easy 1..........

hehe ok manipal............u deserved it

ok i have a more simpler proof given \small |\beta |=1\\ so,\beta =\frac{1}{\bar{\beta }}\\ let\ Z=\frac{\beta -\alpha }{1-\bar{\alpha }\beta } \\ or, Z=\frac{1-\alpha \bar{\beta }}{\bar{\beta }-\bar{\alpha }}\\ taking\ bar\ on\ both\ side,\\ \bar{Z}=\frac{\beta -\alpha }{1-\bar{\alpha }\beta } \\ so\ from\ both\ equation\ we\ have, Z\bar{Z}=1\\ or\\ |Z|=\left| \frac{\beta -\alpha }{1-\bar{\alpha }\beta }\right|=1 & it is not necessary that |@|=1 !!!

......

philip kahan se woh smiley mila????????????

NO ONE GIVES ME PINK [17][2]
I DECIDED TO HAVE MIN[1][3][4]

yup and |α| has to be 1.........

TAKE ZZ=|Z|²...wher Z is d given expression..........
so......
.(β-α)(β-α)/(1-αβ)(1-αβ)
=
(β-α)(1-α)/(1-αβ)(1-α)
=
(β-αβ-α-αα)/(1-α-αβ-αβα)

do more simplification and ull get 1....

nothing given about mod a, sri shankara charya...

if mod α is one

then the value is one

ans=1