Liouville's and Roth's theorems for complex algebraic numbers Liouville's theorem says that  If $\alpha$ is an irrational number which is the root of a polynomial $p$ of degree $d > 0$ with integer coefficients, then there exists a real number $C > 0$ depending only on $\alpha$ such that, for all integers $p, q$, with $q > 0$,
$\left| \alpha - \frac{p}{q}  \right | > \frac{C}{q^d}$
Does this hold for the case when $\alpha \in \mathbb{C}$ and $p/q \in \mathbb{Q}[i]$? Can someone point me out to a reference for either a proof or a counter example? Is there an analogous result for Roth's theorem as well?
 A: Certainly this is false for general $\alpha\in\Bbb C$, the whole point of Liouville's theorem is to get at transcendental numbers, and there are plenty of such for which there is no such constant. Take any Liouville number for instance.
If you demand $\alpha$ be algebraic--but not necessarily real--of course, you can easily extend to $\Bbb Q(i)$ simply by noting that $\alpha= x+iy, \,$ and ${p\over q}=a+bi$ then

$$\left|\alpha-{p\over q}\right|=|(x-a)+i(y-b)|>|x-a|>Cq^{-d}.$$

Ultimately extending to non-real algebraic numbers is uninteresting, since the real and imaginary parts are both algebraic themselves, which is easily seen as
$$\begin{cases} \Re(\alpha)={1\over 2}(\alpha+\overline{\alpha}) \\ \Im(\alpha) ={1\over 2}(\alpha-\overline{\alpha})\end{cases}$$
and you always have $|z|>|\Re(z)|$ for any complex $z$.
Similarly for Roth's theorem, you again use the inequality $|\Re(z)|<|z|$, to see that if $\left|\alpha-{p\over q}\right|<C(\alpha,\epsilon)q^{-2-\epsilon}$ has a solution, then also $\left|\Re(\alpha)-\Re\left({p\over q}\right)\right|<C(\alpha,\epsilon)q^{-2-\epsilon}$, and of course the real case shows there are only finitely many such things. You may object that perhaps there are infinitely many imaginary parts which work with the finitely many real parts, but of course $|\Im(z)<|z|$ is also a fact, so this is not the case.
