ye mai bhi laya... par misyake kaha hain...
I have no clue what mistake i am making.. but i guess someone who can find the mistake in this question? !
If mod(z)=1, then prove that the points represented by√(1+z)/(1-z) lie on one or other of two fixed perpendicular lines
(1+z)/(1-z)
subst z=eiθ
(1+z)/(1-z) = (1+cosθ+isinθ)/(1-cosθ-isinθ)
in the numerator, cosθ=2sin2θ/2-1
in the denominaotr, cosθ=1-2cos2θ/2
also, sinθ=2sinθ/2.cosθ/2
we get 1+z/1-z= {cosθ/2.(cosθ/2+isinθ/2)}/{sinθ/2(sinθ/2-icosθ/2)}
see carefully the RHS gets reduced to cotθ/2.i
Thus, √(1+z)/(1-z) = √cot(θ/2) . √i
so square root fo i will have two roots.. one will be 1/√2+1/√2i and -1/√2-1/√2i
I have no clue what mistake i am making.. but i guess someone who can find the mistake in this question? !
ye mai bhi laya... par misyake kaha hain...
I have no clue what mistake i am making.. but i guess someone who can find the mistake in this question? !
so what we can say is that the two roots of i are perpendicular
isnt it?
the slopes are 1 and -1
:)