complex geometry

(1+z)/(1-z)

subst z=e^iθ

(1+z)/(1-z) = (1+cosθ+isinθ)/(1-cosθ-isinθ)

in the numerator, cosθ=2sin²Î¸/2-1
in the denominaotr, cosθ=1-2cos²Î¸/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? !

so what we can say is that the two roots of i are perpendicular

isnt it?

the slopes are 1 and -1

:)