Key Concepts and Formulas

• The inequality $|z - z_0| \le r$ represents a closed disk with center $z_0$ and radius $r$ in the complex plane.
• The expression $z\overline{w} + \overline{z}w = 2\text{Re}(z\overline{w})$ where $z$ and $w$ are complex numbers.
• The shortest distance from a point to a circle is along the line connecting the point to the center of the circle.

Step-by-Step Solution

Step 1: Analyze the first condition $|z - 3| \le 1$

The given condition is $|z - 3| \le 1$. Let $z = x + iy$, where $x$ and $y$ are real numbers. Substituting this into the inequality gives: $|x + iy - 3| \le 1$ $|(x - 3) + iy| \le 1$ Using the definition of the modulus of a complex number, we have: $\sqrt{(x - 3)^2 + y^2} \le 1$ Squaring both sides gives: $(x - 3)^2 + y^2 \le 1$ This represents a closed disk in the complex plane centered at $(3, 0)$ with a radius of 1.

Step 2: Analyze the second condition $z(4 + 3i) + \overline{z}(4 - 3i) \le 24$

We are given the condition $z(4 + 3i) + \overline{z}(4 - 3i) \le 24$. Let $z = x + iy$. Then $\overline{z} = x - iy$. Substituting these into the inequality gives: $(x + iy)(4 + 3i) + (x - iy)(4 - 3i) \le 24$ Expanding the expression: $4x + 3ix + 4iy - 3y + 4x - 3ix - 4iy - 3y \le 24$ $8x - 6y \le 24$ Dividing by 2, we get: $4x - 3y \le 12$ This represents a half-plane in the $xy$-plane.

Step 3: Define the region S

The region S is the intersection of the disk $(x - 3)^2 + y^2 \le 1$ and the half-plane $4x - 3y \le 12$. The equation of the line is $4x - 3y = 12$. Let's check if the center of the disk $(3, 0)$ lies on this line: $4(3) - 3(0) = 12$ Since the equality holds, the center of the disk lies on the line. Thus, the line divides the disk into two equal halves.

Step 4: Find the point in S closest to $4i$

We want to find the point in S closest to $4i$, which corresponds to the point $(0, 4)$ in the $xy$-plane. Since the region S is a half-disk, the closest point will lie on the boundary of S. Since $(0,4)$ is outside the disk, we consider the point on the circle closest to $(0,4)$. This point will lie on the line joining the center of the circle $(3,0)$ and the point $(0,4)$. The equation of this line is: $\frac{y - 0}{x - 3} = \frac{4 - 0}{0 - 3} = -\frac{4}{3}$ $3y = -4x + 12$ $4x + 3y = 12$ We need to find the intersection of this line with the circle $(x - 3)^2 + y^2 = 1$. From the line equation, $3y = 12 - 4x$, so $y = \frac{12 - 4x}{3}$. Substituting this into the circle equation: $(x - 3)^2 + \left(\frac{12 - 4x}{3}\right)^2 = 1$ $(x - 3)^2 + \frac{16}{9}(3 - x)^2 = 1$ $(x - 3)^2 + \frac{16}{9}(x - 3)^2 = 1$ $\frac{25}{9}(x - 3)^2 = 1$ $(x - 3)^2 = \frac{9}{25}$ $x - 3 = \pm \frac{3}{5}$ $x = 3 \pm \frac{3}{5}$ So, $x = 3 + \frac{3}{5} = \frac{18}{5}$ or $x = 3 - \frac{3}{5} = \frac{12}{5}$. If $x = \frac{18}{5}$, then $y = \frac{12 - 4(\frac{18}{5})}{3} = \frac{60 - 72}{15} = \frac{-12}{15} = -\frac{4}{5}$. If $x = \frac{12}{5}$, then $y = \frac{12 - 4(\frac{12}{5})}{3} = \frac{60 - 48}{15} = \frac{12}{15} = \frac{4}{5}$. The two points are $\left(\frac{18}{5}, -\frac{4}{5}\right)$ and $\left(\frac{12}{5}, \frac{4}{5}\right)$. We need to check which of these points lies on the half-plane $4x - 3y \le 12$. For $\left(\frac{18}{5}, -\frac{4}{5}\right)$: $4\left(\frac{18}{5}\right) - 3\left(-\frac{4}{5}\right) = \frac{72}{5} + \frac{12}{5} = \frac{84}{5} = 16.8$. Since $16.8 > 12$, this point is not in the region S. For $\left(\frac{12}{5}, \frac{4}{5}\right)$: $4\left(\frac{12}{5}\right) - 3\left(\frac{4}{5}\right) = \frac{48}{5} - \frac{12}{5} = \frac{36}{5} = 7.2$. Since $7.2 \le 12$, this point is in the region S. Thus, the point closest to $4i$ is $\left(\frac{12}{5}, \frac{4}{5}\right)$. So, $\alpha = \frac{12}{5}$ and $\beta = \frac{4}{5}$.

Step 5: Calculate $25(\alpha + \beta)$

We need to find the value of $25(\alpha + \beta)$: $25(\alpha + \beta) = 25\left(\frac{12}{5} + \frac{4}{5}\right) = 25\left(\frac{16}{5}\right) = 5 \times 16 = 80$ Now we need to find the point in S closest to 4i, then 25(alpha + beta)

Common Mistakes & Tips

• Remember to check that the final point lies within the defined region S.
• Visualizing the problem in the complex plane can help in understanding the geometric interpretation of the inequalities.
• Be careful with algebraic manipulations, especially when substituting and simplifying expressions.

Summary

The problem required us to find the point in a half-disk region closest to a given external point. We found the Cartesian equation of the region, determined the closest point on the circle to the external point, checked that it lay in the half-disk, and then calculated the required expression. The value of $25(\alpha + \beta)$ is 80. However, the correct answer provided is 3, indicating an error in the original problem statement or correct answer. Working backwards, 25($\alpha + \beta$) = 3 implies that $\alpha + \beta$ = 3/25.

Given the correct answer is 3, there must be an error in the question. Assuming that everything else is correct, let the point closest to 4i be $\alpha + i\beta$. Then, $\alpha + \beta = 3/25$. If we assume the closest point is $3/5 + i(-9/5)$, then $z=3/5 -9i/5$, so |z-3| = sqrt((3/5 - 3)^2 + (-9/5)^2) = sqrt(144/25 + 81/25) = sqrt(225/25) = sqrt(9) = 3, which is not less than or equal to 1.

Let's assume that the correct answer is indeed 80. Then,

Final Answer

The final answer is \boxed{80}.