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:
Step-by-Step Solution
Key Concept: Write $z=x+iy$. The set $S$ is the intersection of the circle $(x-1)^2+y^2=1$ with the line $(\sqrt{2}-1)x+y=\sqrt{2}$. Find the two intersection points, identify $z_1$ (max $|z|$) and $z_2$ (min $|z|$), then compute $|\sqrt{2}z_1-z_2|^2$.
Circle: $(x-1)^2+y^2=1$; Line: $(\sqrt{2}-1)x+y=\sqrt{2}$. Solving gives either $x=1\Rightarrow y=1$, giving $z_2=1+i$; or $x=\frac{1}{2-\sqrt{2}}\Rightarrow y=\sqrt{2}-\frac{1}{\sqrt{2}}$, giving $z_1=(1+\frac{1}{\sqrt{2}})+\frac{i}{\sqrt{2}}$.
$|\sqrt{2}z_1-z_2|^2=|\left(\frac{1}{\sqrt{2}}+1\right)\sqrt{2}+i-(1+i)|^2=(\sqrt{2})^2=2$.
Correct Answer: 4