Fundamentals - easy 1..........

1

Akand ·Mar 16 '09 at 9:26

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

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sriraghav ·Mar 17 '09 at 8:47

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dimensions (dimentime) ·Mar 17 '09 at 8:36

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 !!!

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Subash ·Mar 16 '09 at 10:19

image link

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tapanmast Vora ·Mar 16 '09 at 10:15

website mili but how to insert that ROTFloor smiley here?

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MATRIX ·Mar 16 '09 at 10:02

......

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Mani Pal Singh ·Mar 16 '09 at 9:57

YEH LE

http://www.artofproblemsolving.com/Forum/images/smiles/rotfl.gif

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iitimcomin ·Mar 16 '09 at 9:54

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

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Mani Pal Singh ·Mar 16 '09 at 9:52

@ AKAND

FROM NOW DONT ASK 4 PINKS

PINK UR ANSWER BY UR OWN

IF U THINK IT DESERVES!!!!!!!!!!!!!!!!!!

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Philip Calvert ·Mar 16 '09 at 9:38

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Akand ·Mar 16 '09 at 9:34

waaaaaaaaahhhhhh bhaiyya......u dint giv pink waaaaaaaahhhhhhh

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Lokesh Verma ·Mar 16 '09 at 9:27

haha manipal..

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Dr.House ·Mar 14 '09 at 10:01

is it a number?

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Mani Pal Singh ·Mar 16 '09 at 9:23

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

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Akand ·Mar 16 '09 at 9:20

bhaiyya no pink over here????

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Akand ·Mar 14 '09 at 10:15

yup and |Î±| has to be 1.........

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Akand ·Mar 14 '09 at 10:14

TAKE ZZ=|Z|²...wher Z is d given expression..........
so......
.(Î²-Î±)(Î²-Î±)/(1-Î±Î²)(1-Î±Î²)
=
(Î²-Î±)(1-Î±)/(1-Î±Î²)(1-Î±)
=
(Î²-Î±Î²-Î±-Î±Î±)/(1-Î±-Î±Î²-Î±Î²Î±)