If ` alpha, beta ` are the roots of the equations `x^(2) - 2 x + 4 = 0 , " then " alpha ^(n) + beta^(n) = 2 ^(n + 1) cos ( n phi // 3)`

Answer

To solve the problem, we need to find the expression for \( \alpha^n + \beta^n \) where \( \alpha \) and \( \beta \) are the roots of the equation \( x^2 - 2x + 4 = 0 \).

### Step 1: Find the roots of the quadratic equation
The roots of the equation \( x^2 - 2x + 4 = 0 \) can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -2 \), and \( c = 4 \).

Calculating the discriminant:
\[ b^2 - 4ac = (-2)^2 - 4 \cdot 1 \cdot 4 = 4 - 16 = -12 \]
Since the discriminant is negative, the roots are complex.

Now substituting into the quadratic formula:
\[ x = \frac{2 \pm \sqrt{-12}}{2} = \frac{2 \pm 2\sqrt{3}i}{2} = 1 \pm \sqrt{3}i \]
Thus, the roots are:
\[ \alpha = 1 + \sqrt{3}i \quad \text{and} \quad \beta = 1 - \sqrt{3}i \]

### Step 2: Express the roots in polar form
We can express \( \alpha \) and \( \beta \) in polar form. The modulus \( r \) of \( \alpha \) is:
\[ r = \sqrt{(1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \]
The argument \( \phi \) can be calculated as:
\[ \phi = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3} \]
Thus, we can write:
\[ \alpha = 2\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right) = 2e^{i\frac{\pi}{3}} \]
Similarly, for \( \beta \):
\[ \beta = 2\left(\cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\right) = 2e^{-i\frac{\pi}{3}} \]

### Step 3: Calculate \( \alpha^n + \beta^n \)
Using the polar forms:
\[ \alpha^n = \left(2e^{i\frac{\pi}{3}}\right)^n = 2^n e^{i\frac{n\pi}{3}} \]
\[ \beta^n = \left(2e^{-i\frac{\pi}{3}}\right)^n = 2^n e^{-i\frac{n\pi}{3}} \]
Now, adding these two:
\[ \alpha^n + \beta^n = 2^n e^{i\frac{n\pi}{3}} + 2^n e^{-i\frac{n\pi}{3}} = 2^n \left(e^{i\frac{n\pi}{3}} + e^{-i\frac{n\pi}{3}}\right) \]
Using Euler's formula, we know:
\[ e^{i\theta} + e^{-i\theta} = 2\cos\theta \]
Thus,
\[ \alpha^n + \beta^n = 2^n \cdot 2 \cos\left(\frac{n\pi}{3}\right) = 2^{n+1} \cos\left(\frac{n\pi}{3}\right) \]

### Final Result
We conclude that:
\[ \alpha^n + \beta^n = 2^{n+1} \cos\left(\frac{n\pi}{3}\right) \]