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Consider the quadratic equation `az^(2)+bz+c=0` where a,b,c and non-zero complex numbers. Now answer the following: Q. The condition that the equation has one complex root `alpha` such that `|alpha|=1`, isA. `(overline(b)c-boverline(a))/(aoverline(a)-coverline(c))=(aoverline(a)+coverline(c))/(overline(c)b+aoverline(b))`B. `(overline(b)(c)+boverline(a))/(aoverline(a)+coverline(c))=(aoverline(a)+boverline(a))/(aoverline(a)+coverline(c))=(aoverline(a)+coverline(c))/(overline(c)b+aoverline(b))`C. `(overline(b)c-boverline(a))(overline(c)b-aoverline(b))=(aoverline(a)-coverline(c))^(2)`D. `(overline(b)c+boverline(a))/(aoverline(a)+coverline(c))=(aoverline(a)+coverline(c))/(overline(c)b-aoverline(b))` |
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Answer» Correct Answer - C Let `alpha` be one root of equation then `overline(alpha)=-alpha` Given equation is `az^(2)+bz+c=0` `becauseaalpha^(2)+balpha+c=0` Taking conjugate we get `becauseoverline(alpha)overline(alpha)^(2)+overline(b).overline(c)=0` `overline(a).alpha^(2)-overline(b).apha+overline(c)=0` and `overline(a)z^(2)-overline(b)z+overline(c)=0` So `(overline(a))/(a)=(overline(b))/(b)=(overline(c))/(c)` if `|alpha|=1` then `overline(alpha)=(1)/(alpha)` as `aalpha^(2)+balpha+c=0` ..(i) `impliesoverline(a)overline(alpha)^(2)+overline(b)overlinealpha)+overline(c)=0` `implies(overline(a))/(alpha^(2))+(overline(b))/(alpha)+overline(c)=0` So `overline(c).alpha^(2)+overline(b).alpha+overline(a)=0` .(ii) Now appying condition for one common root for (1) and (@) we get `(aoverline(b)-boverline(c))(boverline(a)-coverline(b))=(aoverline(a)-coverline(c))^(2)` |
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