Let z = a+bi
Since lzl = 1, a2+b2=1.
l 1+z l2 + l 1-z l2 = l (1+a) + bi l + l (1-a) - bi l
= (1+a)2 + b2 + (1-a)2 + (-b)2
= 1+2a+a2+1-2a+a2+2b2
= 2(a2+b2+1)
= 4 (since, by assumption, a2 + b2 = 1)