For any two complex numbers $z_{1}$ and $z_{2},$ prove that $\operatorname{Re}(z_{1} z_{2})=\operatorname{Re} z_{1} \operatorname{Re} z_{2}-\operatorname{Im} z_{1} \operatorname{Im} z_{2}.$

Let $z_{1}=x_{1}+i y_{1}$ and $z_{2}=x_{2}+i y_{2}.$
Then,$z_{1} z_{2}=(x_{1}+i y_{1})(x_{2}+i y_{2}).$
Expanding the product,we get $z_{1} z_{2}=x_{1} x_{2}+i x_{1} y_{2}+i y_{1} x_{2}+i^{2} y_{1} y_{2}.$
Since $i^{2}=-1,$ we have $z_{1} z_{2}=x_{1} x_{2}+i x_{1} y_{2}+i y_{1} x_{2}-y_{1} y_{2}.$
Grouping the real and imaginary parts,$z_{1} z_{2}=(x_{1} x_{2}-y_{1} y_{2})+i(x_{1} y_{2}+y_{1} x_{2}).$
The real part is $\operatorname{Re}(z_{1} z_{2})=x_{1} x_{2}-y_{1} y_{2}.$
Since $\operatorname{Re} z_{1}=x_{1}, \operatorname{Re} z_{2}=x_{2}, \operatorname{Im} z_{1}=y_{1},$ and $\operatorname{Im} z_{2}=y_{2},$
we get $\operatorname{Re}(z_{1} z_{2})=\operatorname{Re} z_{1} \operatorname{Re} z_{2}-\operatorname{Im} z_{1} \operatorname{Im} z_{2}.$
Hence,proved.