Complex Functions using polar form Given that Z1.Z2 $\ne$ 0
use the polar form to prove that
Re($Z_1\bar Z_2$)= |$Z_1$||$Z_2$| iif $\theta_1 - \theta_2 = 2n\pi$ 
, n = ±1,±2,...,±n
and $\theta_1 = Arg(Z_1) ,  \theta_2 = Arg(Z_2)$

Hint: given that $\operatorname{Re} z = \frac{1}{2}(z+\bar z)$, $\operatorname{Im} z = \frac{1}{2i}(z-\bar z)$ and $|z|^2=z \bar z\,$:
$$
\begin{align}
\operatorname{Re}(z_1\bar z_2) = |z_1| |z_2| \;\;& \iff\;\;(z_1 \bar z_2+\bar z_1 z_2)^2 = 4\, z_1 \bar z_1 z_2 \bar z_2 \\
 & \iff\;\; (z_1 \bar z_2-\bar z_1 z_2)^2 = 0 \\
 & \iff\;\;  \operatorname{Im}(z_1 \bar z_2) = 0 \\
 & \iff\;\; \arg(z_1 \bar z_2) = \arg(z_1) - \arg(z_2) = k\,\pi
\end{align}
$$
Since $\operatorname{Re}(z_1\bar z_2) = |z_1| |z_2| \gt 0$ it follows that $\arg(z_1) - \arg(z_2) = 2\,n\,\pi$.
