If `|z|=1` and let `omega=((1-z)^2)/(1-z^2)` , then prove that the locus of `omega` is equivalent to `|z-2|=|z+2`

Explanation

Given
`omega=((1-z)^(2))/(1-z^(2))=(1-z)/(1+z)`
`implies" "omega=(zbar(z)-z)/(zbar(z)+z)=(bar(z)-1)/(bar(z)+1)=-((bar(1-z))/(1+z))=-bar(omega)`
`:." "omega+bar(omega)=0`
`implies omega" is purely imaginary. Hence, "omega" lies on the y-axis."`
Also `|z-2|=|z+2|impliesz` lies on perpendicular bisector of line segment joning 2 and -2, which is the imaginary axis.