$$ \text{If } \alpha \text{ and } \beta \text{ are the roots of the equation } 2z^2-3z-2i=0 ,\; \text{ where } i=\sqrt{-1}, \text{ then }  16\cdot \operatorname{Re}\!\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right) \cdot \operatorname{Im}\!\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right) \text{ is equal to:} $$

The expression $$E$$ simplifies to:

$$E = \frac{\alpha^{11}(\alpha^8+1) + \beta^{11}(\beta^8+1)}{\alpha^{15}+\beta^{15}}$$.

From $$2z^2 - 3z - 2i = 0$$, we have $$\alpha + \beta = \frac{3}{2}$$ and $$\alpha\beta = -i$$.

Notice that $$(\alpha\beta)^4 = (-i)^4 = 1$$. This type of symmetry usually suggests $$E$$ simplifies to a value related to the roots' properties.

Through substitution and simplification of the symmetric powers:

$$E = \alpha^4 + \beta^4$$ (as $$\alpha^8 = (\alpha\beta)^4 \frac{\alpha^4}{\beta^4} = \frac{\alpha^4}{\beta^4}$$).

Using $$S_1 = \alpha+\beta = 1.5$$ and $$P = \alpha\beta = -i$$:

$$\alpha^2+\beta^2 = S_1^2 - 2P = \frac{9}{4} + 2i$$.

$$\alpha^4+\beta^4 = (\alpha^2+\beta^2)^2 - 2P^2 = (\frac{9}{4} + 2i)^2 - 2(-1) = (\frac{81}{16} - 4 + 9i) + 2 = \frac{17}{16} + 2 + 9i = \frac{49}{16} + 9i$$.

$$\text{Re}(E) = \frac{49}{16}, \text{Im}(E) = 9$$.

Result: $$16 \cdot \frac{49}{16} \cdot 9 = 49 \cdot 9 = \mathbf{441}$$. (Option A)