Prove the solutions to the equation $z^n=1$. Fix a positive integer $n$. Prove that the solutions to the equation $z^n=1$ are precisely
$$z=e^{2\pi i \frac{m}{n}}$$
where $m \in \mathbb Z$. 
$Hint:$ To show that every solution of $z^n=1$ is of this form, first prove that it must be of the form $z=e^{2\pi i \frac{a}{n}}$ for some $a \in \mathbb R$, then write $a=m+b$ for some integer $m$ and some real number $0 \leq b <1$, and then argue $b=0$.
I am confused about the hint, since by writing out
$$1=e^{i2m\pi}$$
where $m \in \mathbb Z$ we can get $$z=e^{2\pi i \frac{m}{n}}$$ immediately. So what does the hint mean? Thank you for any help!
 A: The hint must have a typo, I think it should be " first show that the solution has the form $e^{2\pi i a}$ (instead of $e^{2\pi i \frac{a}{n}}$) for some $a\in\mathbb{R}$".
Here is how you should do it:
Let $w$ be a solution, then $w^n=1$ in particular $|w|=1$.
So, there exists $t\in \mathbb{R}$ such that $w=e^{2\pi i t}$, since $w^n=1$ we have that $nt\in\mathbb{Z}$ so $t=\frac{m}{n}$ for an integer $m\in\mathbb{Z}$.
A: I encountered this problem too. I feel that the hint is not wrong and it really helped as well. After giving it much thought, I was wondering if this question should be done in this way :)
We let $z=e^{2\pi i\frac an}$, where $a \in \mathbb R$ and $n \in \mathbb Z$ . Certainly, $z^n=e^{2\pi i a}$. Now consider
$|z^n|=|e^{2\pi i a}$|. Re-writing the exponential form into polar form will give us
$|z^n|=\sqrt {cos^2 (2\pi i a)+sin^2 (2\pi ia)}=1$ by using the Pythagorean's trigonometric identity. This tells us that the solution must be in the form $z=e^{2\pi i\frac an}$ simply because $z^n = 1$ so $|z^n|=1$.
Now we let $a=b+m$, where $m \in \mathbb Z$, $0 \le b \lt 1$.
We prove by contradiction. Suppose that $0 \lt b \lt 1$, in other words, $b \neq 0$. This shows that $z^n=z^{2 \pi i (b+m)}=z^{2 \pi i b}.z^{2 \pi i m}$.
Since $m \in \mathbb Z$, we know that $e^{2 \pi i m}=1$. (You can write in polar form and show that it is indeed equals to $1$). So we have simplified our equation to be $z^n = e^{2 \pi i b}$, where $0 \lt b \lt 1$. Now clearly, $e^{2 \pi i b} \neq 1$ because $b$ has not even rotated one complete round. So $z^n \neq 1$, which is a contradiction to our statement.
Hence, $b = 0$. Therefore, $z^n = z^{2 \pi i (b+m)} = z^{2 \pi i m}$. Since $m, n \in \mathbb Z$, then $z = e^{2 \pi i \frac mn}$.
Hope it helps!
