# Sequence on a circle

1. Dec 11, 2005

### symplectic_manifold

Hello!
I've got some questions to the following task.

Let $r\in\mathbb{Q}$. Determine the partial limits of the complex sequence $(z_n)$ defined by $z_n=e^{i2\pi{rn}}$.

There is also a hint given: Set $r=\frac{p}{q}$ with $p\in\mathbb{Z},q\in\mathbb{N}\setminus{\{0\}$, so that the fraction $\frac{p}{q}$ is irreducible. From $\frac{p}{q}\alpha\in\mathbb{Z},\alpha\in\mathbb{Z}$ then follows $\frac{\alpha}{q}\in\mathbb{Z}$.

Every complex number is a point of the complex plane and can be written as:
$cos(t)+i\\sin(t)=e^{it}$, so the defined sequence elements can be rewritten as: $z_n={e^{i2\pi{rn}}=cos(n\cdot{2\pi\frac{p}{q}})+i\\sin(n\cdot{2\pi\frac{p}{q})=(cos({2\pi\frac{p}{q}})+i\\sin(2\pi\frac{p}{q}))^n$
From the equation it's pretty clear that we're dealing with the q-th roots of 1 in the complex, and that p<q, because we have q points in the complex plane, whose radius-vectors are at angle fewer or equals than 2*pi*p. If we add the factor n, we increase the angle of each radius-vector of a particular point (with given p/q) n-fold. Is it right?
Now, I don't know at the moment how to get to the partial sequences. I can't see how the hint could help. It says that alpha is greater-equals q, but why should it be so in our case (I mean q=3/4 for example, then alpha could still be 1,2,3, couldn't it?...we would still get some points)? If it must in fact be greater-equals q and alpha/q is an integer, then does it mean that we could take all elements with indices, which are multiple of q, to get a subsequence?

Last edited: Dec 11, 2005
2. Dec 11, 2005

### matt grime

All you need to do in general is to write the sequence as a disjoint union of convergent sequences*. Of course in general that is disingenuous since that might be hard.

But in this case, let's take p=1 q=4 for this example. What is the sequence? i,-1,-i,1,i,-1,-i,1,....

so the only limit of subsequences are 1,-1,i,-i. Notice that the behaviour is going to be repeated....

* proof of claim. If we can split z_n into distinct convergent subsequences each converging to *different* things and such that every term of z_n is in one of the convergent subsequences, then the limits of these subsequences are all of the accumulation points. Why: if x_m is a convergent subsequence it must eventually lie wholly in one of these distinct subsequences otherwise infinitely many of the x_m's would lie in one subsequence and infinitely many would lie in another, ie x_m would have two subsequences converging to different things, contradicting the assumption x_m is convergent.

3. Dec 11, 2005

### symplectic_manifold

I have a subtle feeling that the accumulation points are in fact i,-1,-i,1, for if we take p/q sufficently large...say p/q<p'/q', for the same n, arg(z') with p'/q' will be larger than arg(z) with p/q. Is it right? But this means p/q must grow together with n, doesn't it?....which doesn't fit in together with the given assumption that r is arbitrary but fixed...Have messed everything up?

...something else:
in my first post below I actually identified alpha with n, though I didn't say this as it occurs to me now...sorry for that. So the hint says that n should divide q. And I can't get to what exent it should bring us further.

Last edited: Dec 11, 2005
4. Dec 11, 2005

### matt grime

p and q are fixed; they can't grow with n. What are you getting at? Are you saying that you think the accumulation points are ALWAYS 1,-1,i,-i? cos they aren't, just think of the case r=1, when the series is the constant series 1,1,1,1..

Look, just try a few examples for different r (ie different p and q) and see what's going on try fixing p=1, q=2,3,4,5,6 to begin with and try spotting a pattern. Hint: remember that any n in Z can be written uniquely as aq+b for 0<=b<q (assuming q is positive, which we can).