If `az^(2)+bz+1=0` where `a,b, in C` and `|a|=(1)/(2)` have a root `alpha` such that `|alpha|=1` then `4|avecb-b|` is equal to

Answer

3

Explanation

`aalpha^(2)+balpha+1=0`
`impliesb=-(1)/(alpha)-aalpha`
`b=-overline(alpha)-aalpha`
`overline(b)=-alpha-overline(a)overline(alpha)`
`aoverline(b)=-aalpha-|overline(a)|^(2)overline(alpha)`
`=-aalpha-(1)/(4)overline(alpha)`
`aoverline(b)-b=-aalpha-(1)/(4)overline(alpha),overline(alpha)+aalpha=(3)/(4)overlie(alpha)`
`|aoverline(b)-b|=(3)/(4)|overline(alpha)|=(3)/(4)`