If `alpha` is a complex number such that `alpha^(2)+alpha+1=0,` then what is `alpha^(31)` equal to ?

Options

A. `alpha`
B. `alpha^(2)`
C. 0
D. 1

Explanation

To solve the problem, we need to find \( \alpha^{31} \) given that \( \alpha \) is a complex number satisfying the equation \( \alpha^2 + \alpha + 1 = 0 \).

### Step 1: Solve the quadratic equation
We start with the equation:
\[ \alpha^2 + \alpha + 1 = 0 \]
Using the quadratic formula \( \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 1, c = 1 \):
\[ \alpha = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm \sqrt{-3}}{2} \]
This simplifies to:
\[ \alpha = \frac{-1 \pm \sqrt{3}i}{2} \]

### Step 2: Identify the roots
The roots of the equation are:
\[ \alpha_1 = \frac{-1 + \sqrt{3}i}{2}, \quad \alpha_2 = \frac{-1 - \sqrt{3}i}{2} \]
These roots are known as the non-real cube roots of unity, often denoted as \( \omega \) and \( \omega^2 \), where:
\[ \omega = \frac{-1 + \sqrt{3}i}{2}, \quad \omega^2 = \frac{-1 - \sqrt{3}i}{2} \]

### Step 3: Use properties of cube roots of unity
The cube roots of unity satisfy the property:
\[ \omega^3 = 1 \quad \text{and} \quad \omega^2 + \omega + 1 = 0 \]
This means that any power of \( \omega \) can be reduced modulo 3.

### Step 4: Calculate \( \alpha^{31} \)
Now, we need to find \( \alpha^{31} \). Since \( \alpha \) could be either \( \omega \) or \( \omega^2 \), we will calculate both cases.

1. **If \( \alpha = \omega \):**
\[ \alpha^{31} = \omega^{31} \]
To reduce \( 31 \) modulo \( 3 \):
\[ 31 \mod 3 = 1 \]
Thus:
\[ \omega^{31} = \omega^1 = \omega \]

2. **If \( \alpha = \omega^2 \):**
\[ \alpha^{31} = (\omega^2)^{31} = \omega^{62} \]
Reducing \( 62 \) modulo \( 3 \):
\[ 62 \mod 3 = 2 \]
Thus:
\[ \omega^{62} = \omega^2 \]

### Conclusion
In both cases, whether \( \alpha = \omega \) or \( \alpha = \omega^2 \):
\[ \alpha^{31} = \alpha \]

Thus, the final answer is:
\[ \alpha^{31} = \alpha \]