If `alpha` is nonreal and `alpha= (1)^(1/5)` then the find the value of `2^(|1+alpha+alpha^2+alpha^-2-alpha^-1|)`

Options

A. 4
B. 2
C. 1
D. None of these

Answer

To solve the problem, we need to find the value of \( 2^{|1 + \alpha + \alpha^2 + \alpha^{-2} - \alpha^{-1}|} \) given that \( \alpha \) is a non-real fifth root of unity.

### Step-by-step Solution:

1. **Understanding Alpha**:
Since \( \alpha = 1^{1/5} \), we recognize that \( \alpha \) is a fifth root of unity. The fifth roots of unity are given by:
\[ \alpha_k = e^{2\pi ik/5} \quad \text{for } k = 0, 1, 2, 3, 4 \]
Here, \( \alpha \) is non-real, so we can take \( \alpha = e^{2\pi i/5} \) (one of the non-real roots).

2. **Using the Property of Roots of Unity**:
The sum of all fifth roots of unity is zero:
\[ 1 + \alpha + \alpha^2 + \alpha^3 + \alpha^4 = 0 \]
Therefore, we can express \( \alpha^3 + \alpha^4 \) as:
\[ \alpha^3 + \alpha^4 = - (1 + \alpha + \alpha^2) \]

3. **Finding \( \alpha^{-1} \) and \( \alpha^{-2} \)**:
The inverse of \( \alpha \) can be expressed as:
\[ \alpha^{-1} = e^{-2\pi i/5} = \alpha^4 \]
Similarly,
\[ \alpha^{-2} = e^{-4\pi i/5} = \alpha^3 \]

4. **Substituting Values**:
Now, substituting these into the expression \( 1 + \alpha + \alpha^2 + \alpha^{-2} - \alpha^{-1} \):
\[ 1 + \alpha + \alpha^2 + \alpha^3 - \alpha^4 \]
We can rearrange this as:
\[ 1 + \alpha + \alpha^2 + (-1 - \alpha - \alpha^2) = 0 \]

5. **Calculating the Modulus**:
Thus, we have:
\[ |1 + \alpha + \alpha^2 + \alpha^{-2} - \alpha^{-1}| = |0| = 0 \]

6. **Final Calculation**:
Now we substitute this back into the expression we need to evaluate:
\[ 2^{|1 + \alpha + \alpha^2 + \alpha^{-2} - \alpha^{-1}|} = 2^0 = 1 \]

### Conclusion:
The value of \( 2^{|1 + \alpha + \alpha^2 + \alpha^{-2} - \alpha^{-1}|} \) is \( 1 \).