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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|=`A. `1//4`B. `1//2`C. `5//4`D. `3//4` |
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Answer» Correct Answer - D `(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)` |
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