For what $c \in \mathbb{C}$ is there a entire function such that $f(1/k) = c^k$?

Question

For what $c \in \mathbb{C}$ is there a entire function $f$ such that $f(1/k) = c^k$?

I would like to use the identity theorem. However, the obvious choice of $c^{1/z} = e^{\log(c)/z}$ is not analytic at the limit point $0$.

If $|c|>1$, by continuity we get $f(0) \to \infty$, but that contradicts that $f$ is entire.

If $|c|<1$, we can get $f \equiv 0$. Suppose not, and choose $n$ to be such that $f^{(n)}(0) \neq 0$.Then by looking at the series expansion of $f$ one can see $\frac{1}{z^n}f(z)$ does not go to $0$ as $z \to 0$. But, $f(1/k) = k^nc^k \to 0$ as $k \to \infty$.

What about $|c|=1$? Well, if $c=1$ then $f \equiv 1$ satisfies the condition. If $c = e^{i\frac{p}{q}\pi}$ for rational $p/q$, I can choose a subsequence of $f(1/k)$ that equals $1$ and apply the identity theorem to get $f \equiv 1$.

I have an idea on how to handle the other $|c|=1$, but it is not pretty, and I would like others' thoughts.

Answer 1

It is well known that $\{c^{k}:k \geq 1\}$ is dense in the unit circle if $c$ is not a root of unity (i.e. $c=e^{2\pi it}$ with $t$ irrational). In this case we get a contradiction to the fact that $f(\frac 1 k) \to f(0)$. If $c$ is a root of unity then $f(\frac 1 k)=1$ for infinitely many $n$, so $f \equiv 1$.

Answer 2

Given $c \in \mathbb{C}\setminus\{0\}$, let $f$ be an entire function that satisfies $f(1/k) = c^k$ for all $k = 1, 2, \cdots$. Write $f(z) = z^m g(z)$ for some $m \geq 0$ and entire function $g(z)$ with $g(0) \neq 0$. Then

$$ c^k = f(1/k) = k^{-m} g(1/k) $$

and hence

$$ 1 = \lim_{k\to\infty} \frac{g(\frac{1}{k+1})}{g(\frac{1}{k})} = c. $$

So $f(1/k) = 1$ for all $k = 1, 2, \cdots$ and by the identity theorem, $f \equiv 1$.

Answer 3

If $|c|=1,$ then $\lim c^k$ exists iff $c=1.$ Proof: We only need to prove the forward direction. Let $L = \lim c^k.$ Then $L\ne 0$ and we have

$$c=\frac{c^{k+1}}{c^k} \to \frac{L}{L}=1.$$