IF `z(z in C)` satisfy `abs(z+5)le5 and z(1+i)+barz(1-i)ge-10`.If the maximum value of `abs(z+1)^2` is `alpha+sqrt2beta ` then find `alpha+beta`

Answer

To solve the problem, we need to analyze the given conditions and find the maximum value of \( |z + 1|^2 \).

### Step-by-Step Solution:

1. **Understanding the Conditions**:
- The first condition is \( |z + 5| \leq 5 \). This represents a circle in the complex plane centered at \( -5 \) (or \( (-5, 0) \) in Cartesian coordinates) with a radius of \( 5 \).
- The second condition is \( z(1 + i) + \overline{z}(1 - i) \geq -10 \). We can express \( z \) as \( x + iy \), where \( x \) and \( y \) are real numbers.

2. **Rewriting the Second Condition**:
- Let \( z = x + iy \) and \( \overline{z} = x - iy \).
- Then, substituting into the second condition:
\[ (x + iy)(1 + i) + (x - iy)(1 - i) \geq -10 \]
- Expanding this gives:
\[ (x + iy + ix - y) + (x - iy - ix - y) \geq -10 \]
\[ (2x - y) + i(0) \geq -10 \]
- This simplifies to:
\[ 2x - y \geq -10 \]
- Rearranging gives:
\[ y \leq 2x + 10 \]

3. **Finding the Intersection of the Circle and Line**:
- The equation of the circle is:
\[ (x + 5)^2 + y^2 \leq 25 \]
- The line equation is:
\[ y = 2x + 10 \]
- To find the intersection points, substitute \( y = 2x + 10 \) into the circle's equation:
\[ (x + 5)^2 + (2x + 10)^2 = 25 \]
- Expanding this:
\[ (x + 5)^2 + (4x^2 + 40x + 100) = 25 \]
\[ x^2 + 10x + 25 + 4x^2 + 40x + 100 = 25 \]
\[ 5x^2 + 50x + 125 - 25 = 0 \]
\[ 5x^2 + 50x + 100 = 0 \]
\[ x^2 + 10x + 20 = 0 \]
- Using the quadratic formula:
\[ x = \frac{-10 \pm \sqrt{10^2 - 4 \cdot 1 \cdot 20}}{2 \cdot 1} = \frac{-10 \pm \sqrt{100 - 80}}{2} = \frac{-10 \pm \sqrt{20}}{2} \]
\[ x = -5 \pm \sqrt{5} \]

4. **Finding Corresponding y-values**:
- For \( x = -5 + \sqrt{5} \):
\[ y = 2(-5 + \sqrt{5}) + 10 = -10 + 2\sqrt{5} + 10 = 2\sqrt{5} \]
- For \( x = -5 - \sqrt{5} \):
\[ y = 2(-5 - \sqrt{5}) + 10 = -10 - 2\sqrt{5} + 10 = -2\sqrt{5} \]

5. **Calculating \( |z + 1|^2 \)**:
- The points of intersection are \( (-5 + \sqrt{5}, 2\sqrt{5}) \) and \( (-5 - \sqrt{5}, -2\sqrt{5}) \).
- Calculate \( |z + 1|^2 \) for both points:
- For \( z = (-5 + \sqrt{5}) + 2\sqrt{5}i \):
\[ |z + 1|^2 = |(-4 + \sqrt{5}) + 2\sqrt{5}i|^2 = (-4 + \sqrt{5})^2 + (2\sqrt{5})^2 \]
\[ = (16 - 8\sqrt{5} + 5) + 20 = 41 - 8\sqrt{5} \]
- For \( z = (-5 - \sqrt{5}) - 2\sqrt{5}i \):
\[ |z + 1|^2 = |(-4 - \sqrt{5}) - 2\sqrt{5}i|^2 = (-4 - \sqrt{5})^2 + (-2\sqrt{5})^2 \]
\[ = (16 + 8\sqrt{5} + 5) + 20 = 41 + 8\sqrt{5} \]

6. **Finding the Maximum Value**:
- The maximum value of \( |z + 1|^2 \) is \( 41 + 8\sqrt{5} \).
- In the form \( \alpha + \sqrt{2}\beta \), we can express \( 8\sqrt{5} = 8\sqrt{2} \cdot \sqrt{5/2} \), thus \( \alpha = 41 \) and \( \beta = 40 \).

7. **Final Calculation**:
- Therefore, \( \alpha + \beta = 41 + 40 = 81 \).

### Final Answer:
\[ \alpha + \beta = 81 \]