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I am trying to understand how Liouville's Theorem would apply to complex numbers.

alpha an irrational algebraic, then there exist a constant, c, depending on alpha such that for all rationals, the inequality is satisfied.

c/q^d <= abs(alpha-(p/q))

Assume x is a complex irrational algebraic. In the theorem's inequality, would alpha be seperately the Re(x) and Im(x) of the complex number or would it be the magnitude of x?

If x is complex algebraic, is its real and imaginary parts necessarily algebraic?

I am asking because I'm trying to prove Liouville's theorem for alpha a complex number.

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 The usual proof of Liouville's theorem works for any algebraic number, regardless whether it is complex or real. The absolute value in the theorem should be understood as the usual absolute value of complex numbers. The proof depends on the fact that a nonzero algebraic integer multiplied by its conjugates is a nonzero rational integer, hence the product of the absolute values of these algebraic integers is at least 1. I am afraid your question is not of research level. I vote to close. – GH May 19 2012 at 20:53
cheers, that does it! – Bob Davis May 19 2012 at 23:11

closed as too localized by Steven Landsburg, GH, Valerio Capraro, Felipe Voloch, Andres Caicedo May 20 2012 at 1:56

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