Let $$\alpha, \beta$$ be real and $$z$$ be a complex number. If $$z^2 + \alpha z + \beta = 0$$ has two distinct roots on the line $$\mathrm{Re}\, z = 1$$, then it is necessary that:

Let the two roots lie on the vertical line $$\mathrm{Re}\,z = 1$$.
Hence each root can be written as $$z = 1 + it$$ where $$t \in \mathbb{R}$$.

Because the quadratic equation $$z^{2} + \alpha z + \beta = 0$$ has real coefficients $$\alpha, \beta$$, the non-real roots must occur in conjugate pairs.
Therefore the two distinct roots are $$z_1 = 1 + it, \;\; z_2 = 1 - it,$$ with $$t \neq 0$$ to keep them distinct.

Apply Vieta’s relations for the quadratic $$z^{2} + \alpha z + \beta = 0$$:

• Sum of roots: $$z_1 + z_2 = -\alpha.$$
Compute the sum: $$z_1 + z_2 = (1 + it) + (1 - it) = 2.$$
Hence $$-\alpha = 2 \;\;\Longrightarrow\;\; \alpha = -2.$$

• Product of roots: $$z_1 z_2 = \beta.$$
Compute the product: $$z_1 z_2 = (1 + it)(1 - it) = 1 + t^{2}.$$
Since $$t \neq 0,$$ we have $$t^{2} \gt 0,$$ giving $$\beta = 1 + t^{2} \gt 1.$$

Thus a necessary condition is $$\beta \in (1,\infty).$$

Among the given options, only Option C matches this requirement.

Option C which is: $$\beta \in (1, \infty)$$