Answer
The standard form of the expression $\left( 2-3i \right)\left( 1-i \right)-\left( 3-i \right)\left( 3+i \right)$ is $-11-5i$.
Work Step by Step
Consider the expression,$\left( 2-3i \right)\left( 1-i \right)-\left( 3-i \right)\left( 3+i \right)$
Use the FOIL method.
$\begin{align}
& \left( 2-3i \right)\left( 1-i \right)-\left( 3-i \right)\left( 3+i \right)=\left( 2-2i-3i+3{{i}^{2}} \right)-\left( 9+3i-3i-{{i}^{2}} \right) \\
& =\left( 2-5i+3{{i}^{2}} \right)-\left( 9-{{i}^{2}} \right)
\end{align}$
Replace the value ${{i}^{2}}=-1$.
$\begin{align}
& \left( 2-3i \right)\left( 1-i \right)-\left( 3-i \right)\left( 3+i \right)=\left( 2-5i+3\left( -1 \right) \right)-\left( 9-\left( -1 \right) \right) \\
& =\left( 2-5i-3 \right)-\left( 9+1 \right) \\
& =-1-5i-10 \\
& =-11-5i
\end{align}$
Therefore, the standard form of the expression $\left( 2-3i \right)\left( 1-i \right)-\left( 3-i \right)\left( 3+i \right)$ is $-11-5i$.
