## Precalculus (6th Edition) Blitzer

The standard form of the expression${{\left( 2-\sqrt{-3} \right)}^{2}}$ is $1-4i\sqrt{3}$.
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 $1-4i\sqrt{3}$.