Diophantine approximations of ratios of transcendental numbers

I am looking for good diophantine approximations for a specific class of irrational numbers.

Let $e^{2 \pi i \theta}$ be a complex algebraic number. I would like a result to the effect that $\theta$ can be approximated well; more specifically, for any constant $k$, I would like for the inequality

$|n \theta -m| < \frac{1}{k n}$

to have infinitely many integer solutions in $n$ and $m$.

What I know is that Hurwitz's theorem guarantees a value of $k$ of at least $\sqrt{5}$, and that Khinchin's theorem asserts that, for any given $k$, the inequality $|n \alpha -m| < \frac{1}{k n}$ will have infinitely many solutions for $\textit{almost}$ $\textit{all}\$ irrational numbers $\alpha$.

Are there any other relevant results I can use here? And is it plausible to conjecture that irrational numbers of the form $\theta$ are somehow mysteriously guaranteed to have good approximations (i.e., with any value of $k$) as given above?

Your inequality $\left| \theta - \dfrac{m}{n}\right| < \dfrac{1}{kn^2}$ has infinitely many integer solutions iff the continued fraction of $\theta$ has unbounded elements. In particular, this would not be true if $\theta$ was algebraic of degree $2$ (i.e. a root of a quadratic polynomial over the integers). However, by the Gelfond-Schneider theorem, any irrational $\theta$ such that $e^{2\pi i \theta} = (-1)^{2\theta}$ is algebraic must be transcendental. But there are also uncountably many transcendental numbers whose continued fractions have bounded elements.
• @Robert: Thanks a lot, that does seem useful. Let me make sure I fully understand. I we write $k(\theta)$ to denote the sup of the constants $k$ such that the approximation inequality has infinitely many (n,m)-solutions, I believe that what you are saying is that $k(\theta)$ is infinite iff $\theta$ has unbounded elements in its continued fraction representation. Is that correct? And if so, supposing that the elements in the continued fraction are bounded by $B(\theta)$, say, is there a relationship between $k(\theta)$ and $B(theta)$? Any reference you could point me to? – Joel Ouaknine Jun 2 '12 at 3:28
• @Felipe, re. what does OP want? I would ideally like someone to say, "Yes, for that class of transcendental numbers, for any value of $k$ there are always infinitely many $(n,m)$-solutions to the inequality", or "No, sorry, here's a counterexample that has $k$ bounded by 17 (say)". Preferably with pointers to the literature... :) – Joel Ouaknine Jun 2 '12 at 3:42
• If $a_k$ are the elements and $p_k/q_k$ the convergents of the simple continued fraction of $\theta$, then $$\dfrac{1}{q_k^2 (a_{k+1}+2)} < \left| \theta - \dfrac{p_k}{q_k}\right| \le \dfrac{1}{q_k^2 a_{k+1}}$$ Moreover, every irreducible rational fraction $p/q$ with $|\theta - p/q| < 1/(2q^2)$ is a convergent. – Robert Israel Jun 3 '12 at 6:12