Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 2 - Section 2.1 - Complex Numbers - Exercise Set - Page 315: 80


The statement ${{x}^{2}}+{{y}^{2}}=\left( x+yi \right)\left( x-yi \right)$ is true.

Work Step by Step

Consider the expression, $\left( x+yi \right)\left( x-yi \right)$ Use the distributive law $a\left( b+c \right)=ab+ac$. $\begin{align} & \left( x+yi \right)\left( x-yi \right)={{x}^{2}}+x\left( -yi \right)+yi\cdot x+yi\left( -yi \right) \\ & ={{x}^{2}}-xyi+xyi-{{y}^{2}}{{i}^{2}} \\ & ={{x}^{2}}-{{y}^{2}}{{i}^{2}} \end{align}$ Replace the value ${{i}^{2}}=-1$. $\begin{align} & \left( x+yi \right)\left( x-yi \right)={{x}^{2}}-{{y}^{2}}\left( -1 \right) \\ & ={{x}^{2}}+{{y}^{2}} \end{align}$ Therefore, the statement ${{x}^{2}}+{{y}^{2}}=\left( x+yi \right)\left( x-yi \right)$ is true.
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