# Cyclic Sequence of Angles

1. Feb 20, 2008

### e(ho0n3

[SOLVED] Cyclic Sequence of Angles

Fix an angle $\theta$. Let n be a positive integer and define $\theta_n = n\theta \bmod 2\pi$.

The sequence $\theta_1, \theta_2, \ldots$ is cyclic if if it starts repeating itself at some point, i.e. the sequence has the form $\theta_1, \ldots, \theta_k, \theta_1 \ldots$.

What I would like to find out is: For which angles $\theta$ is the sequence $\{\theta_n\}$ cyclic? If for some integer m > 1, $\theta_1 = \theta_m \equiv \theta = m\theta$, then $m\theta = \theta + 2\pi x$ for some non-negative integer x. Solving for $\theta$, I get $\theta = 2\pi x / (m - 1)$. So it seems that any rational multiple of $\pi$ will create a cyclic sequence. Is this correct?

2. Feb 20, 2008

### Dick

3. Feb 21, 2008

### e(ho0n3

This all began when I started contemplating about the limit as n approaches infinity of zn, z being complex with |z| < 1. If I represent z as r(cos t + isin t), zn = rn(cos nt + isin nt). If {nt} is cyclic, then I could break up the sine and cosine terms, multiply through by rn and apply the limit on each term. Each term goes to 0 because r < 1 so the limit is 0. Right?

4. Feb 21, 2008

### Dick

Yes, but you don't have to worry about 'cyclic' |cos nt+i*sin nt| is bounded, since |cos|<=1 and |sin|<=1. Regardless of the arguments. So if you multiply by r^n with r<1, the result certainly goes to 0.

5. Feb 21, 2008

### e(ho0n3

That makes sense. So for any complex z with |z| < 1, zn goes to 0 as n goes to infinity. I began feeling paranoid about this when I was trying to compute the limit of nzn as n goes to infinity. I rewrote this as n/z-n and applied l'Hopital's rule to get -zn/log z. I wasn't sure about the limit of zn here, but now I am. Thanks.

Last edited: Feb 21, 2008