1.

If p =a + bomega + comega^(2),q = b + comega+ aomega^(2), and r =c + aomega +bomega^(2), where a,b,c ne 0 and omega is the complex cuberoot of unity , then .

Answer»

If p,Q,r lie on the CIRCLE|z|=2, the trinagleformed by these POINTIS equilateral.
`p^(2)+q^(2)+r^(2) =a^(2)+b^(2)+c^(2)`
`p^(2)+q^(2) + r^(2) = 2 (pq+qr + rp)`
NONE of these

Solution :`p+q+r=a + bomega + comega^(2) + b + aomega^(2)+ c + aomega + bomega^(2)`
`therefore p +q+ r = (a+b+c)(1+ omega + omega^(2))=0`
p,q,r lie on the circle `|z|=2`, whosecircumcenter is origin. Also `(p+q+ r)//3=0` . Hence the cenroidwith cicumcenter. So, the triangle is equilateral.
Now ,`(P +q+ r)^(2)= 0`
`rArr p^(2) +q^(2) + r^(2) = -2pqr[(1)/(p)+(1)/(q) +(1)/(r)]`
`=-2pqr[(1)/(a+bomega+comega^(2))+(1)/(omega(bomega^(2)+c+aomega^(2)))+(1)/(c+aomega +bomega^(2))]`
`=2pqr[(1)/(omega^(2)(aomega +bomega^(2)+c))+(1)/(omega(bomega^(2) + c+ aomega))+(1)/(c+aomega+bomega^(2))]`
`(-2pqr)/(aomega+ bomega^(2)+c)[(1)/(omega^(2))+(1)/(omega)+ (1)/(1)]= 0""(2)`
Hence `p^(2)+q^(2) + r^(2) = 2` (pq + qr + rp)`


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