Precalculus (6th Edition) Blitzer

The statement ${{x}^{2}}+{{y}^{2}}=\left( x+yi \right)\left( x-yi \right)$ is true.
Consider the expression, $\left( x+yi \right)\left( x-yi \right)$ Use the distributive law $a\left( b+c \right)=ab+ac$. \begin{align} & \left( x+yi \right)\left( x-yi \right)={{x}^{2}}+x\left( -yi \right)+yi\cdot x+yi\left( -yi \right) \\ & ={{x}^{2}}-xyi+xyi-{{y}^{2}}{{i}^{2}} \\ & ={{x}^{2}}-{{y}^{2}}{{i}^{2}} \end{align} Replace the value ${{i}^{2}}=-1$. \begin{align} & \left( x+yi \right)\left( x-yi \right)={{x}^{2}}-{{y}^{2}}\left( -1 \right) \\ & ={{x}^{2}}+{{y}^{2}} \end{align} Therefore, the statement ${{x}^{2}}+{{y}^{2}}=\left( x+yi \right)\left( x-yi \right)$ is true.