1hehe ok manipal............u deserved it
23 Answers
1ok 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 !!!
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11YEH LE
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11@ AKAND
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1TAKE 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....
1ncert book is also visible in 2nd pic.............further, do u remember the Q only or ans too???????[7]
