Solve (z-i)+(z+i)+(z-1)+(z-1)

Since we are dealing with complex numbers and taking its modulus, we can rewrite (z-i)=((-1)(i-z))=(i-z) doing the same for (z-1)=(1-z) we get (i-z)+(z+i)+(1-z)+(z-1)=(i+i+z-z+1+1+z-z) =(2i+2)=4 as we are taking its modulus.