If z is a complex number such that `z+z^(-1)=1,` then what is the value of `z^(99)+z^(-99)` ?

Options

A. 1
B. `-1`
C. 2
D. `-2`

Explanation

To solve the problem where \( z \) is a complex number such that \( z + z^{-1} = 1 \), we need to find the value of \( z^{99} + z^{-99} \).

### Step-by-step Solution:

1. **Start with the given equation**:
\[ z + \frac{1}{z} = 1 \]

2. **Multiply both sides by \( z \)** to eliminate the fraction:
\[ z^2 + 1 = z \]

3. **Rearrange the equation** to form a standard quadratic equation:
\[ z^2 - z + 1 = 0 \]

4. **Use the quadratic formula** \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = -1, c = 1 \):
\[ z = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
\[ z = \frac{1 \pm \sqrt{1 - 4}}{2} \]
\[ z = \frac{1 \pm \sqrt{-3}}{2} \]
\[ z = \frac{1 \pm i\sqrt{3}}{2} \]

5. **Identify the roots**:
\[ z_1 = \frac{1 + i\sqrt{3}}{2}, \quad z_2 = \frac{1 - i\sqrt{3}}{2} \]

6. **Recognize the roots as complex numbers** in exponential form. These can be expressed in terms of \( \omega \), where \( \omega = e^{i\frac{\pi}{3}} \) (the cube root of unity):
\[ z_1 = \omega, \quad z_2 = \omega^2 \]

7. **Use the property of roots of unity**: Since \( \omega^3 = 1 \), we know \( z^{99} \) can be simplified:
\[ z^{99} = (\omega)^{99} = \omega^{99 \mod 3} = \omega^0 = 1 \]
\[ z^{-99} = (\omega^2)^{99} = (\omega^{99})^2 = 1^2 = 1 \]

8. **Combine the results**:
\[ z^{99} + z^{-99} = 1 + 1 = 2 \]

### Final Answer:
\[ \boxed{2} \]