If z is a complex number satisfying `z + z^-1 = 1` then `z^n + z^-n , n in N`, has the value

Options

A. ` 2(-1)^(n)` when n is a multiple of 3
B. `(-1)^(n-1)` , when n is not a multiple of 3
C. `(-1)^(n+1)` ,when n is a mutliple of 3.
D. 0 when n is not a multiple of 3.

Answer

To solve the problem where \( z \) is a complex number satisfying \( z + z^{-1} = 1 \) and we need to find the value of \( z^n + z^{-n} \) for \( n \in \mathbb{N} \), we can follow these steps:

### Step 1: Rewrite the equation
Starting from the equation:
\[ z + z^{-1} = 1 \]
we can multiply both sides by \( z \) (assuming \( z \neq 0 \)):
\[ z^2 + 1 = z \]
Rearranging gives:
\[ z^2 - z + 1 = 0 \]

### Step 2: Solve the quadratic equation
Now we can use the quadratic formula \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = -1, c = 1 \):
\[ z = \frac{1 \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{1 \pm \sqrt{1 - 4}}{2} = \frac{1 \pm \sqrt{-3}}{2} \]
This simplifies to:
\[ z = \frac{1 \pm i\sqrt{3}}{2} \]
Thus, the two roots are:
\[ z_1 = \frac{1 + i\sqrt{3}}{2}, \quad z_2 = \frac{1 - i\sqrt{3}}{2} \]

### Step 3: Convert to polar form
Next, we convert these complex numbers to polar form. The modulus \( r \) of \( z_1 \) is:
\[ r = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1 \]
The argument \( \theta \) is:
\[ \theta = \tan^{-1}\left(\frac{\sqrt{3}/2}{1/2}\right) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \]
Thus, we can express \( z_1 \) in polar form as:
\[ z_1 = e^{i\pi/3} \]
Similarly, for \( z_2 \):
\[ z_2 = e^{-i\pi/3} \]

### Step 4: Calculate \( z^n + z^{-n} \)
Now we can find \( z^n + z^{-n} \):
\[ z^n = (e^{i\pi/3})^n = e^{i n \pi/3}, \quad z^{-n} = (e^{-i\pi/3})^n = e^{-i n \pi/3} \]
Thus,
\[ z^n + z^{-n} = e^{i n \pi/3} + e^{-i n \pi/3} = 2 \cos\left(\frac{n \pi}{3}\right) \]

### Conclusion
Therefore, the value of \( z^n + z^{-n} \) is:
\[ \boxed{2 \cos\left(\frac{n \pi}{3}\right)} \]