Let $$z$$ be those complex numbers which satisfy $$|z + 5| \leq 4$$ and $$z(1 + i) + \bar{z}(1 - i) \geq -10$$, $$i = \sqrt{-1}$$. If the maximum value of $$|z + 1|^2$$ is $$\alpha + \beta\sqrt{2}$$, then the value of $$(\alpha + \beta)$$ is

Let $$z = x + iy$$. The condition $$|z + 5| \leq 4$$ represents a closed disk centered at $$(-5, 0)$$ with radius 4, i.e., $$(x + 5)^2 + y^2 \leq 16$$.

For the second condition: $$z(1 + i) + \bar{z}(1 - i) = (x + iy)(1 + i) + (x - iy)(1 - i) = 2x - 2y$$. So the condition becomes $$2x - 2y \geq -10$$, or $$x - y \geq -5$$.

We need to maximize $$|z + 1|^2 = (x + 1)^2 + y^2$$.

The distance from the center $$(-5, 0)$$ of the disk to the line $$x - y + 5 = 0$$ is $$\frac{|-5 - 0 + 5|}{\sqrt{2}} = 0$$. So the line passes through the center of the disk, and the feasible region is a semicircular disk.

Parametrize the boundary circle as $$x = -5 + 4\cos\theta$$, $$y = 4\sin\theta$$. The condition $$x - y \geq -5$$ becomes $$4\cos\theta - 4\sin\theta \geq 0$$, i.e., $$\cos\theta \geq \sin\theta$$, which gives $$\theta \in [-\frac{3\pi}{4}, \frac{\pi}{4}]$$.

Now $$|z + 1|^2 = (x + 1)^2 + y^2 = (-4 + 4\cos\theta)^2 + 16\sin^2\theta = 16 - 32\cos\theta + 16\cos^2\theta + 16\sin^2\theta = 32 - 32\cos\theta$$.

This is maximized when $$\cos\theta$$ is minimized. Over $$\theta \in [-\frac{3\pi}{4}, \frac{\pi}{4}]$$, the minimum of $$\cos\theta$$ occurs at $$\theta = -\frac{3\pi}{4}$$, where $$\cos\left(-\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}$$.

Therefore, the maximum value of $$|z + 1|^2 = 32 - 32\left(-\frac{\sqrt{2}}{2}\right) = 32 + 16\sqrt{2}$$.

So $$\alpha = 32$$ and $$\beta = 16$$, giving $$\alpha + \beta = 48$$.