1but its given ! dat both of them are equal!
If α and β are roots of z +1/z =2(cosθ+isinθ) ,0<θ<π
then show that mod(α-i)=mod(β-1)
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17 Answers
1wen u urself r having doubt , then how do u expect others to undersatnd it ??
sorry but pls dun mind,,,, i din understand ur soln either.
1Isnt'it??
I did not use any complex thing .....
instead of keeping eiθ
i mean to say that the value of eiθ in this case is1
thats it
and more over the problems on complex numbers should be done in a complex way!!!
haha!!
1sri. u r not getting me.......see the
the value of z+1/z can never be less than 2
but the value of (cosθ+isinθ) can never be greater than 2
so compulsarily both of them should be equal.............
got the point??
that's what I mean to say is the value of θ should be 0
1yaar y are u making this so complex...
woh asaan sa .. bachcha type method din work or wat ?
(my first post..)
dat din work ??
1give some or the other reply..........
how can I understand whether u agree with me or not??
1well , what u r trying is possibily this x + 1/x = 2 . which is true iff x=1 . but here it's 2.eiθ . I don't think that approach works here. that becomes a particular case I think.
1I just doubt because there are some conceptual errors...
in the problem itself...
so I am asking for open discussion.......
just we have to use some logic for this question
just brood over the soln what I gaVE
1because the maximum value ofmod cosθ+isinθis 1
because it is cos2θ naaaa
so
since it is equal θshould be either0 or pi......
1I got it
there is nothing to do with the eqn
the min value of x+1/x=2
so 2(cisθ) sholud be min so θshould be 0.......
so α+β=2
αβ=1
.........carry on.....
1I dont know about the soln!! I am trying but one step what sky gave is wrong
sum of the roots is 2ei0
1z2 + 1 = 2z eiθ
=> z2 + 1 - 2z eiθ = 0
now,
α+β = 2 eiθ (edited)
αβ= 1
try with this...