If you think more geometrically, then it's not terribly complex (sorry for the pun.)
Multiplication of complex numbers is counter-clockwise rotation and scaling. Division is clockwise rotation and inverse scaling.
In polar, you have \$A=r_a e^{I\theta_a}=r_a\angle\theta_a\$ and \$\beta=r_b e^{I\theta_b}=r_b\angle\theta_b\$. Then multiplication just produces \$r_ar_b\:\angle\: \theta_a+\theta_b\$.
So your equation is:
$$\frac{r_a\angle\theta_a}{1+r_ar_b\:\angle\: \theta_a+\theta_b}$$
So long as \$r_ar_b\$ is very much larger than one, this reduces (via clockwise rotation due to division) to:
$$\frac{r_a\angle\theta_a}{r_ar_b\:\angle\: \theta_a+\theta_b}=\frac{r_a}{r_ar_b}\angle \theta_a-\left(\theta_a+\theta_b\right)=\frac1{r_b}\angle -\theta_b=\frac1{r_b\angle \theta_b}=\frac1{\beta}$$
You can keep this in Euler form instead as:
$$\frac{r_a e^{I\theta_a}}{1+r_ar_b e^{I\left(\theta_a+\theta_b\right)}}$$
And again, so long as \$r_ar_b\$ is very much larger than one:
$$\frac{r_a e^{I\theta_a}}{r_ar_b e^{I\left(\theta_a+\theta_b\right)}}=\frac{ e^{I\theta_a}}{r_b e^{I\left(\theta_a+\theta_b\right)}}=\frac1{r_b e^{I\left(\theta_a+\theta_b\right)}e^{-I\theta_a}}=\frac1{r_b e^{I\theta_b}}=\frac1{\beta}$$
If you insist, you can \$A=a +bI\$ and \$\beta=c+dI\$. Then \$\mid A\mid=\sqrt{a^2 +b^2}\$ and \$\mid \beta\mid=\sqrt{c^2 +d^2}\$. Then your absolute value approach yields:
$$\begin{align*} &\frac{\mid A\mid}{\sqrt{1+a^2c^2+a^2d^2+b^2c^2+b^2d^2+2\left(ac-bd\right)}}\\\\&=\frac{\mid A\mid}{\sqrt{1+\mid \beta\mid^2\cdot\mid A\mid^2+2\left(ac-bd\right)}} \end{align*}$$
And I think you can readily see the reduction here, too.