Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 4 - Quadratic Functions and Equations - 4-8 Complex Numbers - Practice and Problem-Solving Exercises - Page 253: 32

Answer

$1-\dfrac{3}{2}i$

Work Step by Step

Using $(a+b)^2=a^2+2ab+b^2$ or the square of a binomial, the given expression, $ \dfrac{3+2i}{(1+i)^2} ,$ is equivalent to \begin{align*} & \dfrac{3+2i}{(1)^2+2(1)(i)+(i)^2} \\\\&= \dfrac{3+2i}{1+2i+i^2} \\\\&= \dfrac{3+2i}{1+2i+(-1)} &\text{ (use $i^2=-1$)} \\\\&= \dfrac{3+2i}{2i} .\end{align*} Multiplying both the numerator and the denominator of the expression above by the denominator, then \begin{align*}\require{cancel} & \dfrac{3+2i}{2i}\cdot\dfrac{2i}{2i} \\\\&= \dfrac{3(2i)+2i(2i)}{2i(2i)} &\text{ (use Distributive Property)} \\\\&= \dfrac{6i+4i^2}{4i^2} \\\\&= \dfrac{6i+4(-1)}{4(-1)} &\text{ (use $i^2=-1$)} \\\\&= \dfrac{6i-4}{-4} \\\\&= \dfrac{-\cancel{4}^2+\cancel6^3i}{-\cancel4^2} &\text{ (divide by $2$)} \\\\&= \dfrac{-2+3i}{-2} \\\\&= \dfrac{-2}{-2}+\dfrac{3}{-2}i \\\\&= 1-\dfrac{3}{2}i .\end{align*} Hence, the simplified form of the given expression is $ 1-\dfrac{3}{2}i $.
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