If the least and the largest real values of $$\alpha$$, for which the equation $$z + \alpha|z - 1| + 2i = 0$$ ($$z \in C$$ and $$i = \sqrt{-1}$$) has a solution, are $$p$$ and $$q$$ respectively; then $$4(p^2 + q^2)$$ is equal to ______.

Let

$$z=x+iy$$

Then,

$$x+iy+\alpha\sqrt{(x-1)^2+y^2}+2i=0$$

Equating real and imaginary parts,

$$x+\alpha\sqrt{(x-1)^2+y^2}=0 \quad\cdots(1)$$ and $$y+2=0$$

Hence,

$$y=-2$$

Substitute into (1):

$$x+\alpha\sqrt{(x-1)^2+4}=0$$

$$\alpha=-\frac{x}{\sqrt{(x-1)^2+4}}$$

Therefore,

$$\alpha^2=\frac{x^2}{(x-1)^2+4}$$

$$=\frac{x^2}{x^2-2x+5}$$

Let

$$f(x)=\frac{x^2}{x^2-2x+5}$$

Differentiate:

$$f'(x)=\frac{2x(x^2-2x+5)-x^2(2x-2)}{(x^2-2x+5)^2}$$

$$=\frac{2x(5-x)}{(x^2-2x+5)^2}$$

Hence critical points are $$x=0,\qquad x=5$$

Now,

$$f(0)=0$$

and

$$f(5)=\frac{25}{20}$$

$$=\frac54$$

Thus,

$$0\le\alpha^2\le\frac54$$

Hence,

$$-\frac{\sqrt5}{2}\le\alpha\le\frac{\sqrt5}{2}$$

Therefore,

$$p=-\frac{\sqrt5}{2},\qquad q=\frac{\sqrt5}{2}$$

Now,

$$4(p^2+q^2)=4\left(\frac54+\frac54\right)$$

$$=4\cdot\frac52$$

$$=10$$

Therefore, the required value is

$$\boxed{10}$$