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.



entire functions of one complex variable with prescribed value and order.
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 $a \in A, 0\le n \le m(a)$. 
> 
> 
> 


