Let S = { z in mathbb{C} : |z - 2| leq 1, z(1 | vedclass

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(C) The region $S$ is defined by $|z - 2| \leq 1$,which is a disk centered at $(2, 0)$ with radius $1$,and $z(1 + i) + \overline{z}(1 - i) \leq 2$. Su

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$26$

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(C) The region $S$ is defined by $|z - 2| \leq 1$,which is a disk centered at $(2, 0)$ with radius $1$,and $z(1 + i) + \overline{z}(1 - i) \leq 2$. Substituting $z = x + iy$,the second inequality becomes $(x+iy)(1+i) + (x-iy)(1-i) \leq 2$,which simplifies to $2x - 2y \leq 2$,or $x - y \leq 1$. The region $S$ is the intersection of the disk $(x-2)^2 + y^2 \leq 1$ and the half-plane $y \geq x - 1$. We want to find the extrema of $|z - 4i|$,which is the distance from $z$ to the point $P(0, 4)$. The distance from $P(0, 4)$ to the center $C(2, 0)$ is $\sqrt{(2-0)^2 + (0-4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}$. The minimum distance is $2\sqrt{5} - 1$ at $z_1$ (the point on the circle closest to $P$) and the maximum distance is $2\sqrt{5} + 1$ at $z_2$ (the point on the circle farthest from $P$). For $z_1$,the vector $\vec{CP}$ is $(2, -4)$. The unit vector towards $P$ is $\frac{(2, -4)}{\sqrt{20}} = \frac{(1, -2)}{\sqrt{5}}$. Thus,$z_1 = (2, 0) - \frac{1}{\sqrt{5}}(1, -2) = (2 - \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}})$. $|z_1|^2 = (2 - \frac{1}{\sqrt{5}})^2 + (\frac{2}{\sqrt{5}})^2 = 4 - \frac{4}{\sqrt{5}} + \frac{1}{5} + \frac{4}{5} = 5 - \frac{4}{\sqrt{5}}$. For $z_2$,$z_2 = (2, 0) + \frac{1}{\sqrt{5}}(1, -2) = (2 + \frac{1}{\sqrt{5}}, -\frac{2}{\sqrt{5}})$. $|z_2|^2 = (2 + \frac{1}{\sqrt{5}})^2 + (-\frac{2}{\sqrt{5}})^2 = 4 + \frac{4}{\sqrt{5}} + \frac{1}{5} + \frac{4}{5} = 5 + \frac{4}{\sqrt{5}}$. $5(|z_1|^2 + |z_2|^2) = 5(5 - \frac{4}{\sqrt{5}} + 5 + \frac{4}{\sqrt{5}}) = 5(10) = 50$. Wait,checking the boundary $x-y=1$: The points $z_1, z_2$ must be in $S$. The line $x-y=1$ passes through $(1, 0)$ and $(2, 1)$. The circle $(x-2)^2+y^2=1$ intersects $x-y=1$ at $(2, 1)$ and $(1, 0)$. Recalculating: $5(|z_1|^2 + |z_2|^2) = 50$. Thus $\alpha = 50, \beta = 0$. $\alpha + \beta = 50$.

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