As for your first question, it seems hard to characterize all such curves. There two types of families I have thought of. Of course you can multiply curves that satisfy the property, so let us only talk about minimal curves with respect to the property.

• As you mention, the real line.

• Genus 0 or 1 curves with a rational point of infinite order and the required property (mentioned in comments) that $y$ appears only with even powers. This uses the simple fact that conjugate numbers in $\mathbb{Q}(\sqrt{-1})$ have minimal polynomial over the rationals. If we want to extend this to higher genus, there is the following question: Given a rational class in the jacobian of a curve, does it have infinitely many multiples such that $a(x)$, in its representation as a reduced divisor $[(a(x), b(x,y)]$, has only real roots? If so, then the property holds for all curves with a rational point on the jacobian of infinite order.

• The unit circle fits into the above. But as you mention, there is another way, which we can generalize: curves of the form $||f(z)||=1, z=x+iy$. The question is now: how many points in the union of all CM fields does this equation have? If I had to guess, infinitely many, so I think these probably satisfy the property.

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Yes. There exist many such curves:

For any rational monic polynomial without multiple roots if $x+iy$ is a root, then $x-iy$ is a root as well. So if your curve $C$ satisfies $y\not =0,\ (x,y)\in C \rightarrow (x,-y)\not\in C$ then it cannot contain the rootset of any polynomial with complex roots. To finish, make the curve have finitely many real roots. For example: $$C: x-y^3=0$$x-y=0$$1 Yes. There exist many such curves: For any rational monic polynomial without multiple roots if x+iy is a root, then x-iy is a root as well. So if your curve C satisfies y\not =0,\ (x,y)\in C \rightarrow (x,-y)\not\in C then it cannot contain the rootset of any polynomial with complex roots. To finish, make the curve have finitely many real roots. For example:$$C: x-y^3=0