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If $\alpha$ and $\beta$ are different complex numbers with $|\beta|=1$, then $\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right|$ is equal to
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Since, $|\beta|=1 \quad \therefore|\beta|^{2}=\beta \bar{\beta}=1$
$\begin{aligned} \therefore \quad\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right| &=\left|\frac{\beta-\alpha}{\beta \bar{\beta}-\bar{\alpha} \beta}\right| \\ &=\frac{|\beta-\alpha|}{|\beta||\bar{\beta}-\bar{\alpha}|} \\ &=\frac{|\beta-\alpha|}{1 \cdot|\overline{\beta-\alpha}|} \quad[\because|\overline{\mathrm{z}}|=|\mathrm{z}|] \\ &=1 \end{aligned}$
$\begin{aligned} \therefore \quad\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right| &=\left|\frac{\beta-\alpha}{\beta \bar{\beta}-\bar{\alpha} \beta}\right| \\ &=\frac{|\beta-\alpha|}{|\beta||\bar{\beta}-\bar{\alpha}|} \\ &=\frac{|\beta-\alpha|}{1 \cdot|\overline{\beta-\alpha}|} \quad[\because|\overline{\mathrm{z}}|=|\mathrm{z}|] \\ &=1 \end{aligned}$
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