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

Answer

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

### Step-by-Step Solution:

1. **Identify the Roots of the Equation**:
The equation \( \alpha^2 + \alpha + 1 = 0 \) can be recognized as related to the cube roots of unity. The roots of this equation are the complex numbers \( \omega \) and \( \omega^2 \), where \( \omega = e^{2\pi i / 3} \) and \( \omega^2 = e^{-2\pi i / 3} \). These satisfy \( \omega^3 = 1 \).

2. **Use the Property of Cube Roots of Unity**:
Since \( \alpha \) satisfies \( \alpha^2 + \alpha + 1 = 0 \), we know that \( \alpha \) is either \( \omega \) or \( \omega^2 \). Both of these roots satisfy the property \( \alpha^3 = 1 \).

3. **Express \( \alpha^{31} \)**:
We can express \( \alpha^{31} \) in terms of \( \alpha^3 \):
\[ \alpha^{31} = \alpha^{30} \cdot \alpha = (\alpha^3)^{10} \cdot \alpha \]
Since \( \alpha^3 = 1 \):
\[ \alpha^{30} = (1)^{10} = 1 \]
Therefore:
\[ \alpha^{31} = 1 \cdot \alpha = \alpha \]

4. **Conclusion**:
Thus, \( \alpha^{31} = \alpha \).

### Final Answer:
\[ \alpha^{31} = \alpha \]