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: 58


The simplified form of the expression ${{x}^{2}}-2x+5$ for $x=1-2i$ is $0$.

Work Step by Step

Consider the expression, ${{x}^{2}}-2x+5$ Substitute $x=1-2i$ in the expression ${{x}^{2}}-2x+5$. ${{x}^{2}}-2x+5={{\left( 1-2i \right)}^{2}}-2\left( 1-2i \right)+5$ Use the formula ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ and apply distributive law. $\begin{align} & {{x}^{2}}-2x+5=\left( 1-4i+4{{i}^{2}} \right)-\left( 2-4i \right)+5 \\ & =1-4i+4{{i}^{2}}-2+4i+5 \end{align}$ Replace the value ${{i}^{2}}=-1$. \[{{x}^{2}}-2x+5=1-4i+4\left( -1 \right)-2+4i+5\] Combine the real and imaginary parts. \[\begin{align} & {{x}^{2}}-2x+5=1+5-4-2+4i-4i \\ & =\left( 6-6 \right)+\left( 4-4 \right)i \\ & =0 \end{align}\] Therefore, the simplified form of the expression ${{x}^{2}}-2x+5$ for $x=1-2i$ is $0$.
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