If $$\frac{z - \alpha}{z + \alpha}$$ $$(\alpha \in R)$$ is a purely imaginary number and $$|z| = 2$$, then a value of $$\alpha$$ is:

Let us denote the unknown complex number by $$z=x+iy$$ where $$x,\,y\in\mathbb R$$.

We are given that the modulus of $$z$$ equals $$2$$, so by the definition of modulus we have

$$|z|=\sqrt{x^{2}+y^{2}}=2\; \Longrightarrow\; x^{2}+y^{2}=4.$$

The expression to be examined is

$$\frac{z-\alpha}{\,z+\alpha\,}, \qquad \text{with } \alpha\in\mathbb R,$$

and it is stated to be a purely imaginary number. A complex number is purely imaginary precisely when its real part equals zero. Hence we require

$$\operatorname{Re}\!\left(\frac{z-\alpha}{\,z+\alpha\,}\right)=0.$$

First we write the numerator and denominator in terms of $$x$$ and $$y$$:

$$z-\alpha=(x-\alpha)+iy, \qquad z+\alpha=(x+\alpha)+iy.$$

To eliminate the complex denominator, we multiply the fraction by the conjugate of the denominator divided by itself:

$$\frac{z-\alpha}{\,z+\alpha\,}= \frac{(x-\alpha)+iy}{(x+\alpha)+iy}\; \frac{(x+\alpha)-iy}{(x+\alpha)-iy}.$$

The denominator now becomes real:

$$[(x+\alpha)+iy][(x+\alpha)-iy]=(x+\alpha)^{2}+y^{2}.$$

Expanding the numerator gives

$$$\begin{aligned} &[(x-\alpha)+iy][(x+\alpha)-iy] \\[2pt] &=(x-\alpha)(x+\alpha)+y^{2}+i\bigl[y(x+\alpha)-y(x-\alpha)\bigr]. \end{aligned}$$$

Compute each part separately:

Real part of numerator: $$(x-\alpha)(x+\alpha)+y^{2}=x^{2}-\alpha^{2}+y^{2}.$$

Imaginary part of numerator: $$y(x+\alpha)-y(x-\alpha)=y(x+\alpha-x+\alpha)=2\alpha y.$$

Thus

$$\frac{z-\alpha}{\,z+\alpha\,}= \frac{\bigl(x^{2}+y^{2}-\alpha^{2}\bigr)\;+\;i\,(2\alpha y)}{(x+\alpha)^{2}+y^{2}}.$$

The denominator $$(x+\alpha)^{2}+y^{2}$$ is a positive real number, so the real part of the entire fraction equals the real part of the numerator divided by this positive quantity. Therefore, for the whole expression to be purely imaginary we must have

$$x^{2}+y^{2}-\alpha^{2}=0.$$

But from the modulus condition we already know that $$x^{2}+y^{2}=4,$$ hence

$$4-\alpha^{2}=0 \quad\Longrightarrow\quad \alpha^{2}=4.$$

Taking square roots (and remembering $$\alpha\in\mathbb R$$) gives

$$\alpha=\pm 2.$$

Among the options listed, only $$\alpha=2$$ appears.

Hence, the correct answer is Option D.