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 \( \alpha^{21} + \beta^{21} \) where \( \alpha \) and \( \beta \) are the roots of the equation \( x^2 - \sqrt{3}x + 1 = 0 \).

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

Calculating the discriminant:
\[ b^2 - 4ac = (-\sqrt{3})^2 - 4 \cdot 1 \cdot 1 = 3 - 4 = -1 \]
Since the discriminant is negative, the roots are complex.

Now substituting into the quadratic formula:
\[ x = \frac{\sqrt{3} \pm \sqrt{-1}}{2} = \frac{\sqrt{3} \pm i}{2} \]
Thus, 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 polar form
We can express \( \alpha \) and \( \beta \) in polar form using Euler's formula. The modulus \( r \) and argument \( \theta \) can be calculated as follows:
- The modulus \( r \) is given by:
\[ r = \sqrt{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4} + \frac{1}{4}} = \sqrt{1} = 1 \]
- The argument \( \theta \) for \( \alpha \) is:
\[ \theta = \tan^{-1}\left(\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \]
Thus, we can write:
\[ \alpha = e^{i\frac{\pi}{6}}, \quad \beta = e^{-i\frac{\pi}{6}} \]

### Step 3: Use De Moivre's Theorem
According to De Moivre's theorem:
\[ \alpha^n = \left(e^{i\frac{\pi}{6}}\right)^n = e^{i\frac{n\pi}{6}}, \quad \beta^n = \left(e^{-i\frac{\pi}{6}}\right)^n = e^{-i\frac{n\pi}{6}} \]
Thus,
\[ \alpha^{21} = e^{i\frac{21\pi}{6}}, \quad \beta^{21} = e^{-i\frac{21\pi}{6}} \]

### Step 4: Calculate \( \alpha^{21} + \beta^{21} \)
Now we need to find:
\[ \alpha^{21} + \beta^{21} = e^{i\frac{21\pi}{6}} + e^{-i\frac{21\pi}{6}} = 2 \cos\left(\frac{21\pi}{6}\right) \]
We simplify \( \frac{21\pi}{6} \):
\[ \frac{21\pi}{6} = \frac{21}{6} \cdot \pi = \frac{7\pi}{2} \]
Now, we can reduce \( \frac{7\pi}{2} \) to an equivalent angle within the range \( [0, 2\pi) \):
\[ \frac{7\pi}{2} - 3\pi = \frac{7\pi - 6\pi}{2} = \frac{\pi}{2} \]
Thus,
\[ \cos\left(\frac{21\pi}{6}\right) = \cos\left(\frac{\pi}{2}\right) = 0 \]
Therefore,
\[ \alpha^{21} + \beta^{21} = 2 \cdot 0 = 0 \]

### Final Answer
\[ \alpha^{21} + \beta^{21} = 0 \]