We are given that $\alpha$ and $\beta$ are the roots of the quadratic equation:

$x^2 - 2x + 3 = 0.$

To solve for $\alpha$ and $\beta$, we can use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$ where $a = 1$, $b = -2$, and $c = 3$.

Substitute the values of $a$, $b$, and $c$ into the formula:

$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(3)}}{2(1)} = \frac{2 \pm \sqrt{4 - 12}}{2} = \frac{2 \pm \sqrt{-8}}{2}.$

Since $\sqrt{-8} = 2i\sqrt{2}$, we have:

$x = \frac{2 \pm 2i\sqrt{2}}{2} = 1 \pm i\sqrt{2}.$

Thus, the roots are $\alpha = 1 + i\sqrt{2}$ and $\beta = 1 - i\sqrt{2}$.

Step 1: Find $\frac{1}{(2 - \alpha)^4} + \frac{1}{(2 - \beta)^4}$

We need to find:

$\frac{1}{(2 - \alpha)^4} + \frac{1}{(2 - \beta)^4}.$

First, simplify $2 - \alpha$ and $2 - \beta$:

$2 - \alpha = 2 - (1 + i\sqrt{2}) = 1 - i\sqrt{2},$ $2 - \beta = 2 - (1 - i\sqrt{2}) = 1 + i\sqrt{2}.$

Now, calculate $(2 - \alpha)^4$ and $(2 - \beta)^4$. Since $2 - \alpha = \overline{2 - \beta}$, the two terms are conjugates.

Let $z = 1 - i\sqrt{2}$. Then:

$(2 - \alpha)^4 = z^4, \quad (2 - \beta)^4 = \overline{z}^4.$

Using De Moivre's theorem, we calculate the fourth power of a complex number. Express $z$ in polar form:

$z = 1 - i\sqrt{2} \quad \Rightarrow \quad |z| = \sqrt{1^2 + (\sqrt{2})^2} = \sqrt{3}.$

The argument $\theta$ is given by:

$\theta = \tan^{-1}\left(\frac{-\sqrt{2}}{1}\right).$

Thus, $z^4$ will have magnitude $|z|^4 = 3^2 = 9$ and argument $4\theta$.

Since $z^4$ and $\overline{z}^4$ have the same magnitude, both terms yield the same result, and:

$\frac{1}{(2 - \alpha)^4} + \frac{1}{(2 - \beta)^4} = \frac{2}{81}.$

The final answer is:

$\boxed{\frac{2}{81}}.$

Would you like further details or have any questions?

Here are 5 related questions to expand your understanding:

1. How does the quadratic formula apply to finding complex roots?
2. What is De Moivre's theorem, and how is it used with powers of complex numbers?
3. Can you derive the polar form of a complex number from its rectangular form?
4. How do you calculate the magnitude and argument of a complex number?
5. What are the properties of complex conjugates and how do they simplify calculations?

Tip: When dealing with complex numbers, using polar form simplifies raising them to powers or finding roots.