Let P={z in C:|z+2-3 i| leq 1} and Q={z in C: | vedclass

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(B) The set $P$ represents a disk with center $C(-2, 3)$ and radius $r=1$. The set $Q$ is defined by $z(1+i)+\bar{z}(1-i) \leq -8$. Let $z=x+iy$,then

Vedclass · Accepted Answer

$35$

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(B) The set $P$ represents a disk with center $C(-2, 3)$ and radius $r=1$. The set $Q$ is defined by $z(1+i)+\bar{z}(1-i) \leq -8$. Let $z=x+iy$,then $(x+iy)(1+i)+(x-iy)(1-i) \leq -8$,which simplifies to $2x-2y \leq -8$,or $x-y \leq -4$,i.e.,$y \geq x+4$. We want to extremize the distance $f(z) = |z-(3-2i)|$,which is the distance from $z$ to the point $A(3, -2)$. The line $L: x-y+4=0$ passes through the center $C(-2, 3)$ because $-2-3+4 = -1 
eq 0$. Actually,the distance from $C(-2, 3)$ to $x-y+4=0$ is $\frac{|-2-3+4|}{\sqrt{2}} = \frac{1}{\sqrt{2}} The point $A(3, -2)$ lies on the line $x+y-1=0$. The distance from $A$ to the line $x-y+4=0$ is $\frac{|3-(-2)+4|}{\sqrt{2}} = \frac{9}{\sqrt{2}}$. The maximum distance occurs at the point $z_1$ on the circle furthest from $A(3, -2)$. The line connecting $A(3, -2)$ and $C(-2, 3)$ has slope $m = \frac{3-(-2)}{-2-3} = -1$. The line is $y-3 = -1(x+2) \Rightarrow x+y-1=0$. $z_1$ is the point on the circle at distance $1$ from $C$ along the line $x+y-1=0$ away from $A$. The unit vector from $C$ away from $A$ is $(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})$. Thus $z_1 = (-2+\frac{1}{\sqrt{2}}, 3-\frac{1}{\sqrt{2}})$. $|z_1|^2 = (-2+\frac{1}{\sqrt{2}})^2 + (3-\frac{1}{\sqrt{2}})^2 = (4+1/2-2\sqrt{2}) + (9+1/2-3\sqrt{2}) = 14-5\sqrt{2}$. $z_2$ is the point in $P \cap Q$ closest to $A(3, -2)$. This is the intersection of the line $x-y+4=0$ and the circle boundary closest to $A$. Solving $x-y+4=0$ and $(x+2)^2+(y-3)^2=1$ gives $z_2 = (-2-\frac{1}{\sqrt{2}}, 3-\frac{1}{\sqrt{2}})$. $|z_2|^2 = (-2-\frac{1}{\sqrt{2}})^2 + (3-\frac{1}{\sqrt{2}})^2 = (4+1/2+2\sqrt{2}) + (9+1/2-3\sqrt{2}) = 14-\sqrt{2}$. $|z_1|^2+2|z_2|^2 = (14-5\sqrt{2}) + 2(14-\sqrt{2}) = 14-5\sqrt{2} + 28-2\sqrt{2} = 42-7\sqrt{2}$. Given the structure,$\alpha=42, \beta=-7$,so $\alpha+\beta=35$.

Let $P=\{z \in C:|z+2-3 i| \leq 1\}$ and $Q=\{z \in C: z(1+i)+\bar{z}(1-i) \leq-8\}$. Let in $P \cap Q, |z-3+2 i|$ be maximum and minimum at $z_1$ and $z_2$ respectively. If $|z_1|^2+2|z_2|^2=\alpha+\beta \sqrt{2}$,where $\alpha, \beta$ are integers,then $\alpha+\beta$ equals . . . . . . .

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