Answer
The standard form of $\left( 1+i \right)\left( 4-3i \right)$ is $7+i$.
Work Step by Step
Consider the expression,
$\left( 1+i \right)\left( 4-3i \right)$
Apply FOIL method on the expression $\left( 1+i \right)\left( 4-3i \right)$.
$\begin{align}
& \left( 1+i \right)\left( 4-3i \right)=1\left( 4 \right)+1\left( -3i \right)+i\left( 4 \right)+i\left( -3i \right) \\
& =4-3i+4i-3{{i}^{2}}
\end{align}$
As ${{i}^{2}}=-1$
Therefore,
$\begin{align}
& 4-3i+4i-3{{i}^{2}}=4-3i+4i-3\left( -1 \right) \\
& =4-3i+4i+3
\end{align}$
Combine the real part and imaginary part separately and either add or subtract as required.
$\begin{align}
& 4-3i+4i+3=\left( 4+3 \right)+\left( -3i+4i \right) \\
& =7+\left( -3+4 \right)i \\
& =7+i
\end{align}$
Thus,
$\left( 1+i \right)\left( 4-3i \right)=7+i$
Hence, the standard form of the expression $\left( 1+i \right)\left( 4-3i \right)$ is $7+i$.
