Let $$z \in C$$ be such that $$|z| \lt 1$$. If $$\omega = \frac{5 + 3z}{5(1 - z)}$$, then:

Let us write the unknown complex number in rectangular form as $$z = x + iy,$$ where $$x, y \in \mathbb R$$ and the given condition $$|z|\lt 1$$ becomes

$$x^{2} + y^{2} \lt 1.$$

The expression for $$\omega$$ is

$$\omega = \dfrac{5 + 3z}{5(1 - z)}.$$

Substituting $$z = x + iy$$ we have

Numerator: $$5 + 3z \;=\; 5 + 3(x + iy) \;=\; (5 + 3x) + i\,3y.$$

Denominator: $$5(1 - z) \;=\; 5\bigl(1 - (x + iy)\bigr) \;=\; 5\bigl[(1 - x) - iy\bigr].$$

To separate real and imaginary parts we multiply the fraction by the conjugate of the denominator and use the standard identity

For any $$a+ib$$ and $$c+id,$$

$$(a+ib)(c-id) = (ac + bd) + i(bc - ad).$$

First write

$$\omega = \dfrac{(5 + 3x) + i\,3y}{5\bigl[(1 - x) - iy\bigr]} \;=\; \dfrac{\bigl[(5 + 3x) + i\,3y\bigr]\bigl[5(1 - x) + i\,5y\bigr]} {\,\bigl|5(1 - z)\bigr|^{2}}.$$

The conjugate of $$5\bigl[(1 - x) - iy\bigr]$$ is $$5\bigl[(1 - x) + iy\bigr].$$ Multiplying the numerators and applying the formula stated above gives

Real part of the new numerator:

$$$ (5 + 3x)\,5(1 - x) \;-\; (3y)\,5y \;=\; 5\,(5 + 3x)(1 - x) \;-\; 15y^{2}. $$$

Imaginary part of the new numerator:

$$$ (5 + 3x)\,5y \;+\; (3y)\,5(1 - x) \;=\; 5y\bigl[(5 + 3x) + 3(1 - x)\bigr] \;=\; 5y\,[\,5 + 3x + 3 - 3x\,] \;=\; 40y. $$$

Thus the expanded numerator is

$$\bigl[\,5(5 + 3x)(1 - x) - 15y^{2}\bigr] \;+\; i\,40y.$$

The squared modulus of the denominator is

$$$ \bigl|5(1 - z)\bigr|^{2} = 25\,\bigl|(1 - x) - i\,y\bigr|^{2} = 25\bigl[(1 - x)^{2} + y^{2}\bigr]. $$$ \]

Hence

$$$ \omega = \dfrac{\,\bigl[\,5(5 + 3x)(1 - x) - 15y^{2}\bigr] + i\,40y\,} {25\bigl[(1 - x)^{2} + y^{2}\bigr]}. $$$

Therefore

$$$ \operatorname{Re}(\omega) = \dfrac{5(5 + 3x)(1 - x) - 15y^{2}} {25\bigl[(1 - x)^{2} + y^{2}\bigr]},\qquad \operatorname{Im}(\omega) = \dfrac{40y} {25\bigl[(1 - x)^{2} + y^{2}\bigr]}. $$$

Now we consider the quantity mentioned in option A:

$$$ 5\,\operatorname{Re}(\omega) = \dfrac{5\Bigl[\,5(5 + 3x)(1 - x) - 15y^{2}\Bigr]} {25\bigl[(1 - x)^{2} + y^{2}\bigr]} = \dfrac{5(5 + 3x)(1 - x) - 15y^{2}} {5\bigl[(1 - x)^{2} + y^{2}\bigr]}. $$$

Simplifying the numerator:

$$$ (5 + 3x)(1 - x) = 5(1 - x) + 3x(1 - x) = 5 - 5x + 3x - 3x^{2} = 5 - 2x - 3x^{2}. $$$

Hence

$$$ 5(5 + 3x)(1 - x) - 15y^{2} = 5(5 - 2x - 3x^{2}) - 15y^{2} = 25 - 10x - 15x^{2} - 15y^{2}. $$$

So

$$$ 5\,\operatorname{Re}(\omega) = \dfrac{25 - 10x - 15x^{2} - 15y^{2}} {5\bigl[(1 - x)^{2} + y^{2}\bigr]}. $$$

Expand the denominator factor:

$$$ 5\bigl[(1 - x)^{2} + y^{2}\bigr] = 5\bigl[1 - 2x + x^{2} + y^{2}\bigr] = 5 - 10x + 5x^{2} + 5y^{2}. $$$

To prove the inequality $$5\operatorname{Re}(\omega) \gt 1,$$ we compare numerator and denominator:

$$$ \text{Numerator} - \text{Denominator} = \bigl[25 - 10x - 15(x^{2} + y^{2})\bigr] - \bigl[5 - 10x + 5(x^{2} + y^{2})\bigr]. $$$

Simplifying term by term, the $$-10x$$ parts cancel, and we get

$$$ 25 - 5 - 15(x^{2} + y^{2}) - 5(x^{2} + y^{2}) = 20 - 20(x^{2} + y^{2}) = 20\bigl[1 - (x^{2} + y^{2})\bigr]. $$$

But $$x^{2} + y^{2} = |z|^{2} \lt 1,$$ so $$1 - (x^{2} + y^{2}) \gt 0.$$ Therefore

$$$ \text{Numerator} - \text{Denominator} \gt 0 \quad\Longrightarrow\quad \text{Numerator} \;\gt \; \text{Denominator}, $$$

which means

$$$ \dfrac{\text{Numerator}}{\text{Denominator}} \gt 1 \quad\Longrightarrow\quad 5\,\operatorname{Re}(\omega) \gt 1. $$$

Thus the inequality in option A is satisfied for every complex number $$z$$ with $$|z|\lt 1.$$

The remaining options do not hold in general: by choosing suitable values of $$x$$ and $$y$$ (for example, $$z = 0.8i$$ or $$z = -0.9$$) one can check that options B, C, and D are violated, while option A always remains true.

Hence, the correct answer is Option A.