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 problem asks for the value of \( \alpha + \beta \), where \( \alpha \) and \( \beta \) are integers derived from the expression \( |z_1|^2 + 2|z_2|^2 = \alpha + \beta\sqrt{2} \). The complex numbers \( z_1 \) and \( z_2 \) are the points in the region \( P \cap Q \) where the distance \( |z - 3 + 2i| \) is maximum and minimum, respectively. (Note: There appears to be a typo in the original question image. The expression should be \( |z_1|^2 + 2|z_2|^2 \), not \( |z_1|^2 + 2|z|^2 \), as \( z \) is a variable. We will proceed with the corrected expression.)

Concept Used:

The solution involves the geometric interpretation of complex numbers in the Argand plane.

1. Set P: \( |z - z_0| \le r \) represents a closed disk with center \( z_0 \) and radius \( r \).

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

3. Intersection \( P \cap Q \): This is the geometric intersection of the disk and the half-plane, which forms a segment of a circle.

4. Extremum of \( |z - z_A| \): This represents the maximum or minimum distance from a point \( z \) in the region \( P \cap Q \) to a fixed point \( z_A \). These extrema will occur at the boundary points of the region \( P \cap Q \).

Step-by-Step Solution:

Step 1: Characterize the sets P and Q geometrically.

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

For set Q: Let \( z = x+iy \). The inequality is \( (x+iy)(1+i) + (x-iy)(1-i) \le -8 \).

\[ (x - y + i(x+y)) + (x - y - i(x+y)) \le -8 \] \[ 2(x-y) \le -8 \implies x - y \le -4 \implies y \ge x+4 \]

This represents a half-plane on and above the line \( L: y = x+4 \).

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

We need to find the points in the disk \( (x+2)^2 + (y-3)^2 \le 1 \) that also satisfy \( y \ge x+4 \). Let's check the distance from the disk's center \( C_p(-2, 3) \) to the line \( x-y+4=0 \).

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

Since the distance \( d = 1/\sqrt{2} < r = 1 \), the line intersects the circle. The region \( P \cap Q \) is a segment of the disk.

Step 3: Identify the points \( z_1 \) and \( z_2 \) for maximum and minimum distance from \( A = 3 - 2i \).

We need to find the points in \( P \cap Q \) that are farthest from and closest to the point \( A(3, -2) \). Let the point corresponding to \( z \) be \( Z(x, y) \).

The point \( z_1 \) (maximum distance) will be the point on the circle's boundary \( |z+2-3i|=1 \) that is furthest from A, provided it lies in the region Q. This point lies 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 is \( \hat{u} = \frac{-1+i}{\sqrt{2}} \).

\[ z_1 = C_p + r \cdot \hat{u} = (-2+3i) + 1 \cdot \left(\frac{-1+i}{\sqrt{2}}\right) = \left(-2 - \frac{1}{\sqrt{2}}\right) + i\left(3 + \frac{1}{\sqrt{2}}\right) \]

Let's check if \( z_1 \) is in Q. For \( z_1 = x+iy \), we check if \( y \ge x+4 \).

\[ 3 + \frac{1}{\sqrt{2}} \ge \left(-2 - \frac{1}{\sqrt{2}}\right) + 4 \implies 3 + \frac{1}{\sqrt{2}} \ge 2 - \frac{1}{\sqrt{2}} \implies 1 \ge -\frac{2}{\sqrt{2}} \]

This is true, so \( z_1 \) is in the region \( P \cap Q \).

The point \( z_2 \) (minimum distance) will be the point in \( P \cap Q \) closest to A. The point on the circle closest to A is \( C_p - r\hat{u} \), but this point does not satisfy \( y \ge x+4 \). Therefore, the minimum distance must be to a point on the line segment boundary of the region, which is the chord \( y = x+4 \). The closest point on this line to A is the perpendicular projection of A onto the line. Let this point be M.

The line is \( x-y+4=0 \). The projection M is \( (x_M, y_M) \). Let's find its coordinates. The projection of \( (3,-2) \) on \( x-y+4=0 \) is \( \frac{x_M-3}{1} = \frac{y_M-(-2)}{-1} = -\frac{3-(-2)+4}{1^2+(-1)^2} = -\frac{9}{2} \). \( x_M = 3 - 9/2 = -3/2 \). \( y_M = -2 + 9/2 = 5/2 \). This point \( M(-1.5, 2.5) \) lies on the chord, so \( z_2 = -3/2 + i(5/2) \).

Final Computation & Result:

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

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

And for \( z_2 \):

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

Step 5: Compute 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 this with \( \alpha + \beta\sqrt{2} \), we have \( \alpha = 31 \) and \( \beta = 5 \).

\[ \alpha + \beta = 31 + 5 = 36 \]

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