Let S = {z in mathbb{C} : |z-3| leq 1 text{ a | vedclass

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(C) The condition $|z-3| \leq 1$ represents a disk with radius $1$ centered at $(3, 0)$.The condition $z(4+3i) + \bar{z}(4-3i) \leq 24$ can be simplif

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$80$

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(C) The condition $|z-3| \leq 1$ represents a disk with radius $1$ centered at $(3, 0)$.The condition $z(4+3i) + \bar{z}(4-3i) \leq 24$ can be simplified by substituting $z = x + iy$:$(x+iy)(4+3i) + (x-iy)(4-3i) \leq 24$$(4x - 3y + i(3x + 4y)) + (4x - 3y - i(3x + 4y)) \leq 24$$8x - 6y \leq 24 \Rightarrow 4x - 3y \leq 12$.We want to find the point $(\alpha, \beta)$ in the region $S$ closest to $(0, 4)$.The line $4x - 3y = 12$ passes through $(3, 0)$ and $(0, -4)$.The distance from $(0, 4)$ to the line $4x - 3y - 12 = 0$ is $d = \frac{|4(0) - 3(4) - 12|}{\sqrt{4^2 + (-3)^2}} = \frac{|-24|}{5} = 4.8$.Since the circle is centered at $(3, 0)$ with radius $1$,the closest point on the circle to $(0, 4)$ lies on the line segment connecting $(0, 4)$ and $(3, 0)$.The line passing through $(0, 4)$ and $(3, 0)$ is $\frac{x}{3} + \frac{y}{4} = 1$,or $4x + 3y = 12$.Solving the system:$4x + 3y = 12$$(x-3)^2 + y^2 = 1$From the first,$x = 3 - \frac{3}{4}y$. Substituting into the circle equation:$(3 - \frac{3}{4}y - 3)^2 + y^2 = 1$$\frac{9}{16}y^2 + y^2 = 1$ $\Rightarrow \frac{25}{16}y^2 = 1$ $\Rightarrow y^2 = \frac{16}{25}$ $\Rightarrow y = \pm \frac{4}{5}$.For the point closest to $(0, 4)$,we take $y = \frac{4}{5}$.Then $x = 3 - \frac{3}{4}(\frac{4}{5}) = 3 - \frac{3}{5} = \frac{12}{5}$.Thus,$\alpha = \frac{12}{5}$ and $\beta = \frac{4}{5}$.$25(\alpha + \beta) = 25(\frac{12}{5} + \frac{4}{5}) = 25(\frac{16}{5}) = 5 	imes 16 = 80$.

Let $S = \{z \in \mathbb{C} : |z-3| \leq 1 \text{ and } z(4+3i) + \bar{z}(4-3i) \leq 24\}$. If $\alpha + i\beta$ is the point in $S$ which is closest to $4i$,then $25(\alpha + \beta)$ is equal to

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