Rate of convergence of an algebraic irrational rotation Let $\alpha \in \mathbb{S}^1$ be an algebraic number with $\mathop{\mathrm{arg}}(\alpha)/\pi$ irrational. Is it possible for the rotation by $\alpha$ to converge exponentially fast to a $\xi \in \mathbb{S}^1$? That is, may there exist $\xi \in \mathbb{S}^1$ and $A > 1$ such that $\liminf_{n \to \infty} A^n |\alpha^n - \xi| = 0$?
This answer in MathOverflow cites a result of Feldman that implies that if $\xi$ is algebraic and $\xi \neq \alpha^q$ for every $q \in \mathbb{Q}$, then $|n \log \alpha - \log \xi| > C n^{-N}$, where $C > 0$ and $N \geq 1$ are effectively computable numbers depending only on $\alpha$, showing that the rate cannot be exponential in this case.
What about the case where $\xi$ is not algebraic and the case where $\xi = \alpha^q$ for some $q \in \mathbb{Q}$?
Thanks in advance.
 A: The problem concerns lower bounds for expressions of the type
$$|m 2 \pi i + n \log(\alpha) +  \log(\xi)|,$$
where $\alpha$ and $\xi$ are fixed. 
Lower bounds for such expressions (which rule out exponential convergence of the form you seek) follow immediately from Baker's Theorem when $\alpha$ and $\xi$ are algebraic. The statement of Baker's theorem has the assumption that the logarithms in question are linearly independent, so this seems to rule out exactly the case you are interested in, namely: the case when $\xi^p = \alpha^q$ for some $p/q \in \mathbf{Q}$ is exactly the case when this linear independence doesn't hold. However, the problem then reduces to the (easier) case of Baker's theorem in two variables, since one now seeks lower bounds on
$$ |m \cdot 2 \pi i + n \cdot \log(\alpha) +  \log(\xi)| = \frac{1}{p}
| m p \cdot 2 \pi i + n p  \cdot \log(\alpha) +  p \cdot \log(\xi)|$$
$$= \frac{1}{p} | m p \cdot 2 \pi i + (n p + q) \cdot \log(\alpha)|.$$
and now we can apply Baker for two variables, as long as $n p + q \ne 0$, which can happen at most once (the case when $\alpha^n = \xi$).
Finally, if $\xi$ can be transcendental (and $\log(\alpha)/(2 \pi i)$ is not rational) then there will always exist $\xi$ for which the limit inferior is zero. Simply choose $\xi$ as follows:
Step 1: Let $B > A  > 1$ be arbitrary.
For some (arbitrary) value $n = n_0$ of $n \ge 1$, let $S_0$ denote the closed ball in $S^1$ of radius $1/B^{n_0}$ around $\alpha^{n_0}$. We see that, for any $\xi \in S_0$, the bound:
$$B^{n} |\alpha^n - \xi| < 2,$$
holds for $n =n_0$.
Step 2: Because the powers of $\alpha$ will be dense in $S^1$, we will eventually find a value $n = n_1$ of $n$ such that $\alpha^{n_1}$ also lies in $S_0$. We now choose $S_1$ to be the intersection of $S_0$ and the ball in $S^1$ of radisu $1/B^{n_1}$ around $\alpha^{n_1}$. This is a closed non-empty set which is contained in $S_0$.
Step 3,4,5,...: We repeat this procedure, namely, given a tiny set $S_{k-1}$, we find $n = n_k$ such that $\alpha^{n_k}$ is contained in $S_{k-1}$, and then let $S_k$ be the intersection of $S_{k-1}$ with a ball in $S^1$ of radius $1/B^{n_k}$ around $\alpha^{n_k}$.
Conclusion:
Now we note that:


*

*The intersection of $S_0 \supset S_1 \supset S_2 \ldots$ is non-empty, and consists of a single point $\xi$. This is basically the compactness of $[0,1]$.

*By construction, $\xi \in S_{n_k}$, and so
$$B^n |\alpha^n - \xi| < 2$$
for all $n = n_k$, and in particular there are infinitely many such $n$ for which this inequality holds.

*Since $A < B$, one trivially deduces that
$$\liminf A^n |\alpha^n - \xi| = 0.$$
