Fields Having Algebraic Numbers as Subfield In the following text, we call a power series $\sum_{n=0}^{+\infty}c_nz^n$ rational if it satisfies $\forall n, c_n \in \Bbb{Q}(i)$ and finite if it is a polynomial.
Let $\Bbb{A}$ be the set of complex zeropoints of all rational finite series (which is equivalent to all complex algebraic numbers), $\Bbb{A}^\omega$ be the set of complex zeropoints of all rational series that have infinite convergence radius, $\Bbb{A}^\infty$ be the set of complex zeropoints of all rational series that have non-zero convergence radius.
It is easy to show that $\Bbb{A} \subseteq \Bbb{A}^\omega \subseteq \Bbb{A}^\infty \subseteq \Bbb{C}$ and $\Bbb{A}$ is countable, so what about the size of $\Bbb{A}^\omega$ and $\Bbb{A}^\infty$? $\Bbb{A}$ is obviously a proper subset of $\Bbb{A}^\omega$ because $\ln 2$ is in the latter not the former, but the relation between $\Bbb{A}^\omega$ and $\Bbb{A}^\infty$, $\Bbb{A}^\infty$ and $\Bbb{C}$ is not clear for me:
For the first pair, I suspect that $\Bbb{A}^\omega = \Bbb{A}^\infty$ or weaker $\operatorname{Gal}(\operatorname{Frac}(\Bbb{A}^\infty) / \operatorname{Frac}(\Bbb{A}^\omega))$ is finite/countable but I have no reason -- it is so hard to determine if a given number is in $\Bbb{A}^\omega$.
For the second pair, I suspect that $\Bbb{A}^\infty$ is a proper subset of $\Bbb{C}$ because I think numbers in $\Bbb{A}^\infty$ is computable but there are uncomputable numbers in $\Bbb{C}$. However, I don't know much about computation and I hope someome would correct my fault.
Edit: Now that computable numbers are countable, I tend to conjecture that $\Bbb{A}^\infty = \Bbb{C}$, but I can not even prove that $\Bbb{A}^\omega$ is uncountable.
 A: Claim:$\;$For all $u\in\Bbb{C}$ there exist $c_0,c_1,c_2,...\in\Bbb{Q}(i)$, not all zero, such that the power series
$$
f(z)=\sum_{k=0}^\infty c_kz^k
$$
has $u$ as a zero and has infinite radius of convergence.

Proof:

Let $u\in\Bbb{C}$.

If $u=0$ we can just take $f(z)=z$.

Next suppose $u\ne 0$.

Let
$$
f(z)=\sum_{k=0}^\infty c_kz^k
$$
where $c_0,c_1,c_2,...\in\Bbb{Q}(i)$ with $c_0\ne 0$ are chosen in succession so as to satisfy
$$
\left|\,\sum_{k=0}^n c_ku^k\right|
\,<\,
\frac{|u|^{n+1}}{2{\,\cdot\,}(n+1)!}
$$
for all nonnegative integers $n$.

Then by the squeeze theorem, it follows that $f(u)=0$. 

To show that $f(z)$ has infinite radius of convergence, it suffices to show that for all integers $n\ge 1$ we have
$$
|c_n| < \frac{b}{n!}
$$
where $b=\max(|u|,1)$.

Fix $n\ge 1$.$\;$Then
\begin{align*}
|c_n|
&=
\frac{|c_nu^n|}{|u|^n}
\\[4pt]
&=
\frac
{
\left|\,
\left(
{\displaystyle{
\sum_{k=0}^n c_ku^k
}}
\right)
-
{\displaystyle{
\left(\sum_{k=0}^{n-1} c_ku^k\right)
}}
\right|
}
{|u|^n}
\\[4pt]
&\le
\frac
{
\left|\,
{\displaystyle{
\sum_{k=0}^n c_ku^k
}}
\right|
+
\left|\,
{\displaystyle{
\sum_{k=0}^{n-1} c_ku^k
}}
\right|
}
{|u|^n}
\\[4pt]
&=
\frac
{
\left|\,
{\displaystyle{
\sum_{k=0}^n c_ku^k
}}
\right|
}
{|u|^n}
+
\frac
{
\left|\,
{\displaystyle{
\sum_{k=0}^{n-1} c_ku^k
}}
\right|
}
{|u|^n}
\\[4pt]
&<
\frac
{
\left(
{\Large{
\frac{|u|^{n+1}}{2{\,\cdot\,}(n+1)!}
}}
\right)
}
{|u|^n}
+
\frac
{
\left(
{\Large{
\frac{|u|^n}{2{\,\cdot\,}n!}
}}
\right)
}
{|u|^n}
\\[4pt]
&=
\frac{|u|}{2{\,\cdot\,}(n+1)!}
+
\frac{1}{2{\,\cdot\,}n!}
\\[4pt]
&\le
\frac{b}{2{\,\cdot\,}(n+1)!}
+
\frac{b}{2{\,\cdot\,}n!}
\\[4pt]
&<
\frac{b}{2{\,\cdot\,}n!}
+
\frac{b}{2{\,\cdot\,}n!}
\\[4pt]
&=
\frac{b}{n!}
\\[4pt]
\end{align*}
which completes the proof.
