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Consider the quadratic equation `az^(2)+bz+c=0` where `a,b,c` are non-zero complex numbers. Now answer the following. The condition that the equation has both roots purely imaginary isA. `(bar(a))/a=(bar(b))/b=(bar(c))/c`B. `(bar(a))/a=-(bar(b))/b=(bar(c))/c`C. `(bar(a))/a=(bar(b))/b=-(bar(c))/c`D. `-(bar(a))/a=(bar(b))/b=(bar(c))/c` |
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Answer» Correct Answer - B Let `alpha` be one root of equation then `bar(alpha)=-alpha` Given equation is `az^(2)+bz+c=0` `:.aalpha^(2)+balpha+c=0` Taking conjugate we get `:.bar(a).bar(alpha)^(2)+bar(b).bar(alpha)+bar(c)=0` `bar(a).alpha^(2)-bar(b).alpha+bar(c)=0` so `alpha.beta` are roots of equation `az^(2)+bz+c=0` and `bar(a)z^(2)-bar(b)z+bar(c)=0` So `(bar(a))/a=-(bar(b))/b=(bar(c))/c` if `|alpha|=1` then `bar(alpha)=1/(alpha).` As `aalpha^(2)+balpha+c=0`..........1 `impliesbar(a)bar(alpha)^(2)+bar(b)bar(alpha)+bar(c)=0` `implies(bar(a))/(alpha^(2))+(bar(b))/(alpha)+bar(c)=0` So `bar(c).alpha^(2)+bar(b).alpha+bar(a)=0`.........2 Now applying condition for one common root for 1 and 2 we get `(a bar(b)-b bar(c))(b bar(a)-c bar(b))=(a bar(a)-c bar(c))^(2)` |
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