holomorphic function with integral coefficients

Question

I'm trying to prove that an holomorphic function on $\{Z, |Z|<1\}$ and continuous on $\{Z, |Z|\leq 1\}$ with coefficients in $\mathbb Z$ is polynomial. I have tried to establish some partial reciprocal theorems of the radial Abel theorem in ordre to prove it but it hasn't be useful.

Answer 1

Let $f(z)=\sum_{n=0}^\infty a_nz^n$, then for every $r\in(0,1)$. $$ a_n=\frac{1}{2\pi i}\int_{|z|=r}\frac{f(z)\,dz}{z^{n+1}}, $$ But as $f$ is continuous in the closed disc, then $\frac{1}{2\pi i}\int_{|z|=1}\frac{f(z)\,dz}{z^{n+1}}$ is also well-defined and $$ \frac{1}{2\pi i}\int_{|z|=1}\frac{f(z)\,dz}{z^{n+1}}=\lim_{r\to 1^-}\frac{1}{2\pi i}\int_{|z|=r}\frac{f(z)\,dz}{z^{n+1}}=a_n. $$ But $$ a_n=\frac{1}{2\pi i}\int_{|z|=1}\frac{f(z)\,dz}{z^{n+1}}=\frac{1}{2\pi }\int_0^{2\pi}f(\mathrm{e}^{i\vartheta})\,\mathrm{e}^{-in\vartheta}\,d\vartheta, $$ and the right-hand side tends to zero, as $n$ tends to infinity since $\mathrm{e}^{-in\vartheta}/\sqrt{2\pi}$, $n\in\mathbb N$, is orthonormal in $L^2[0,2\pi]$. Thus, if the $a_n$'s are integers, they will vanish eventually.