Alan Sokal proved that chromatic roots are dense in the whole complex plane. I.e., if $P(G;z)$ denotes the chromatic polynomial of a finite simple graph $G$ evaluated at $z \in \mathbb{C}$, then $$\bigcup_G \big\{z \in \mathbb{C}:P(G;z)=0\big\}$$ is a dense subset of $\mathbb{C}$. Generalizing this...

Question: Is, for all fixed $c \in \mathbb{C}$, $$\bigcup_G \big\{z \in \mathbb{C}:P(G;z)=c\big\}$$ is a dense subset of $\mathbb{C}$?

Comments:

For any $c$, the subset will be a countable subset of $\mathbb{C}$.

It doesn't seem straightforward to edit a graph $G$ with chromatic root $z$, to give a graph $G'$ with $P(G';z)=c$. (Although, maybe I'm missing something.) This would rule out an obvious approach. (It still might be possible to modify Sokal's method to find an answer, but this would be lengthy.)

Real chromatic roots are not dense in $\mathbb{R}$. So probably there's a big difference between the two cases ($\mathbb{C}$ and $\mathbb{R}$) here too.

isdense. I wonder: can we argue naively that $U$ is closed? Open? $\endgroup$