If a is a non-real fourth root of unity, then the value of `alpha^(4n-1)+alpha^(4n-2)+alpha^(4n-3), n in N` is

Options

A. 0
B. `-1`
C. 3
D. none of these

Explanation

To solve the problem, we need to evaluate the expression \( \alpha^{4n-1} + \alpha^{4n-2} + \alpha^{4n-3} \) where \( \alpha \) is a non-real fourth root of unity.

### Step-by-Step Solution:

1. **Identify the Fourth Roots of Unity**:
The fourth roots of unity are the solutions to the equation \( x^4 = 1 \). These roots are:
\[ 1, -1, i, -i \]
Since \( \alpha \) is a non-real fourth root of unity, we can choose \( \alpha = i \) or \( \alpha = -i \).

2. **Use the Property of Roots of Unity**:
We know that any fourth root of unity raised to the power of 4 equals 1:
\[ \alpha^4 = 1 \]
Therefore, we can reduce any exponent of \( \alpha \) modulo 4.

3. **Reduce the Exponents**:
We can express \( \alpha^{4n-1} \), \( \alpha^{4n-2} \), and \( \alpha^{4n-3} \) in terms of lower powers:
\[ \alpha^{4n-1} = \alpha^{-1}, \quad \alpha^{4n-2} = \alpha^{-2}, \quad \alpha^{4n-3} = \alpha^{-3} \]

4. **Combine the Terms**:
Now, we can rewrite the expression:
\[ \alpha^{4n-1} + \alpha^{4n-2} + \alpha^{4n-3} = \alpha^{-1} + \alpha^{-2} + \alpha^{-3} \]

5. **Substituting for \( \alpha \)**:
If we take \( \alpha = i \), we can calculate:
\[ \alpha^{-1} = \frac{1}{i} = -i, \quad \alpha^{-2} = \frac{1}{i^2} = -1, \quad \alpha^{-3} = \frac{1}{i^3} = i \]

6. **Calculate the Sum**:
Now we can sum these values:
\[ \alpha^{-1} + \alpha^{-2} + \alpha^{-3} = -i - 1 + i \]
The \( -i \) and \( i \) cancel each other out:
\[ -1 \]

7. **Final Result**:
Thus, the value of \( \alpha^{4n-1} + \alpha^{4n-2} + \alpha^{4n-3} \) is:
\[ \boxed{-1} \]