TITLE: A transcendence question involving the exponential function
QUESTION [5 upvotes]: Let $(z_n)$ be a sequence of complex numbers satisfying $|z_n|\to +\infty$ and such that $\{e^{z_n}\mid n \in \mathbb{N}\}$ is infinite.
Is it always true that $\{(z_n,e^{z_n})\mid n \in\mathbb{N}\}$ is Zariski-dense in $\mathbb{C}^2$? In other words, if $p(x,y) \in \mathbb{C}[x,y]$ is a polynomial such that $p(z_n,e^{z_n})=0$ for every $n \in \mathbb{N}$, is it always the case that $p = 0$?
This is clearly true if $z_n$ are taken to be real numbers by an "order of growth" argument. But in general I'm not even completely sure if this should be true. Any ideas?

REPLY [9 votes]: I claim that there are many occasions where a sequence $(z_n,e^{z_n})$ with $z_n\to \infty$ is NOT Zariski dense.
Take for instance the complex function $f(w)=e^{1/w}-1/w$. It has an essential singularity at 0, so by Picard's great theorem we know that for all but one (maybe) values $c$ there exists a sequence $w_n$ converging to 0 with $f(w_n)=c$.
So let's define the polynomial $p(x,y)=y-x-c$ (for a non-stupid choice of $c$) and the sequence $z_n=1/w_n$ (for the corresponding sequence $w_n$).
Note that $z_n\to \infty$, $p(z_n,e^{z_n})=0$ and $e^{z_n}=z_n+c$ is not constant.

REPLY [7 votes]: It is false. Equation $\sin x=1/x$ has infinitely many real solutions $(x_n)$, take $z_n=ix_n$, we get $z_n(e^{2z_n}-1)=-2e^{z_n}$.