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