Answer
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 $0$.
