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: 57


The simplified form of the expression ${{x}^{2}}-2x+2$ for $x=1+i$ is $0$.

Work Step by Step

Consider the expression, ${{x}^{2}}-2x+2$ Substitute $x=1+i$ in the expression${{x}^{2}}-2x+2$. ${{x}^{2}}-2x+2={{\left( 1+i \right)}^{2}}-2\left( 1+i \right)+2$ Use the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ and apply distributive law. $\begin{align} & {{x}^{2}}-2x+2=\left( 1+2i+{{i}^{2}} \right)-\left( 2+2i \right)+2 \\ & =1+2i+{{i}^{2}}-2-2i+2 \end{align}$ Replace the value ${{i}^{2}}=-1$. ${{x}^{2}}-2x+2=1+2i+\left( -1 \right)-2-2i+2$ Combine the real and imaginary parts. $\begin{align} & {{x}^{2}}-2x+2=\left( 1-1-2+2 \right)+\left( 2-2 \right)i \\ & =0 \end{align}$ Therefore, the simplified form of the expression ${{x}^{2}}-2x+2$ for $x=1+i$ is $0$.
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