Assume that $f : ℂ→ℂ$ is a non-constant non polynomial and entire function and there exist $a∈ℂ$ such that the fiber $f⁻¹(a)$ is finite.
My question is: How to prove that $a$ is unique.
Assume that $f : ℂ→ℂ$ is a non-constant non polynomial and entire function and there exist $a∈ℂ$ such that the fiber $f⁻¹(a)$ is finite.
My question is: How to prove that $a$ is unique.
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Using Picard's (big) theorem, the proof is rather short:
Suppose $f$ is an entire function such that the fibres $f^{-1}(a)$ and $f^{-1}(b)$ are finite, where $a\neq b$. Then there is an $R > 0$ such that $f(z) \notin \{a,b\}$ for $\lvert z\rvert > R$. But then $f$ omits two values in the punctured neighbourhood $U = \{ z\in\mathbb{C} : \lvert z\rvert > R\}$ of $\infty$, hence by Picard's theorem $f$ has a pole or a removable singularity in $\infty$, and not an essential singularity. Thus $f$ is a polynomial (and either is constant, or all fibres are finite).
I don't yet see a proof without Picard's theorem.