2i}{4+3i} ,$ is equivalent to \begin{array}{l}\require{cancel} \dfrac{3+2i}{4+3i}\cdot\dfrac{4-3i}{4-3i} \\\\= \dfrac{3(4)+3(-3i)+2i(4)+2i(-3i)}{(4)^2-(3i)^2} \\\\= \dfrac{12-9i+8i-6i^2}{16-9i^2} \\\\= \dfrac{12-9i+8i-6(-1)}{16-9(-1)} \\\\= \dfrac{12-9i+8i+6}{16+9} \\\\= \dfrac{(12+6)+(-9i+8i)}{16+9} \\\\= \dfrac{18-i}{25} \end{array}

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.