### Instruction: Let $\omega$ be a complex number such that
\[\omega + \frac{1}{\omega} = 1.\]Find all possible values of
\[\omega^n + \frac{1}{\omega^n},\]where $n$ is a positive integer.

Enter all possible values, separated by commas.### Response: I notice that the given equation is quadratic in $\omega$, so I can try to solve for $\omega$ by multiplying both sides by $\omega$ and rearranging.
I get $\omega^2 - \omega + 1 = 0$, which has solutions $\omega = \frac{1 \pm i\sqrt{3}}{2}$ by the quadratic formula.
These are complex numbers in polar form with magnitude $1$ and argument $\pm \frac{\pi}{3}$, so I can write them as $\omega = e^{\pm i\pi/3}$.
This means that $\omega^n = e^{\pm in\pi/3}$ and $\frac{1}{\omega^n} = e^{\mp in\pi/3}$, so I can use the addition formula for complex exponentials to find $\omega^n + \frac{1}{\omega^n}$.
I get $\omega^n + \frac{1}{\omega^n} = e^{\pm in\pi/3} + e^{\mp in\pi/3} = 2\cos(n\pi/3)$, where I used the fact that $\cos(\theta) = \frac{e^{i\theta} + e^{-i\theta}}{2}$.
Now I just need to find the possible values of $2\cos(n\pi/3)$ for positive integers $n$.
I know that $\cos(n\pi/3)$ is periodic with period $6$, so I only need to check the values of $n$ from $1$ to $6$.
I get $\cos(\pi/3) = \frac{1}{2}$, $\cos(2\pi/3) = -\frac{1}{2}$, $\cos(\pi) = -1$, $\cos(4\pi/3) = -\frac{1}{2}$, $\cos(5\pi/3) = \frac{1}{2}$, and $\cos(2\pi) = 1$.
Multiplying by $2$, I get the possible values of $2\cos(n\pi/3)$ as $1, -1, -2, 2$.
Therefore, the possible values of $\omega^n + \frac{1}{\omega^n}$ are $1, -1, -2, 2$.