Consider $\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right|^2=\left(\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right)\left(\frac{\overline{\beta-\alpha}}{1-\bar{\alpha} \beta}\right)$
$\begin{aligned}
&=\left(\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right)\left(\frac{\bar{\beta}-\bar{\alpha}}{1-\alpha \bar{\beta}}\right) \quad\left(\because|\mathrm{z}|^2=z . \bar{z}\right) \\
&=\frac{\beta \bar{\beta}-\bar{\alpha} \beta-\alpha \bar{\beta}+\alpha \bar{\alpha}}{1-\bar{\alpha} \beta-\alpha \bar{\beta}+(\alpha \bar{\alpha})(\beta \bar{\beta})}=\frac{|\beta|^2-\bar{\alpha} \beta-\alpha \bar{\beta}+|\alpha|^2}{1-\bar{\alpha} \beta-\alpha \bar{\beta}+|\alpha|^2|\beta|^2}
\end{aligned}$
But $|\beta|^2=1$
$\therefore\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right|^2=\frac{1-\alpha \bar{\beta}-\bar{\alpha} \beta+|\alpha|^2}{1-\alpha \bar{\beta}-\bar{\alpha} \beta+|\alpha|^2}=1 \text {. }$
Hence $\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right|=1$