If `z^2+z+1=0` where `z` is a complex number, then the value of `(z+1/z)^2+(z^2+1/z^2)^2+....+(z^6+1/z^6)^2` is

Options

A. 54
B. 6
C. 12
D. 18

Answer

C

Explanation

We know that `omega` and `omega^(2)` are roots of `z^(2)+z+1=0` such that `omega=1/omega^(2)`, So, the value of the given expression remains same for z=`omega` and, `z=1/omega^(2)`. Also, `1+omega+omega^(2)=0` and `omega^(3)=1`.
Putting, `z=omega`, we obtain.
`(z+1/z)^(2)+(z^(2)+1/z^(2))+(z^(3)+1/z^(3))^(2)+.......+(z^(6)+1/z^(6))^(2)`
`={(z+1/z)^(2)+(z^(2)+1/z^(2))^(2)+(z^(4)+1/z^(4))^(2)+(z^(5)+1/z^(5))^(2)+(z^(3)=1/z^(3))^(2)+(z^(6)+1/z^(6))^(2)}`
`4(omega+omega^(2))^(2)+(2^(2)+2^(2))=4+8=12`