Prove that every holomorphic function $f:\overline{\mathbb{C}} \rightarrow \overline{\mathbb{C}}$ such that $f(z) \in \mathbb{C} $ for every $z\in \mathbb{C}$ is a polynomial
(I made a similar post, but I thought it was the same exercise)
The hint from my notes say to use the Cauchy integral formula, but I don't see how I can use that.
I am trying to understand the problem, but I am not sure if I comprehend it very well.
$\bullet $ Seeing that hind my first thought was to show that $f^{(n)}$ is zero, but that doesn't work with the Cauchy estimate.
$\bullet $ Second, since $f:\overline{\mathbb{C}} \rightarrow \overline{\mathbb{C}}$ does that mean that $f(\infty)=\infty$ ?, and I can't use Liouville theorem ?
$\bullet $ I think I am a little confused, can someone clarify this for me and give me a hint on how should I go on proving the statement.
There are two possibilities where: either $f(\infty)\in\Bbb C$ or $f(\infty)=\infty$.
If $f(\infty)=\omega\in\Bbb C$, then, by continuity, there is some $R>0$ such that$$|z|>R\implies|f(z)-\omega|<1.$$So, if $K=\max\left\{|f(z)|\,\middle|\,z\in\overline{D(0,R)}\right\}$, you have that $(\forall z\in\Bbb C):|(f(z)|\leqslant\max\{K,1\}$, and therefore $f|_{\Bbb C}$ is bounded and therefore, by the Liouville theorem, it is constant.
And if $f(\infty)=\infty$, let $\varphi(z)=f\left(\frac1z\right)$. Then $0$ is an isolated singularity of $\varphi$. Since $\lim_{z\to\infty}f(z)=\infty$, $\lim_{z\to0}\varphi(z)=\infty$. So, $\varphi$ has a pole at $0$. That is, for some $N\in\Bbb Z$, with $N\leqslant 0$, $\varphi(z)$ can be written as $\sum_{n=N}^\infty a_nz^n$ near $0$. But then$$f(z)=\varphi\left(\frac1z\right)=\sum_{n=-\infty}^{-N}a_{-n}z^n.$$But $0$ belongs to the domain of $f$, which is an holomorphic function. So, $f(z)=\sum_{n=0}^{-N}a_{-n}z^n$, which is a polynomial function.