## Precalculus (6th Edition) Blitzer

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