Ah I understand the heuristic test you had in mind now, thank you for taking the time to add the details!

@WillSawin Would you mind expanding a bit the part "for $n$ large one can heuristically guess if it is close enough to sufficient in practice"? It certainly seems believable, but do you have a specific example of non-algebraic analytic curve where the "non-algebraicity" can be detected as you suggest?

@RichardStanley Thank you for the reference! It seems to contain a lot of information indeed. I will take a closer look.

@D.S.Lipham yes, corrected, thanks!

I have not looked at Rakhmanov's paper. I am simply pointing out that several of your comments and answers include incorrect mathematical statements or mathematical objects that are not well-defined. I do have higher standards for this forum, even though it is indeed not a paper as you pointed out. But let's agree to disagree and put an end to this discussion.

2. In the inequality that you refer to in your answer, you wrote that $F$ is assumed to be continuous up to the boundary. This is not the case here, so you cannot directly apply this inequality to answer your question without adding more justification. More details are necessary. It is possible that the proof can be fixed, but if I were to referee a paper that includes your proof as it is, I would reject the paper. You are welcome to disagree of course, but that is my opinion.

Yes, I do understand that the distance to a compact set is well-defined. That is not the issue here. 1. In your answer you write $\min_{|\rho|=1} |F(\zeta)-F(\rho)|$. This is not well-defined since $F(\rho)$ may not be even defined for all $\rho$ with $|\rho|=1$.

See en.wikipedia.org/wiki/…

This is true if and only if $\mathcal{K}$ is locally connected! If $\mathcal{K}$ is not locally connected then the argument you provide does not work.

$F(\rho)$ may not even be defined...!

I don't understand what you mean by $\min_{|\rho|=1} |F(\zeta)-F(\rho)|$. What are your assumptions on $\mathcal{K}$? Note that in general the uniformizing map of the complement of $\mathcal{K}$ may not extend to the boundary.

Right, sorry about that. I edited my answer accordingly. It may be true that $C=1$ but I don't think the method I outlined can give that. For your purposes do you need $C=1$?