Let S = {z in mathbb{C} : |z-1|=1 text{ and } | vedclass

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(D) Let $z = x + iy$. The condition $|z-1|=1$ implies $(x-1)^2 + y^2 = 1$,which simplifies to $x^2 + y^2 - 2x = 0$. The second condition is $(\sqrt{2}

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(D) Let $z = x + iy$. The condition $|z-1|=1$ implies $(x-1)^2 + y^2 = 1$,which simplifies to $x^2 + y^2 - 2x = 0$. The second condition is $(\sqrt{2}-1)(2x) - i(2iy) = 2\sqrt{2}$,which simplifies to $(\sqrt{2}-1)x + y = \sqrt{2}$,or $y = \sqrt{2} - (\sqrt{2}-1)x$. Substituting $y$ into the first equation: $(x-1)^2 + (\sqrt{2} - (\sqrt{2}-1)x)^2 = 1$. Expanding this: $(x^2 - 2x + 1) + (2 - 2\sqrt{2}(\sqrt{2}-1)x + (\sqrt{2}-1)^2 x^2) = 1$. $(x^2 - 2x + 1) + (2 - (4 - 2\sqrt{2})x + (3 - 2\sqrt{2})x^2) = 1$. $(4 - 2\sqrt{2})x^2 - (6 - 2\sqrt{2})x + 2 = 0$. Dividing by $2$: $(2 - \sqrt{2})x^2 - (3 - \sqrt{2})x + 1 = 0$. Factoring gives $(x-1)((2-\sqrt{2})x - 1) = 0$. Thus,$x = 1$ or $x = \frac{1}{2-\sqrt{2}} = \frac{2+\sqrt{2}}{2} = 1 + \frac{1}{\sqrt{2}}$. For $x=1$,$y = \sqrt{2} - (\sqrt{2}-1)(1) = 1$. So $z_2 = 1+i$,$|z_2|^2 = 2$. For $x = 1 + \frac{1}{\sqrt{2}}$,$y = \sqrt{2} - (\sqrt{2}-1)(1 + \frac{1}{\sqrt{2}}) = \sqrt{2} - (\sqrt{2} + 1 - 1 - \frac{1}{\sqrt{2}}) = \frac{1}{\sqrt{2}}$. So $z_1 = (1 + \frac{1}{\sqrt{2}}) + i\frac{1}{\sqrt{2}}$. Then $|\sqrt{2}z_1 - z_2|^2 = |(\sqrt{2} + 1 + i) - (1+i)|^2 = |\sqrt{2}|^2 = 2$.

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:

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