Correct Answer - C
`(c )` `z^(2)=barz*2^(1-|z|)`
`:.z^(3)=|z|^(2)2^(1-|z|)`…………`(i)`
`implies |z|=2^(1-|z|)` (by taking modulus both sides)
`implies|z|=1`
`impliesz^(3)=1`
`impliesz=1`, `omega`, `omega^(2)`
But `1` is not imaginary
Hence `z=w` or `w^(2)`