Key Concepts and Formulas

• Complex Numbers as Points: Representing $z = x + iy$ as a point $(x, y)$ in the Cartesian plane.
• Modulus and Distance: $|z - z_0|$ represents the distance between complex numbers $z$ and $z_0$.
• Regions in the Complex Plane:
  • $|z - z_0| \leq r$ represents a closed disk centered at $z_0$ with radius $r$.
  • $az + \bar{a}\bar{z} \leq c$ represents a half-plane.

Step-by-Step Solution

Step 1: Analyze Region P

We are given $P = \{z \in \mathbb{C} : |z + 2 - 3i| \leq 1\}$. We want to convert this complex inequality into a Cartesian one to understand the region geometrically. Let $z = x + iy$. Then:

$| (x + iy) + 2 - 3i | \leq 1$ $| (x+2) + i(y-3) | \leq 1$

Using the definition of modulus:

$\sqrt{(x+2)^2 + (y-3)^2} \leq 1$

Squaring both sides:

$(x+2)^2 + (y-3)^2 \leq 1$

This represents a closed disk centered at $(-2, 3)$ with radius $1$. We denote the center as $C_P = (-2, 3)$ and the radius as $r_P = 1$.

Step 2: Analyze Region Q

We are given $Q = \{z \in \mathbb{C} : z(1+i) + \bar{z}(1-i) \leq -8\}$. Again, let $z = x + iy$, so $\bar{z} = x - iy$. We want to convert this into a Cartesian inequality.

$(x+iy)(1+i) + (x-iy)(1-i) \leq -8$ $(x + ix + iy - y) + (x - ix - iy - y) \leq -8$ $2x - 2y \leq -8$ $x - y \leq -4$ $x - y + 4 \leq 0$

This represents a half-plane bounded by the line $x - y + 4 = 0$. We denote this line as $L_2: x - y + 4 = 0$. The region Q consists of all points $(x,y)$ on or to the left of this line.

Step 3: Verify Intersection of P and Q

To ensure that the intersection $P \cap Q$ is non-empty, we calculate the perpendicular distance from the center of the disk $C_P(-2, 3)$ to the line $L_2: x - y + 4 = 0$. The formula for the distance from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ is $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.

$d = \frac{|1(-2) + (-1)(3) + 4|}{\sqrt{1^2 + (-1)^2}} = \frac{|-2 - 3 + 4|}{\sqrt{2}} = \frac{|-1|}{\sqrt{2}} = \frac{1}{\sqrt{2}}$

Since $d = \frac{1}{\sqrt{2}} < 1 = r_P$, the line $L_2$ intersects the disk $P$. Therefore, $P \cap Q$ is non-empty.

Step 4: Identify the Target Function for Maximization/Minimization

We need to find $z_1, z_2 \in P \cap Q$ such that $|z - 3 + 2i|$ is maximum and minimum, respectively. Let $A$ be the point $(3, -2)$ in the complex plane (corresponding to $3 - 2i$). The expression $|z - (3 - 2i)|$ represents the distance from any point $z(x,y)$ to the fixed point $A(3, -2)$. We want to find the points in $P \cap Q$ farthest from and closest to $A(3, -2)$.

Step 5: Finding $z_1$ (Maximum Distance)

The point farthest from $A(3, -2)$ will lie on the boundary of $P \cap Q$. It is likely to be on the circle boundary along the line connecting $A$ and the center of the circle $C_P$. Let's call this line $L_1$. The slope of $L_1$ is $m = \frac{3 - (-2)}{-2 - 3} = \frac{5}{-5} = -1$. Using the point-slope form with $A(3, -2)$:

$y - (-2) = -1(x - 3)$ $y + 2 = -x + 3$ $L_1: x + y - 1 = 0$

Now we find the intersection points of $L_1$ and the circle $(x+2)^2 + (y-3)^2 = 1$. From $L_1$, $y = 1 - x$. Substitute into the circle equation:

$(x+2)^2 + (1-x-3)^2 = 1$ $(x+2)^2 + (-x-2)^2 = 1$ $2(x+2)^2 = 1$ $(x+2)^2 = \frac{1}{2}$ $x+2 = \pm \frac{1}{\sqrt{2}}$ $x = -2 \pm \frac{1}{\sqrt{2}}$

For each $x$ value, find the corresponding $y$ using $y = 1 - x$:

• If $x = -2 + \frac{1}{\sqrt{2}}$, then $y = 1 - \left(-2 + \frac{1}{\sqrt{2}}\right) = 3 - \frac{1}{\sqrt{2}}$. Point $Z_A = \left(-2 + \frac{1}{\sqrt{2}}, 3 - \frac{1}{\sqrt{2}}\right)$.
• If $x = -2 - \frac{1}{\sqrt{2}}$, then $y = 1 - \left(-2 - \frac{1}{\sqrt{2}}\right) = 3 + \frac{1}{\sqrt{2}}$. Point $Z_B = \left(-2 - \frac{1}{\sqrt{2}}, 3 + \frac{1}{\sqrt{2}}\right)$.

