Is the set $\{(z, \bar{z}), z\in\mathbb{C}\} \subseteq \mathbb{C}^2$ Zariski closed? Question is in the title. It feels like the answer should be no, but I don't see a simple way to prove it.
I mean, if it was Zariski closed, then there would exist a polynomial expression in z and $\bar{z}$ that would be 0 for all $z\in\mathbb{C}$. And it's somewhat clear from a complex analysis perspective that this cannot exist (by considering partial derivatives after z and $\bar{z}$). But it feels like it should be more elementary than this. Am I missing something obvious?

Suppose for contradiction that this set is Zariski closed, i.e. there exist polynomials $p_1, \dots, p_k \in \mathbb{C}[x,y]$ such that $p_i(a,b) = 0$ for all $i$ if and only if $a = \overline{b}$.
Now for each $i$, we have that $p_i(z,z) \in \mathbb{C}[z]$ has infinitely many zeros (e.g. any real number), so $p_i(z,z) = 0$.
Finally, for any $z \in \mathbb{C}$, we have $p_i(z,z) = 0$ for all $i$, so $z = \overline{z}$. This is a contradiction.
Edit: made a silly mistake at first (assumed $k = 1$), should be fixed now.
