Accumulation Points of a Complex Sequence Let $z$ be a complex number of absolute value 1: $ z=e^{i\theta},  0 \le \theta \lt 2\pi$.  What are the accumulation points of the sequence $\lbrace z^n \rbrace$?  Distinguish between the case where $\theta$ is a rational multiple of $2\pi$ and the case when it is not.  Be sure to justify all of your assertions.
So far I have the following:
By DeMoivre's theorem, $z_n=\cos n\theta + i\sin n\theta$.
Also, $|z|=1$, so $z_n$ is on the unit circle for all $n$, and any value of $\theta$.  I suppose the thing that is tripping me up is the part about$\theta$ being a rational multiple of $2\pi$.  Is this the case when $\theta = 2\pi p/q$? If so, what difference does it make? 
 A: Hints: 
1) The subgroup $\theta\mathbb{Z}+2\pi\mathbb{Z}$ is dense in $\mathbb{R}$ if and only if $\frac{\theta}{2\pi}\not\in\mathbb{Q}$. 
If $\theta>0$, this can actually be refined as $\theta\mathbb{N}-2\pi\mathbb{N}$ is dense in $\mathbb{R}$ if and only if $\frac{\theta}{2\pi}\not\in\mathbb{Q}$.
In this case, use it to show that every point on the unit circle is an accumulation point of the sequence $(z^n)$.
2) If $\frac{\theta}{2\pi}\in\mathbb{Q}$, say $\theta=\frac{2\pi p}{q}$ with $p,q$ integers such that $0\leq p<q$, given your assumptions. what are the values taken by $z^n$ when $n$ ranges over the positive integers? The sequence is periodic of period $q$. So there are finitely many values. All are $q$ roots of unity. If you arranged to have $p$ and $q$ relatively prime, the sequence hits all the $q$ roots of unity over each period.
A: Suppose $z=e^{i\theta}$, where $\theta = \frac{3}{10}2\pi$ $= \left(\frac{3}{10}\cdot\text{whole circle}\right)$.  Then the powers of $z$ are at points corresponding to the following multiples of the circle:
$$
\frac{3}{10},\quad \frac{6}{10},\quad \frac{9}{10},\quad \frac{12}{10}\equiv\frac{2}{10},\quad \frac{5}{10},\quad \frac{8}{10},\quad \frac{11}{10}\equiv\frac{1}{10},\quad\frac{4}{10},\quad \frac{7}{10},\quad\frac{10}{10}\equiv 0,
$$ 
and then it starts over again.  We've got three times around the whole circle and returned to the starting point in 10 steps.  That's what happens when you have a rational multiple of the whole circle, and that's why it happens. 
The main thing to know to understand what happens when what you have is not a rational multiple of the circle is that what you see above doesn't happen.  Suppose it did: suppose for example that you return to the starting point on the 43rd step after going six times around the circle.  Then you must have had $\left(\theta=\frac{6}{43}\cdot\text{whole circle}\right)$  What would happen if that didn't?
Later note: If $\theta$ is not a rational multiple of $2\pi$, then one never hits the same point twice.  That implies that one hits infinitely many points on the circle.  So you have a set of infinitely many points in a bounded interval.  One can prove that every set of infinitely many points in a bounded interval must have an accumulation point.  However, in this case, one wants to show that every point is an accumulation point.
[to be continued . . . . . ]
A: Hint: Let $\theta \in \mathbb{R}$, and $z = e^{2\pi i \theta}$. We are interested in studying the behavior of the sequence $z^n$.
The case $\theta \in \mathbb{Q}$ has been widely discussed here, so let us assume $\theta \notin \mathbb{Q}$. In that case if $m$ and $n$ are distinct natural numbers, then $z^m \neq z^n$ (why?). 
Now fix $N$, and consider $1,z,z^2, \ldots,z^N$. They are $N+1$ distinct points in the circle, say $z^k = e^{2\pi i \theta_k}$, so there must be $0 \leq m < n \leq N$ such that $|\theta_m - \theta_n| < 1/N$ (why?). Let $w = z^{m-n}$. What can you say about the sequence $w^k$?
Hope this gets you started.    
