$\{a_n\} \subseteq \mathbb C$ discrete set with no limit point . For any sequence $\{z_n\} $ , there is entire $f $ on $\mathbb C$ with $f(a_n)=z_n$?

Question

Let $\{a_n\} \subseteq \mathbb C$ be a discrete set with no limit point . Then for every sequence $\{z_n\}$ of complex numbers , can we find an entire function $f:\mathbb C \to \mathbb C$ such that $f(a_n)=z_n , \forall n \in \mathbb N$ ? I feel I have to use Weierstrass factorization or Mittag-Leffler , but I can't quite crack it . Please help . Thanks in advance

See https://math.stackexchange.com/a/303801/42969. – Martin R Apr 22 '17 at 07:46 — Martin R, Apr 22 '17 at 07:46

score 1 · Answer 1 · answered Apr 12 '17 at 05:16

Hint: Mittag-Leffler allows you to construct a meromorphic function with prescribed principal parts. Weierstrass allows you to construct an entire function with prescribed zeros of any order, and in fact you can take such a function to be a product of elementary factors.

Suppose the zeros of the entire function $g$ coincide with the poles of the meromorphic function $h$. What would the product $g\cdot h$ behave like near these poles/zeros? Can you use this to construct the function you want?

sorry for late response , but could you please provide an elaborate argument ? Thanks in advance — , Apr 21 '17 at 06:36

$\{a_n\} \subseteq \mathbb C$ discrete set with no limit point . For any sequence $\{z_n\} $ , there is entire $f $ on $\mathbb C$ with $f(a_n)=z_n$?

1 Answers1