Answer
The standard form of the expression ${{\left( 4-i \right)}^{2}}-{{\left( 1+2i \right)}^{2}}$ is $18-12i$.
Work Step by Step
Consider the expression,${{\left( 4-i \right)}^{2}}-{{\left( 1+2i \right)}^{2}}$
Use the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ and ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$.
\[\begin{align}
& {{\left( 4-i \right)}^{2}}-{{\left( 1+2i \right)}^{2}}=\left( {{4}^{2}}-2\cdot 4\cdot i+{{i}^{2}} \right)-\left( {{1}^{2}}+2\cdot 1\cdot 2i+{{\left( 2i \right)}^{2}} \right) \\
& =\left( 16-8i+{{i}^{2}} \right)-\left( 1+4i+4{{i}^{2}} \right) \\
& =16-8i+{{i}^{2}}-1-4i-4{{i}^{2}}
\end{align}\]
Combine real and imaginary terms.
\[\begin{align}
& {{\left( 4-i \right)}^{2}}-{{\left( 1+2i \right)}^{2}}=16-1-8i-4i+{{i}^{2}}-4{{i}^{2}} \\
& =15-12i-3{{i}^{2}}
\end{align}\]
Replace the value ${{i}^{2}}=-1$.
\[\begin{align}
& {{\left( 4-i \right)}^{2}}-{{\left( 1+2i \right)}^{2}}=15-12i-3\left( -1 \right) \\
& =15+3-12i \\
& =18-12i
\end{align}\]
Therefore, the standard form of the expression ${{\left( 4-i \right)}^{2}}-{{\left( 1+2i \right)}^{2}}$ is $18-12i$.
