If complex number z=x +iy satisfies the equation `Re (z+1) = |z-1|`, then prove that z lies on `y^(2) = 4x`.

Explanation

We have `Re(z+1) = |z-1|`
`rArr Re (x+iy+1) = |x + iy-1|`
`rArr x + 1 = sqrt((x-1)^(2) + y^(2)))`
`rArr (x+1)^(2) = (x+1)^(2) + y^(2)`
`rArr y^(2) = 4x`
Thus, z lies on `y^(2) = 4x`.