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If p=a+bomega+comega^2, q=b+comega+aomeg...

If `p=a+bomega+comega^2`, `q=b+comega+aomega^2`, and `r=c+aomega+bomega^2`, where `a ,b ,c!=0` and `omega` is the complex cube root of unity, then (a) `p+q+r=a+b+c` (b) `p^2+z^2+r^2=a^2+b^2+c^2` (c) `p^2+z^2+r^2=-2(p q+q r+r p)` (d) none of these

A

If p,q,r lie on the circle |z|=2, the trinagle formed by these point is equilateral.

B

`p^(2)+q^(2)+r^(2) =a^(2)+b^(2)+c^(2)`

C

`p^(2)+q^(2) + r^(2) = 2 (pq+qr + rp)`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
A, C
`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`, whose circumcenter is origin. Also `(p+q+ r)//3=0` . Hence the cenroid with 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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