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.
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.