IIT JEE , AIEEE Forum - Mathematics , Trignometry

VARSHINI

If z=2+t+iSq.root(3-t.t) where t is real and t sq <3,then modulus of (z+1/z-1) whole sq.

ramyani

is the answer 3 ? if my answer is correct, then i 'll post the solution

ramyani

neway, the solution is as follows. if i m wrong, then correct me.

given Z= (2 + t) + i^{[ ]}

(3 - t² )

so, Z + 1 = (3+t) + i^{[ ]}

(3 - t² ) therefore, mod (Z + 1) = ^{[ ]}

[ (3 + t)² + {

(3 - t² )}² ]
= ^{[ ]}

( 12 + 6t )

n, Z - 1 = (1 + t) + i^{[ ]}

(3 - t² ) therefore mod (Z - 1) = ^{[ ]}

[(1 + t)² + {

(3 - t² )}² ]
= ^{[ ]}

( 4 + 2t )

now, mod [(Z + 1)/(Z - 1)] = mod(Z + 1)/ mod (Z - 1)
=

( 12 + 6t ) /

( 4 + 2t )
= ^{[ ]}

(6/2)
= ^{[ ]}

3

so, modulus of (z+1/z-1) whole sq = 3

nadeemoidu

|(z+1)/(z-1)| =

3

So |(z+1)/(z-1)| the whole square = |(z+1)/(z-1)|²= 3

ramyani

ohh!! editing my last step