For the first question, this method is a bit tedious and I'm sure there's an easier way, but here it goes:
Let z = x + iy (I'll be using [ ] as the modulus symbol)
3[(x - 1) + iy]^2 = [(x + 1) + iy]^2
Remember that the modulus squared is equal to the complex no. times its conjugate. (I'll be using [ ] as normal brackets now).
Hence 3[(x - 1) + iy][(x - 1) - iy] = [(x + 1) + iy][(x + 1) - iy]
3[(x - 1)^2 + y^2] = (x+1)^2 + y^2
After performing some rearranging, we arrive at the quadratic
2x^2 - 8x + 2 = 0
Solving this, we get x = 2+sqrt3 and x=2-sqrt3
Since z = x + iy, z= (2+-sqrt3) + 0i
(I'm using [ ] as modulus symbols again)
[z - 2]^2 = [ 2+-sqrt.3 - 2]^2
= [+-sqrt.3]^2
= 3
Hence [z - 2]^2 = 3 for the equation to hold.