Let $$S = z \in \mathbb{C} : z - 1 = 1 \text{ and } (\sqrt{2} - 1)(z + \bar{z}) - i(z - \bar{z}) = 2\sqrt{2}$$. Let $$z_1, z_2 \in S$$ be such that $$z_1 = \max_{z \in S}z$$ and $$z_2 = \min_{z \in S}z$$. Then $$|\sqrt{2}z_1 - z_2^2$$ equals:

A

1

B

4

C

3

D

2