Answer
The standard form of the expression $\left( 8+9i \right)\left( 2-i \right)-\left( 1-i \right)\left( 1+i \right)$ is $23+10i$.
Work Step by Step
Consider the expression,$\left( 8+9i \right)\left( 2-i \right)-\left( 1-i \right)\left( 1+i \right)$
Use the FOIL method.
$\begin{align}
& \left( 8+9i \right)\left( 2-i \right)-\left( 1-i \right)\left( 1+i \right)=\left( 16-8i+18i-9{{i}^{2}} \right)-\left( 1+i-i-{{i}^{2}} \right) \\
& =\left( 16+10i-9{{i}^{2}} \right)-\left( 1-{{i}^{2}} \right)
\end{align}$
Replace the value ${{i}^{2}}=-1$.
$\begin{align}
& \left( 8+9i \right)\left( 2-i \right)-\left( 1-i \right)\left( 1+i \right)=\left( 16+10i-9{{i}^{2}} \right)-\left( 1-{{i}^{2}} \right) \\
& =\left( 16+10i-9\left( -1 \right) \right)-\left( 1-\left( -1 \right) \right) \\
& =25+10i-2 \\
& =23+10i
\end{align}$
Therefore, the standard form of the expression $\left( 8+9i \right)\left( 2-i \right)-\left( 1-i \right)\left( 1+i \right)$ is $23+10i$.
