If `a_(n+1)=1/(1-a_n)` for `n>=1` and `a_3=a_1`. then find the value of `(a_2001)^2001`.

Explanation

To solve the problem, we need to analyze the recurrence relation given by \( a_{n+1} = \frac{1}{1 - a_n} \) and the condition \( a_3 = a_1 \).

### Step-by-Step Solution:

1. **Start with the recurrence relation**:
\[ a_{n+1} = \frac{1}{1 - a_n} \]

2. **Calculate \( a_2 \)**:
\[ a_2 = \frac{1}{1 - a_1} \]

3. **Calculate \( a_3 \)**:
\[ a_3 = \frac{1}{1 - a_2} = \frac{1}{1 - \frac{1}{1 - a_1}} = \frac{1 - a_1}{-a_1} = \frac{1 - a_1}{-a_1} \]

4. **Set \( a_3 = a_1 \)** (given condition):
\[ a_1 = \frac{1 - a_1}{-a_1} \]

5. **Cross-multiply to eliminate the fraction**:
\[ a_1 \cdot (-a_1) = 1 - a_1 \]
\[ -a_1^2 = 1 - a_1 \]

6. **Rearrange the equation**:
\[ a_1^2 - a_1 + 1 = 0 \]

7. **Use the quadratic formula to find \( a_1 \)**:
\[ a_1 = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{1 \pm \sqrt{1 - 4}}{2} = \frac{1 \pm \sqrt{-3}}{2} \]
\[ a_1 = \frac{1 \pm i\sqrt{3}}{2} \]

8. **Since \( a_3 = a_1 \), we have \( a_2 = a_1 \) as well**:
\[ a_1 = a_2 = a_3 = \ldots = a_n \text{ for all } n \]

9. **Determine \( a_{2001} \)**:
\[ a_{2001} = a_1 = \frac{1 + i\sqrt{3}}{2} \text{ or } a_{2001} = \frac{1 - i\sqrt{3}}{2} \]

10. **Calculate \( (a_{2001})^{2001} \)**:
\[ (a_{2001})^{2001} = \left(\frac{1 + i\sqrt{3}}{2}\right)^{2001} \text{ or } \left(\frac{1 - i\sqrt{3}}{2}\right)^{2001} \]

11. **Since both roots are complex conjugates, their powers will yield the same magnitude**:
\[ |a_{2001}| = 1 \quad \text{(as the modulus of both roots is 1)} \]
\[ (a_{2001})^{2001} = (-1)^{2001} = -1 \quad \text{(as the argument will rotate)} \]

### Final Answer:
\[ (a_{2001})^{2001} = -1 \]