If `alpha, beta` are the roots of `x^2 - sqrt(3)x + 1 = 0` then `alpha^(21) + beta^(21)` is:

Options

A. `0`
B. `sqrt(3)`
C. `-sqrt(3)`
D. `1`

Explanation

To solve the problem, we need to find the value of \( \alpha^{21} + \beta^{21} \), where \( \alpha \) and \( \beta \) are the roots of the quadratic equation \( x^2 - \sqrt{3}x + 1 = 0 \).

### Step 1: Find the roots of the quadratic equation
The roots of the quadratic equation \( ax^2 + bx + c = 0 \) can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For our equation, \( a = 1 \), \( b = -\sqrt{3} \), and \( c = 1 \).

Substituting these values into the formula gives:
\[ x = \frac{\sqrt{3} \pm \sqrt{(\sqrt{3})^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
Calculating the discriminant:
\[ (\sqrt{3})^2 - 4 \cdot 1 \cdot 1 = 3 - 4 = -1 \]
Thus, we have:
\[ x = \frac{\sqrt{3} \pm \sqrt{-1}}{2} = \frac{\sqrt{3}}{2} \pm \frac{i}{2} \]
So, the roots are:
\[ \alpha = \frac{\sqrt{3}}{2} + \frac{i}{2}, \quad \beta = \frac{\sqrt{3}}{2} - \frac{i}{2} \]

### Step 2: Express the roots in exponential form
We can express \( \alpha \) and \( \beta \) in terms of trigonometric functions:
\[ \alpha = \cos\left(\frac{\pi}{6}\right) + i \sin\left(\frac{\pi}{6}\right), \quad \beta = \cos\left(\frac{\pi}{6}\right) - i \sin\left(\frac{\pi}{6}\right) \]
This means:
\[ \alpha = e^{i \frac{\pi}{6}}, \quad \beta = e^{-i \frac{\pi}{6}} \]

### Step 3: Calculate \( \alpha^{21} \) and \( \beta^{21} \)
Using the property of exponents:
\[ \alpha^{21} = \left(e^{i \frac{\pi}{6}}\right)^{21} = e^{i \frac{21\pi}{6}} = e^{i \frac{7\pi}{2}} \]
\[ \beta^{21} = \left(e^{-i \frac{\pi}{6}}\right)^{21} = e^{-i \frac{21\pi}{6}} = e^{-i \frac{7\pi}{2}} \]

### Step 4: Simplify \( \alpha^{21} + \beta^{21} \)
Now, we can add \( \alpha^{21} \) and \( \beta^{21} \):
\[ \alpha^{21} + \beta^{21} = e^{i \frac{7\pi}{2}} + e^{-i \frac{7\pi}{2}} \]
Using Euler's formula, this simplifies to:
\[ \alpha^{21} + \beta^{21} = 2 \cos\left(\frac{7\pi}{2}\right) \]

### Step 5: Find \( \cos\left(\frac{7\pi}{2}\right) \)
To find \( \cos\left(\frac{7\pi}{2}\right) \), we can reduce it:
\[ \frac{7\pi}{2} = 3\pi + \frac{\pi}{2} \quad \text{(since } 3\pi \text{ is } 2\pi + \pi\text{)} \]
Thus:
\[ \cos\left(\frac{7\pi}{2}\right) = \cos\left(\frac{\pi}{2}\right) = 0 \]

### Step 6: Final result
Substituting back, we find:
\[ \alpha^{21} + \beta^{21} = 2 \cdot 0 = 0 \]

Thus, the final answer is:
\[ \boxed{0} \]