# |f(z)|≤|z|k\left | f(z) \right | \leq \left | z \right |^{k} then f(z) is a polynomial of degree k. when the domain is restricted

f: C→C \mathbb{C} \rightarrow \mathbb{C} Be an Entire function. Then if ∃A∈R\exists A\in \mathbb{R} and k∈Nk\in \mathbb{N} such that
|f(z)|≤A|z|k\left | f(z) \right | \leq A \left | z \right |^{k} then f(z) is a polynomial of degree k.

The proof of the above theorem follows from the Taylor’s theorem and Cauchy’s estimate. My question is whether this result can be proven when it is the case

f: C∖{z1,z2,..,zn}→C \mathbb{C} \setminus \left \{ z_{1}, z_{2},..,z_{n}\right \} \rightarrow \mathbb{C}

It is given that f is analytic one to one and onto

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1 Answer
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Let j∈{1,…,n}j \in \{1,…,n\}.

Since |f(z)|≤A|z|k|f(z)| \le A|z|^k, ff is bounded in a punctured neighborhood of zjz_j.

Thus zjz_j is a removable singularity of ff. Hence there ist an entire function gg such that

f(z)=g(z)f(z)=g(z) for all z∈C∖{z1,z2,..,zn}z \in\mathbb{C} \setminus \left \{ z_{1}, z_{2},..,z_{n}\right \} .

Thank You Fred.
– Charith Jeewantha
2 days ago