### Instruction: Let $\omega$ be a nonreal root of $z^3 = 1.$  Find the number of different possible values of
\[(\omega + 1)^n,\]where $n$ is a positive integer.### Response: To find the possible values of $(\omega + 1)^n,$ I need to know what $\omega$ is.
I know that $z^3 = 1$ has three roots: $z = 1,$ $z = \frac{-1 + i\sqrt{3}}{2},$ and $z = \frac{-1 - i\sqrt{3}}{2}.$
Since $\omega$ is not real, it must be either $\frac{-1 + i\sqrt{3}}{2}$ or $\frac{-1 - i\sqrt{3}}{2}.$
I can use either one, since they are complex conjugates and have the same modulus and argument.
I will choose $\omega = \frac{-1 + i\sqrt{3}}{2}.$
Then $\omega + 1 = \frac{1 + i\sqrt{3}}{2},$ which is a complex number in polar form with modulus $1$ and argument $\frac{\pi}{3}.$
Therefore, $(\omega + 1)^n = (\frac{1 + i\sqrt{3}}{2})^n = 1^n \cdot e^{i \cdot \frac{\pi}{3} \cdot n} = e^{i \cdot \frac{\pi}{3} \cdot n}.$
This means that $(\omega + 1)^n$ is a complex number on the unit circle with argument $\frac{\pi}{3} \cdot n.$
The argument $\frac{\pi}{3} \cdot n$ can take infinitely many different values, but some of them will repeat as $n$ changes.
For example, $\frac{\pi}{3} \cdot 6 = 2\pi,$ which is the same as $\frac{\pi}{3} \cdot 0 = 0.$
In fact, the argument will repeat every time $n$ is a multiple of $6.$
Therefore, the number of different possible values of $(\omega + 1)^n$ is the same as the number of different remainders when $n$ is divided by $6.$
There are $6$ such remainders: $0, 1, 2, 3, 4, 5.$
So the answer is $6.$