If `z + z^(-1)= 1`, then find the value of `z^(100) + z^(-100)`.

Explanation

To solve the equation \( z + z^{-1} = 1 \) and find the value of \( z^{100} + z^{-100} \), we can follow these steps:

### Step 1: Rewrite the equation
Given:
\[ z + \frac{1}{z} = 1 \]
Multiply both sides by \( z \) (assuming \( z \neq 0 \)):
\[ z^2 + 1 = z \]

### Step 2: Rearrange into standard quadratic form
Rearranging gives:
\[ z^2 - z + 1 = 0 \]

### Step 3: Apply the quadratic formula
Using the quadratic formula \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1 \), \( b = -1 \), and \( c = 1 \):
\[ z = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
\[ z = \frac{1 \pm \sqrt{1 - 4}}{2} \]
\[ z = \frac{1 \pm \sqrt{-3}}{2} \]
\[ z = \frac{1 \pm i\sqrt{3}}{2} \]

### Step 4: Identify the roots
The roots are:
\[ z_1 = \frac{1 + i\sqrt{3}}{2}, \quad z_2 = \frac{1 - i\sqrt{3}}{2} \]

### Step 5: Use De Moivre's Theorem
To find \( z^{100} + z^{-100} \), we note that \( z_1 \) and \( z_2 \) can be expressed in polar form. The modulus \( r \) of \( z_1 \) is:
\[ r = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = 1 \]
The argument \( \theta \) is:
\[ \theta = \tan^{-1}\left(\frac{\sqrt{3}/2}{1/2}\right) = \frac{\pi}{3} \]
Thus, \( z_1 = e^{i\pi/3} \) and \( z_2 = e^{-i\pi/3} \).

### Step 6: Calculate \( z^{100} + z^{-100} \)
Using De Moivre's Theorem:
\[ z_1^{100} = \left(e^{i\pi/3}\right)^{100} = e^{i(100 \cdot \frac{\pi}{3})} = e^{i\frac{100\pi}{3}} = e^{i(33\pi + \frac{\pi}{3})} = e^{i\frac{\pi}{3}} = z_1 \]
\[ z_2^{100} = \left(e^{-i\pi/3}\right)^{100} = e^{-i(100 \cdot \frac{\pi}{3})} = e^{-i\frac{100\pi}{3}} = e^{-i(33\pi + \frac{\pi}{3})} = e^{-i\frac{\pi}{3}} = z_2 \]
Thus:
\[ z^{100} + z^{-100} = z_1 + z_2 \]

### Step 7: Find the sum
From the roots:
\[ z_1 + z_2 = \frac{1 + i\sqrt{3}}{2} + \frac{1 - i\sqrt{3}}{2} = 1 \]

### Conclusion
Therefore, the value of \( z^{100} + z^{-100} \) is:
\[ \boxed{1} \]