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It is well-known that not all complex numbers can be written as a zero to a polynomial with rational coefficients (the transcendental numbers). However, I am wondering if we "extend" polynomials to allow for infinitely many terms (i.e. the power series expansion of a holomorphic function) while still restricting the coefficients to be rational, if every complex number could then be expressed as the zero of such a function?

$\pi$, for instance, is transcendental, but it is a zero of the $\sin$ function, whose power series expansion centered at $z=0$ has only rational coefficients.

reuns
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Richard
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    Yes. Just build the series one term at a time. – Qiaochu Yuan Sep 14 '17 at 23:00
  • Assume $\frac1L < |\alpha| < 1$. Set $f_1(z) = z$ and $f_{n}(z) = f_{n-1}(z)+\frac{b_{n}}{L^n} z^{n}$ where $b_{n} = \displaystyle\arg!!!!!\min_{b \in \mathbb{Z}, |b| \le L^n} |f_{n-1}(\alpha)+\frac{b}{L^n} \alpha^{n}|$. Since the density of ${ n, |\text{arg}(\alpha^n) | < \epsilon}$ is positive, we'll have $f_\infty(\alpha) = 0$ – reuns Sep 14 '17 at 23:07
  • The linked question is completely different, OP's question is not even in the answers. – reuns Sep 15 '17 at 01:11

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