If α and β are different complex numbers with = 1, then find.

Let $\alpha=a+i b$ and $\beta=x+i y$

It is given that, $|\beta|=1$

$\therefore \sqrt{x^{2}+y^{2}}=1$

$\Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}=1$

$\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right|=\left|\frac{(x+i y)-(a+i b)}{1-(a-i b)(x+i y)}\right|$

$=\left|\frac{(x-a)+i(y-b)}{1-(a x+a i y-i b x+b y)}\right|$

$=\left|\frac{(x-a)+i(y-b)}{(1-a x-b y)+i(b x-a y)}\right|$

$=\frac{|(x-a)+i(y-b)|}{|(1-a x-b y)+i(b x-a y)|}$   $\left[\left|\frac{z_{1}}{z_{2}}\right|=\frac{\left|z_{1}\right|}{\left|z_{2}\right|}\right]$

$=\frac{\sqrt{(x-a)^{2}+(y-b)^{2}}}{\sqrt{(1-a x-b y)^{2}+(b x-a y)^{2}}}$

$=\frac{\sqrt{x^{2}+a^{2}-2 a x+y^{2}+b^{2}-2 b y}}{\sqrt{1+a^{2} x^{2}+b^{2} y^{2}-2 a x+2 a b x y-2 b y+b^{2} x^{2}+a^{2} y^{2}-2 a b x y}}$

$=\frac{\sqrt{\left(x^{2}+y^{2}\right)+a^{2}+b^{2}-2 a x-2 b y}}{\sqrt{1+a^{2}\left(x^{2}+y^{2}\right)+b^{2}\left(y^{2}+x^{2}\right)-2 a x-2 b y}}$

$=\frac{\sqrt{1+a^{2}+b^{2}-2 a x-2 b y}}{\sqrt{1+a^{2}+b^{2}-2 a x-2 b y}}$   $[U \sin g(1)]$

$=1$

$\therefore\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right|=1$