I am having trouble with the details of the following question:
Does there exist an entire function such that $f(\frac1n)=e^{-n}$ for all positive integers $n$?
I think this is supposed to be a relatively straightforward application of the uniqueness theorem, but all my approaches fail in some way.
The statement of the uniqueness theorem is as follows:
Let $f,g$ be holomorphic functions on a domain D. Then if there exists a sequence of distinct points $z_1,z_2,...$ converging in $D$ with $f(z_n)=g(z_n)$ for all $z_n$, then $f=g$ on $D$.
On one hand, I want to say that we must have $f(z)=e^{-\frac1z}$ for all $z\in \mathbb{C}\setminus\{0\}$, but the sequence $1/n$ doesn't converge in $\mathbb{C}\setminus\{0\}$, so I can't use uniqueness there.
On the other hand, I want to say that we must have $f(z)=\begin{cases}e^{-\frac1z}\text{ if $z\neq 0$}\\0\text{ if $z=0$}\end{cases}$, but I can't show that the latter function is holomorphic (in fact, it's not even continuous if you approach from the negative real axis), so I can't use uniqueness there either.
I've also tried to use a power series expansion, but the taylor series of $e^{-\frac1z}$ seems incredibly convoluted. But since it converges at $0$, I am suspecting that the taylor series expansion of $e^{-\frac1z}$ at some non-zero number may be the solution, but I don't know how to show this.
Hint: It doesn't follow from the Uniqueness Theorem. $f$ must have zero of some finite order $k$ which implies $|f(z)| \geq c|z|^{k}$ for $|z|$ sufficiently small, for some $c >0$. Get a contradiction to this by putting $z=\frac 1 n$.