A transcendence question involving the exponential function

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?

• What is the problem with taking $z_n=2\pi n i$ and $p(x,y) = y-1$? – Lasse Rempe Apr 22 '16 at 10:26
• Then the set $\{e^{z_n}\}$ is not infinite. That's why I added this hypothesis. – user85435 Apr 22 '16 at 10:40
• Now that is the sort of question that will keep me awake at night and mumbling to myself: "Ich muß wissen! Ich werde wissen!" – Gro-Tsen Apr 22 '16 at 11:27
• I elaborated below, but it seems that Fedor Petrov bitted me with a clearer answer... – Uri Bader Apr 22 '16 at 12:31
• @user89334 For what it's worth, I think the argument with Picard's theorem is better, because it will apply to many similar contexts. – Gro-Tsen Apr 22 '16 at 12:39

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