Question

If $\alpha$ and $\beta$ are different complex number with $|\beta |=1$ then find $\left|\frac{\beta -\alpha }{1-\overline{\alpha }\beta }\right|$.

Answer 1

Given: $|\beta |=1$

$\Rightarrow {\left|\frac{\beta -\alpha }{1-\overline{\alpha }\beta }\right|}^2=\left(\frac{\beta -\alpha }{1-\overline{\alpha }\beta }\right)\overline{\left(\frac{\beta -\alpha }{1-\overline{\alpha }\beta }\right)}=\left(\frac{\beta -\alpha }{1-\overline{\alpha }\beta }\right)\left(\frac{\overline{\beta }-\overline{\alpha }}{1-\alpha \overline{\beta }}\right)$

$\Rightarrow {\left|\frac{\beta -\alpha }{1-\overline{\alpha }\beta }\right|}^2=\frac{\beta \overline{\beta }-\beta \overline{\alpha }-\alpha \overline{\beta }+\alpha \overline{\alpha }}{1-\overline{\alpha }\beta -\alpha \overline{\beta }+\alpha \overline{\alpha }\beta \overline{\beta }}$

$\Rightarrow {\left|\frac{\beta -\alpha }{1-\overline{\alpha }\beta }\right|}^2=\frac{{\left|\beta \right|}^2-\beta \overline{\alpha }-\alpha \overline{\beta }+{\left|\alpha \right|}^2}{1-\overline{\alpha }\beta -\alpha \overline{\beta }+{\left|\alpha \right|}^2{\left|\beta \right|}^2}$

As $\Rightarrow {\left|\beta \right|}^2=1$

$\left|\frac{\beta -\alpha }{1-\overline{\alpha }\beta }\right|={\left(\frac{1-\beta \overline{\alpha }-\alpha \overline{\beta }+{\left|\alpha \right|}^2}{1-\overline{\alpha }\beta -\alpha \overline{\beta }+{\left|\alpha \right|}^2}\right)}^{\frac{1}{2}}={\left(1\right)}^{\frac{1}{2}}=1$