Step 1: Find the roots first.
The roots of $x^2 - 2x + 4 = 0$ are complex, and complex roots become easy when written as an angle.
Step 2: Apply the quadratic formula.
\[ x = \frac{2 \pm \sqrt{4 - 16}}{2} = 1 \pm i\sqrt3 \] So $\alpha = 1 + i\sqrt3$ and $\beta = 1 - i\sqrt3$.
Step 3: Write in polar form.
Each root has size 2 and angle $\frac{\pi}{3}$: \[ \alpha = 2\left(\cos\tfrac{\pi}{3} + i\sin\tfrac{\pi}{3}\right), \quad \beta = 2\left(\cos\tfrac{\pi}{3} - i\sin\tfrac{\pi}{3}\right) \]
Step 4: Raise to the power $n$.
By De Moivre, multiply the angle by $n$ and the size to the power $n$: \[ \alpha^n = 2^n\left(\cos\tfrac{n\pi}{3} + i\sin\tfrac{n\pi}{3}\right) \] and $\beta^n$ has the same form with a minus sign.
Step 5: Add them.
The imaginary parts cancel and the real parts add: \[ \alpha^n + \beta^n = 2^n \cdot 2\cos\tfrac{n\pi}{3} \]
Step 6: Tidy up.
\[ \boxed{ \alpha^n + \beta^n = 2^{n+1}\cos\frac{n\pi}{3} } \]