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In the complex plane, say $a_n \rightarrow \infty$ and $d_n$ and $A_n$ are arbitrary complex numbers. can we find an entire function with $f(a_n)=A_n$ with order $d_n$? (here "order" means $f(z)-A_n$ has a zero with order $d_n$)

If without restrictions on the order, it is an exercise from Ahlfors 3rd edition P197 No.1. Following his hint, an answer is like $\sum_{n} g(z) \frac{A_n}{g^{\prime}(a_n)} \frac{e^{r_n(z-a_n)}}{(z-a_n)}$ for some suitable chosen $r_n$ where $g(z)$ is an entire functions with simple zeros at $a_n$. I am not sure how to do with the requirement on orders. It sounds like a standard result, I greatly appreciate if anyone with an idea or reference to this problem.

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up vote 5 down vote accepted

Yes this is possible. See Theorem 15.13 in Rudin: Real and Complex Analysis:

If $\Omega \subseteq \mathbb C$ is open and $A \subseteq \Omega$ has no limit point in $\Omega$, and to each $a \in A$ there is an associated integer $m(a)$ and complex numbers $w_{n,a}\,(0 \le n \le m(a))$. Then there exists a function $f$, holomorphic on $\Omega$, such that $$f^{(n)}(a) = n!\; w_{n,a}$$ for all $a \in A, 0\le n \le m(a)$.

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See the Weierstrass factorization theorem.

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That only solves when $A_n=0$ –  i707107 May 11 '12 at 23:04
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