For all complex numbers z of the form $$1 + i\alpha$$, $$\alpha \in R$$, if $$z^2 = x + iy$$, then:

Given that $$ z $$ is a complex number of the form $$ 1 + i\alpha $$, where $$ \alpha $$ is a real number, so $$ z = 1 + i\alpha $$. We are told that $$ z^2 = x + iy $$, and we need to find the relation between $$ x $$ and $$ y $$.

First, compute $$ z^2 $$:

$$$ z^2 = (1 + i\alpha)^2 $$$

Expand the square:

$$$ (1 + i\alpha)^2 = 1^2 + 2 \cdot 1 \cdot (i\alpha) + (i\alpha)^2 = 1 + 2i\alpha + i^2 \alpha^2 $$$

Since $$ i^2 = -1 $$, substitute:

$$$ 1 + 2i\alpha + (-1)\alpha^2 = 1 + 2i\alpha - \alpha^2 $$$

Group the real and imaginary parts:

$$$ z^2 = (1 - \alpha^2) + (2\alpha)i $$$

But $$ z^2 = x + iy $$, so comparing real and imaginary parts:

$$$ x = 1 - \alpha^2 \quad \text{(real part)} $$$

$$$ y = 2\alpha \quad \text{(imaginary part)} $$$

We have two equations:

1. $$ x = 1 - \alpha^2 $$

2. $$ y = 2\alpha $$

We need to eliminate $$ \alpha $$ to find a relation between $$ x $$ and $$ y $$. From the second equation, solve for $$ \alpha $$:

$$$ \alpha = \frac{y}{2} $$$

Substitute this into the first equation:

$$$ x = 1 - \left( \frac{y}{2} \right)^2 $$$

Compute the square:

$$$ x = 1 - \frac{y^2}{4} $$$

To eliminate the fraction, multiply both sides by 4:

$$$ 4x = 4 \cdot 1 - 4 \cdot \frac{y^2}{4} $$$

Simplify:

$$$ 4x = 4 - y^2 $$$

Rearrange all terms to one side:

$$$ y^2 + 4x - 4 = 0 $$$

Now, compare this equation with the given options:

A. $$ y^2 - 4x + 4 = 0 $$

B. $$ y^2 + 4x - 4 = 0 $$

C. $$ y^2 - 4x + 2 = 0 $$

D. $$ y^2 + 4x + 2 = 0 $$

Our equation $$ y^2 + 4x - 4 = 0 $$ matches option B.

Hence, the correct answer is Option B.