Proof. Let $z_{1}=x_{1}+i y_{1}$ and $z_{2}=x_{2}+i y_{2}$ with $x_{1}, x_{2}, y_{1}, y_{2} \in \mathbb{R}$. Thus

$$

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

$$

Since $z_{1} z_{2}$ is real and non-zero, $z_{1} \neq 0, z_{2} \neq 0$, and

$$

x_{1} x_{2}-y_{1} y_{2} \neq 0 \quad \text { and } \quad x_{1} y_{2}+y_{1} x_{2}=0

$$

Thus, since $z_{2} \neq 0$, then

$$

\begin{aligned}

\frac{z_{1}}{z_{2}} &=\frac{x_{1}+i y_{1}}{x_{2}-i y_{2}} \cdot \frac{x_{2}+i y_{2}}{x_{2}+i y_{2}} \\

&=\frac{x_{1} x_{2}-y_{1} y_{2}+\left(x_{1} y_{2}+y_{1} x_{2}\right) i}{x_{2}^{2}+y_{2}^{2}} \\

&=\frac{x_{1} x_{2}-y_{1} y_{2}}{x_{2}^{2}+y_{2}^{2}}

\end{aligned}

$$

By setting $r=\frac{x_{1} x_{2}-y_{1} y_{2}}{x_{2}^{2}+y_{2}^{2}}$, we have the result.