Let S = `{z in CC:|z-3|le1and z(4+3i)+bar(z) (4-3i)le24}` .If `alpha + ibeta` is the point in S which is closest to 4i , then `25(alpha +beta)` is equal to ______.

Answer

To solve the problem, we need to find the point \( z = \alpha + i\beta \) in the set \( S \) that is closest to \( 4i \). The set \( S \) is defined by two conditions:

1. \( |z - 3| \leq 1 \)
2. \( z(4 + 3i) + \overline{z}(4 - 3i) \leq 24 \)

### Step 1: Analyze the first condition

The first condition \( |z - 3| \leq 1 \) describes a circle in the complex plane centered at \( 3 + 0i \) (or the point \( (3, 0) \)) with a radius of \( 1 \). Therefore, the points in \( S \) are those that lie within or on the boundary of this circle.

### Step 2: Analyze the second condition

The second condition can be rewritten by substituting \( z = x + iy \) (where \( x \) and \( y \) are the real and imaginary parts of \( z \), respectively):

\[ z(4 + 3i) + \overline{z}(4 - 3i) \leq 24 \]

Substituting \( z = x + iy \) and \( \overline{z} = x - iy \):

\[ (x + iy)(4 + 3i) + (x - iy)(4 - 3i) \leq 24 \]

Expanding this:

\[ (4x - 3y) + i(3x + 4y) + (4x + 3y) - i(3x - 4y) \leq 24 \]

Combining like terms:

\[ (8x) + (0) \leq 24 \]

This simplifies to:

\[ 4x \leq 12 \quad \Rightarrow \quad x \leq 3 \]

### Step 3: Determine the intersection of the conditions

The first condition gives us a circle centered at \( (3, 0) \) with radius \( 1 \), while the second condition restricts \( x \) to be less than or equal to \( 3 \). The intersection of these two conditions will give us the region \( S \).

### Step 4: Find the point closest to \( 4i \)

The point \( 4i \) corresponds to the coordinates \( (0, 4) \). We need to find the point on the boundary of the circle that is closest to \( (0, 4) \). The center of the circle is at \( (3, 0) \).

### Step 5: Parametric representation of the circle

The boundary of the circle can be represented parametrically as:

\[ (x, y) = (3 + \cos \theta, 0 + \sin \theta) \]

### Step 6: Calculate the distance to \( 4i \)

We need to minimize the distance from the point \( (3 + \cos \theta, \sin \theta) \) to \( (0, 4) \):

\[ D = \sqrt{(3 + \cos \theta - 0)^2 + (\sin \theta - 4)^2} \]

### Step 7: Minimize the distance

To find the minimum distance, we can minimize \( D^2 \):

\[ D^2 = (3 + \cos \theta)^2 + (\sin \theta - 4)^2 \]

Expanding this:

\[ D^2 = (9 + 6\cos \theta + \cos^2 \theta) + (\sin^2 \theta - 8\sin \theta + 16) \]

Using \( \sin^2 \theta + \cos^2 \theta = 1 \):

\[ D^2 = 26 + 6\cos \theta - 8\sin \theta \]

### Step 8: Find the critical points

To minimize \( D^2 \), we can take derivatives with respect to \( \theta \) and set them to zero. However, we can also use geometric reasoning since the closest point will be along the line connecting \( (3, 0) \) and \( (0, 4) \).

### Step 9: Find coordinates of the closest point

Using the geometry of the situation, we find that the closest point on the circle to \( (0, 4) \) is:

\[ \left(3 - \frac{3}{5}, 4 - \frac{4}{5}\right) = \left(\frac{12}{5}, \frac{4}{5}\right) \]

### Step 10: Calculate \( 25(\alpha + \beta) \)

Here, \( \alpha = \frac{12}{5} \) and \( \beta = \frac{4}{5} \):

\[ \alpha + \beta = \frac{12}{5} + \frac{4}{5} = \frac{16}{5} \]

Thus,

\[ 25(\alpha + \beta) = 25 \times \frac{16}{5} = 80 \]

### Final Answer

The value of \( 25(\alpha + \beta) \) is \( \boxed{80} \).