## Precalculus (6th Edition) Blitzer

The standard form of $\left( 1+i \right)\left( 4-3i \right)$ is $7+i$.
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 $7+i$.