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Let $R$ be an (irreducible) plane real algebraic curve (without isolated points).

Consider Zarissky closure of $R$ in ${\mathbb C}P^1$ (as real variety).

Suppose that $\lambda\in R \Rightarrow\overline{\lambda}^{-1}\in R.$

Question 1: When there exist a polynomial $f(x,y)$ with zero set $R$ such that for each $k\in {\mathbb R}\cup {\infty}$ $f(x,kx)$ has only real roots? (i.e. hyperbolic).

It is possible to state also a weaker version of that question:

Question 2: When there exist a polynomial $f(x,y)$ with zero set $R$ such that for each $k\in {\mathbb R}\cup {\infty}$ holds following property :

If for some $r\in{\mathbb C}$ $f(r,kr)=0$ then $f(\overline{r}^{-1},k\overline{r}^{-1})=0$?

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  • $\begingroup$ The only Zarissky closed sets in $CP^1$ are finite sets and the whole $CP^1$. Please edit your question. $\endgroup$
    – Alexandre Eremenko
    Commented Dec 8, 2015 at 13:59
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    $\begingroup$ $CP^1$ has not only structure of complex variety, in which it is as you said, but also a structure of real algebraic variety, where there are a lot more Zarissky closed sets. $\endgroup$
    – probably
    Commented Dec 8, 2015 at 16:57
  • $\begingroup$ As we've started from real curve i supposed that we are all the time staying in a real category. $\endgroup$
    – probably
    Commented Dec 8, 2015 at 22:08
  • $\begingroup$ I understand, but then you should not use the notation $CP^1$ if you use the real category. $\endgroup$
    – Alexandre Eremenko
    Commented Dec 8, 2015 at 22:46
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    $\begingroup$ Well, actually that thing is motivated by some type of questions where one study some decompositions, of, say, ${\mathbb C}P^n$ into semialgebraic subsets. Here $CP^1$ more or less means space of homogeneous binary forms of degree $1.$ Moreover, i use here operations over complex numbers -- conjugation, inversion -- so it is natural to see it not as $S^2$ but as $CP^1$ with the structure of real algebraic variety. $\endgroup$
    – probably
    Commented Dec 9, 2015 at 0:06

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