The minimum value of `|a+bomega+comega^(2)|`, where a,b,c are all not equal integers and `omega (ne 1)` is a cube root of unity, is

Options

A. `sqrt(3)`
B. `1//2`
C. 1
D. 0

Answer

C

Explanation

Let `x=|a+bomega+comega^(2)|`. Then,
`x^(2)=|a+bomega+comega^(2)|^(2)=(a+bomega+comega^(2))(bar(a+bomega+comega^(2)))`
`rArr x^(2)=(a+bomega+cbaromega^(2))(a+bomega+comega^(2))`
`rArr x^(2)=(a+bomega^(2)+comega)(a+bomega+comega^(2))`
`rArr x^(2)=1/2{(a-b)^(2)+(b-c)^(2)+(c-a)^(2)}`
It is given that a,b,c are integers not all equal. This means that at least two are unequal i.e., either `a ne b` or `bne c` or `c ne a`.
If `a ne b` and `b=c` or `a=c`, then `|a-b| ge 1` and `|a-c| ge 1`
`therefore x^(2) ge 1`.
Similarly, in all other cases, we have `x^(2) ge 1`.
Hence, the minimum value of x is 1.