TITLE: Existence of some sort of 'infinite algebraicity' of transcendental numbers
QUESTION [13 upvotes]: Given an arbitrary number, say, $\alpha \in \mathbb{C}$, can anyone supply either
(a) a reason for the existence (or general non-existence) of, or
(b) the reverse engineering of
a (convergent) power series $f(z) = \sum a_n z^n$ with rational coefficients for which $f(\alpha) = 0$? (This problem is trivial when the number in question is algebraic, so feel free to assume transcendentality.) I'm looking for a push in the right direction; any input is welcome.
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prototypical example: $\alpha = \pi$ with $f(z) = \sum \frac{1}{(2n+1)!} z^{2n+1}$ = sin z
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I imagine that this question could easily be recast in several different directions. Please add any appropriate tags if you suspect this too.

REPLY [8 votes]: In fact you can do better. Suppose  $\alpha_n$ is a sequence of distinct nonzero complex numbers with $|\alpha_n| \to \infty$ as $n \to \infty$, and $\beta_n$ is any sequence of complex numbers.  Of course a power series with real coefficients satisfies
$f(\overline{z}) = \overline{f(z)}$, so I'll require that if $\alpha_n = \overline{\alpha_m}$ then $\beta_n = \overline{\beta_m}$.  Then there is an entire function $f(z)$, given by a power series $f(z) = \sum_{j=0}^\infty c_j z^j$ with all $c_j$ rational, such that $f(\alpha_n) = \beta_n$ for all $n$.
Note that by a consequence of Weierstrass's and Mittag-Lefler's theorems, for any such sequence $\alpha_n$ and any $N$, there is an entire function taking prescribed values at all $\alpha_n$ and also prescribed values for the first $N$ coefficients of its Maclaurin series.
Start with an entire function (guaranteed by the cited theorem) $f_0$ such that $f_0(\alpha_n) = \beta_n$ and 
$f_0(\overline{\alpha_n}) = \overline{\beta_n}$ for all $n$.  Since $(f_0(z) + \overline{f_0(\overline{z}})/2$ satisfies the same conditions, we may assume $f_0(\overline{z}) = \overline{f_0(z)}$.
Suppose $f_n$ is such a function where in addition the coefficients of $c_0, \ldots, c_{n-1}$ of its Maclaurin series are rational.
Consider $f_{n+1}(z) = f_n(z) + t g_n(z)$ where all $g_n(\alpha_k) = 0$, 
$g_n(\overline{z}) = \overline{g_n(z)}$, and the Maclaurin series of $g_n$ starts
with $z^n$.  There is a dense set of $t \in \mathbb R$ for which the coefficient of $z^n$ in $f_{n+1}(z)$ will be rational.  I will choose such a $t$ small enough that
$|t| \max \{|g_n(z)|: |z| \le n\} < 1/n^2$.  Then the sequence $f_n(z)$ converges uniformly on compact sets to an entire function $f$, which has the required properties.