Question

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

Answer 1

Correct Answer: 36

The objective is to determine the value of \( \alpha + \beta \), where \( \alpha \) and \( \beta \) are integers. These integers are derived from the equation \( |z_1|^2 + 2|z_2|^2 = \alpha + \beta\sqrt{2} \). The complex numbers \( z_1 \) and \( z_2 \) represent points within the intersection region \( P \cap Q \) that exhibit the maximum and minimum distances, respectively, from the point \( 3 - 2i \). (Correction: The problem statement was corrected from \( |z_1|^2 + 2|z|^2 \) to \( |z_1|^2 + 2|z_2|^2 \).)

Concept Used:

The solution employs the geometric representation of complex numbers on the Argand plane.

1. Set P: \( |z - z_0| \le r \) defines a closed disk centered at \( z_0 \) with radius \( r \).

2. Set Q: \( z(1+i) + \bar{z}(1-i) \le -8 \). Substituting \( z = x+iy \) simplifies this inequality to define a half-plane bounded by a straight line.

3. Intersection \( P \cap Q \): This is the region formed by the overlap of the disk and the half-plane, resulting in a circular segment.

4. Extremum of \( |z - z_A| \): This quantifies the maximum or minimum distance from a point \( z \) within the region \( P \cap Q \) to a fixed point \( z_A \). These extreme values occur at the boundaries of the region \( P \cap Q \).

Step-by-Step Solution:

Step 1: Geometric characterization of sets P and Q.

Set P: \( |z - (-2 + 3i)| \le 1 \). This describes a closed disk with center \( C_p = -2 + 3i \) (coordinates \( (-2, 3) \)) and radius \( r = 1 \).

Set Q: For \( z = x+iy \), the inequality \( (x+iy)(1+i) + (x-iy)(1-i) \le -8 \) simplifies to \( (x-y+i(x+y)) + (x-y-i(x+y)) \le -8 \), which further reduces to \( 2(x-y) \le -8 \), or \( x - y \le -4 \), which is equivalent to \( y \ge x+4 \). This defines a half-plane above or on the line \( L: y = x+4 \).

Step 2: Analysis of the intersection region \( P \cap Q \).

The region \( P \cap Q \) consists of points within the disk \( (x+2)^2 + (y-3)^2 \le 1 \) that also satisfy \( y \ge x+4 \). The distance from the disk's center \( C_p(-2, 3) \) to the line \( x-y+4=0 \) is \( d = \frac{|-2 - 3 + 4|}{\sqrt{1^2 + (-1)^2}} = \frac{|-1|}{\sqrt{2}} = \frac{1}{\sqrt{2}} \). Since \( d<r \), the line intersects the circle, and \( P \cap Q \) is a circular segment.

Step 3: Identification of points \( z_1 \) and \( z_2 \) for maximum and minimum distances from \( A = 3 - 2i \).

We seek points in \( P \cap Q \) furthest from and closest to \( A(3, -2) \). The point \( z_1 \) (maximum distance) is expected to be on the boundary of the disk, on the line passing through A and \( C_p \). The vector from A to \( C_p \) is \( \vec{v} = C_p - A = (-2-3) + i(3-(-2)) = -5 + 5i \). The unit vector in this direction is \( \hat{u} = \frac{-1+i}{\sqrt{2}} \). Thus, \( z_1 = C_p + r \cdot \hat{u} = (-2+3i) + 1 \cdot \left(\frac{-1+i}{\sqrt{2}}\right) = \left(-2 - \frac{1}{\sqrt{2}}\right) + i\left(3 + \frac{1}{\sqrt{2}}\right) \). This point satisfies \( y \ge x+4 \) as \( 3 + \frac{1}{\sqrt{2}} \ge \left(-2 - \frac{1}{\sqrt{2}}\right) + 4 \) simplifies to \( 1 \ge -\frac{2}{\sqrt{2}} \), which is true.

The point \( z_2 \) (minimum distance) would typically be \( C_p - r\hat{u} \). However, this point does not satisfy \( y \ge x+4 \). Therefore, the minimum distance must occur on the boundary line segment of \( P \cap Q \). The closest point on the line \( y = x+4 \) to A is its perpendicular projection. The projection M of \( (3,-2) \) onto \( x-y+4=0 \) is found by \( \frac{x_M-3}{1} = \frac{y_M-(-2)}{-1} = -\frac{3-(-2)+4}{1^2+(-1)^2} = -\frac{9}{2} \). This yields \( x_M = 3 - 9/2 = -3/2 \) and \( y_M = -2 + 9/2 = 5/2 \). Thus, \( z_2 = -3/2 + i(5/2) \).

Final Computation & Result:

Step 4: Calculation of \( |z_1|^2 \) and \( |z_2|^2 \).

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

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

Step 5: Computation of the final expression.

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

Comparing \( 31 + 5\sqrt{2} \) with \( \alpha + \beta\sqrt{2} \), we identify \( \alpha = 31 \) and \( \beta = 5 \).

The sum is \( \alpha + \beta = 31 + 5 = 36 \).

The value of \( \alpha + \beta \) is 36.