Let z be a complex number satisfying |z+5| le | vedclass

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(B) Given $|z+5| \leq 4$. Let $z = x+iy$. Then $(x+5)^2 + y^2 \leq 16$ (a disk centered at $(-5, 0)$ with radius $4$).Given $z(1+i)+\bar{z}(1-i) \geq

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$48$

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(B) Given $|z+5| \leq 4$. Let $z = x+iy$. Then $(x+5)^2 + y^2 \leq 16$ (a disk centered at $(-5, 0)$ with radius $4$).Given $z(1+i)+\bar{z}(1-i) \geq -10$. Substituting $z=x+iy$ and $\bar{z}=x-iy$:$(x+iy)(1+i) + (x-iy)(1-i) \geq -10$$(x-y + i(x+y)) + (x-y - i(x+y)) \geq -10$$2(x-y) \geq -10 \implies x-y+5 \geq 0$.We want to maximize $|z+1|^2$,which is the square of the distance from $z$ to the point $P(-1, 0)$.The region is the intersection of the disk $(x+5)^2 + y^2 \leq 16$ and the half-plane $x-y+5 \geq 0$.To find the maximum distance from $P(-1, 0)$ to the region,we check the boundary points. The maximum occurs at the intersection of the line $x-y+5=0$ and the circle $(x+5)^2 + y^2 = 16$.Substituting $y = x+5$ into the circle equation:$(x+5)^2 + (x+5)^2 = 16 \implies 2(x+5)^2 = 16 \implies (x+5)^2 = 8 \implies x+5 = \pm 2\sqrt{2}$.So $x = -5 \pm 2\sqrt{2}$.If $x = -5 - 2\sqrt{2}$,then $y = x+5 = -2\sqrt{2}$. Point $B = (-5-2\sqrt{2}, -2\sqrt{2})$.If $x = -5 + 2\sqrt{2}$,then $y = x+5 = 2\sqrt{2}$. Point $A = (-5+2\sqrt{2}, 2\sqrt{2})$.Calculate squared distances from $P(-1, 0)$:$PB^2 = (-5-2\sqrt{2} - (-1))^2 + (-2\sqrt{2} - 0)^2 = (-4-2\sqrt{2})^2 + 8 = (16 + 8 + 16\sqrt{2}) + 8 = 32 + 16\sqrt{2}$.$PA^2 = (-5+2\sqrt{2} - (-1))^2 + (2\sqrt{2} - 0)^2 = (-4+2\sqrt{2})^2 + 8 = (16 + 8 - 16\sqrt{2}) + 8 = 32 - 16\sqrt{2}$.The maximum value is $32 + 16\sqrt{2}$.Thus $\alpha = 32$ and $\beta = 16$.$\alpha + \beta = 32 + 16 = 48$.

Let $z$ be a complex number satisfying $|z+5| \leq 4$ and $z(1+i)+\bar{z}(1-i) \geq -10$,where $i=\sqrt{-1}$. If the maximum value of $|z+1|^2$ is $\alpha+\beta \sqrt{2}$,then the value of $(\alpha+\beta)$ is ......

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