Complex Functions using polar form

Question

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 $\theta_1 = Arg(Z_1) , \theta_2 = Arg(Z_2)$

Answer 1

HINT:

Let $z_1=r_1e^{i\theta_1}$ and $z_2=r_2e^{i\theta_2}$. Then, $\text{Re}(z_1\bar z_2)=r_1r_2\cos(\theta_1-\theta_2)$.

Answer 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 $\arg(z_1) - \arg(z_2) = 2\,n\,\pi$.