If `az^2+bz+1=0`, where `a,b in C`, `|a|=1/2` and have a root `alpha` such that `|alpha|=1` then `|abarb-b|=`

Options

A. `1//4`
B. `1//2`
C. `5//4`
D. `3//4`

Answer

D

Explanation

`(d)` `aalpha^(2)+balpha+1=0` ……….`(i)`
`implies barabaralpha^(2)+barbbaralpha+1=0`
`implies alpha^(2)+barbalpha+bara=0` (as `|alpha|=alphabaralpha=1`) ……….`(ii)`
From `(i)` and `(ii)`
`(alpha^(2))/(barab-barb)=(alpha)/(1-|a|^(2))=(1)/(abarb-b)implies|abarb-b|=1-|a|^(2)=1-(1)/(4)=(3)/(4)`