# For any two complex numbers z1 and z2, prove that

Question:

For any two complex numbers $z_{1}$ and $z_{2}$, prove that

$\operatorname{Re}\left(z_{1} z_{2}\right)=\operatorname{Re} z_{1} \operatorname{Re} z_{2}-\operatorname{Im} z_{1} \mid m z_{2}$

Solution:

Let $z_{1}=x_{1}+i y_{1}$ and $z_{2}=x_{2}+i y_{2}$

$\therefore z_{1} z_{2}=\left(x_{1}+i y_{1}\right)\left(x_{2}+i y_{2}\right)$

$=x_{1}\left(x_{2}+i y_{2}\right)+i y_{1}\left(x_{2}+i y_{2}\right)$

$=x_{1} x_{2}+i x_{1} y_{2}+i y_{1} x_{2}+i^{2} y_{1} y_{2}$

$=x_{1} x_{2}+i x_{1} y_{2}+i y_{1} x_{2}-y_{1} y_{2}$  $\left[i^{2}=-1\right]$

$=\left(x_{1} x_{2}-y_{1} y_{2}\right)+i\left(x_{1} y_{2}+y_{1} x_{2}\right)$

$\Rightarrow \operatorname{Re}\left(z_{1} z_{2}\right)=x_{1} x_{2}-y_{1} y_{2}$

$\Rightarrow \operatorname{Re}\left(z_{1} z_{2}\right)=\operatorname{Re} z_{1} \operatorname{Re} z_{2}-\operatorname{Im} z_{1} \operatorname{Im} z_{2}$

Hence, proved.