Check which points lie in region $Q$ (i.e., satisfy $x - y + 4 \leq 0$):

• For $Z_A$: $x - y + 4 = \left(-2 + \frac{1}{\sqrt{2}}\right) - \left(3 - \frac{1}{\sqrt{2}}\right) + 4 = -1 + \sqrt{2} > 0$. So $Z_A$ is not in $Q$.
• For $Z_B$: $x - y + 4 = \left(-2 - \frac{1}{\sqrt{2}}\right) - \left(3 + \frac{1}{\sqrt{2}}\right) + 4 = -1 - \sqrt{2} < 0$. So $Z_B$ is in $Q$.

Since $Z_B$ is farther from $A$ and is in $P \cap Q$, $z_1 = \left(-2 - \frac{1}{\sqrt{2}}\right) + i\left(3 + \frac{1}{\sqrt{2}}\right)$.

Step 6: Finding $z_2$ (Minimum Distance)

The minimum distance from $A(3, -2)$ to $P \cap Q$ occurs on the boundary. The closest point on the circle to $A$ is $Z_A$, but it's not in $Q$. Therefore, the minimum distance must occur on the line $L_2$. The closest point on $L_2$ to $A$ is the foot of the perpendicular from $A$ to $L_2$. The line through $A$ perpendicular to $L_2$ has slope $-1$ and equation $x + y - 1 = 0$, which is $L_1$. So we find the intersection of $L_1$ and $L_2$:

$x + y - 1 = 0$ $x - y + 4 = 0$

Adding the equations: $2x + 3 = 0 \implies x = -\frac{3}{2}$. Substituting into $L_1$: $-\frac{3}{2} + y - 1 = 0 \implies y = \frac{5}{2}$. So the intersection point is $Z_C = \left(-\frac{3}{2}, \frac{5}{2}\right)$.

Verify that $Z_C$ is within the disk $P$:

$\left(-\frac{3}{2} + 2\right)^2 + \left(\frac{5}{2} - 3\right)^2 = \left(\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right)^2 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \leq 1$

Since $Z_C$ is in $P$ and on $L_2$, it's in $P \cap Q$. Therefore, $z_2 = -\frac{3}{2} + i\frac{5}{2}$.

Step 7: Calculate $|z_1|^2$ and $2|z_2|^2$

For $z_1 = \left(-2 - \frac{1}{\sqrt{2}}\right) + i\left(3 + \frac{1}{\sqrt{2}}\right)$:

$|z_1|^2 = \left(-2 - \frac{1}{\sqrt{2}}\right)^2 + \left(3 + \frac{1}{\sqrt{2}}\right)^2 = 4 + \frac{4}{\sqrt{2}} + \frac{1}{2} + 9 + \frac{6}{\sqrt{2}} + \frac{1}{2} = 14 + \frac{10}{\sqrt{2}} + 1 = 14 + 5\sqrt{2}$

For $z_2 = -\frac{3}{2} + i\frac{5}{2}$:

$|z_2|^2 = \left(-\frac{3}{2}\right)^2 + \left(\frac{5}{2}\right)^2 = \frac{9}{4} + \frac{25}{4} = \frac{34}{4} = \frac{17}{2}$ $2|z_2|^2 = 2 \cdot \frac{17}{2} = 17$

Step 8: Compute $\left|z_1\right|^2+2\left|z_2\right|^2$ and Determine $\alpha, \beta$

$\left|z_1\right|^2 + 2\left|z_2\right|^2 = (14 + 5\sqrt{2}) + 17 = 31 + 5\sqrt{2}$

Comparing with $\alpha + \beta\sqrt{2}$, we have $\alpha = 31$ and $\beta = 5$. Therefore, $\alpha + \beta = 31 + 5 = 36$.

Common Mistakes & Tips

• Incorrectly Assuming Extrema on the Circle: The minimum distance point isn't necessarily on the circle if the closest point on the unconstrained circle is outside the feasible region.
• Not Checking Intersection: Always verify that the regions intersect before proceeding.
• Careless Calculation: Be extra careful with signs and fractions when expanding and simplifying.

Summary By converting the complex inequalities to Cartesian coordinates, we identified the feasible region as the intersection of a disk and a half-plane. We found the points in this region that maximized and minimized the distance to a given point by considering the geometry and the boundaries of the region. Finally, we calculated the requested expression, obtaining $\alpha + \beta = 36$.

The final answer is \boxed{36}.