Let z = a + ib be a complex number such that `a, b in Q, |z| = 1 , |Z^(2n)-1| =x`. Then `(-1)^(x)` is :

Options

A. positive real number
B. negative real number
C. a complex number not necessarily real
D. can’t decide

Answer

To solve the problem, we need to analyze the given complex number \( z = a + ib \) where \( a, b \in \mathbb{Q} \) and \( |z| = 1 \). We also need to find \( |z^{2n} - 1| = x \) and then determine \( (-1)^x \).

### Step-by-Step Solution:

1. **Understanding the Magnitude Condition**:
Since \( |z| = 1 \), we have:
\[ |z| = \sqrt{a^2 + b^2} = 1 \]
This implies:
\[ a^2 + b^2 = 1 \]

2. **Expressing \( z \) in Exponential Form**:
A complex number on the unit circle can be expressed as:
\[ z = e^{i\theta} \]
where \( \theta \) is a real number. Thus, we can write:
\[ a = \cos(\theta) \quad \text{and} \quad b = \sin(\theta) \]

3. **Calculating \( z^{2n} \)**:
We compute \( z^{2n} \):
\[ z^{2n} = (e^{i\theta})^{2n} = e^{i(2n\theta)} \]

4. **Finding \( |z^{2n} - 1| \)**:
Now we need to calculate \( |z^{2n} - 1| \):
\[ |z^{2n} - 1| = |e^{i(2n\theta)} - 1| \]
Using the formula for the magnitude of a complex number:
\[ |e^{i(2n\theta)} - 1| = \sqrt{(e^{i(2n\theta)} - 1)(e^{-i(2n\theta)} - 1)} = \sqrt{2 - 2\cos(2n\theta)} = 2|\sin(n\theta)| \]
Therefore, we have:
\[ x = 2|\sin(n\theta)| \]

5. **Evaluating \( (-1)^x \)**:
Now we need to evaluate \( (-1)^x \):
- If \( x = 2|\sin(n\theta)| \), then \( x \) is a non-negative real number.
- If \( |\sin(n\theta)| = 0 \), then \( x = 0 \) and \( (-1)^0 = 1 \).
- If \( |\sin(n\theta)| > 0 \), then \( x \) is positive and even, thus \( (-1)^x = 1 \).

6. **Conclusion**:
Since \( |\sin(n\theta)| \) can only take values in the range [0, 1], \( x \) will always be a non-negative even number or zero. Therefore:
\[ (-1)^x = 1 \]

### Final Answer:
\[ (-1)^x = 1 \]