# Real points on a projective curve

Let $$C_d$$ be a smooth curve of degree $$d$$ in $$\mathbb{CP}^2$$. If we pick some homogeneous coordinates $$[z_0:z_1:z_2]$$ on $$\mathbb{CP}^2$$, then $$C_d$$ is the zero set of a generic polynomial of degree $$d$$.

Further, let $$Y \subset \mathbb{CP}^2$$ be the real projective plane on which the coordinates $$z_i$$ are real.

What are known lower bounds on the number of intersection points of $$C_d$$ with $$Y$$?

It looks like the lower bound is $$0$$ if $$d$$ is even and $$1$$ if $$d$$ is odd.
Construction. Suppose $$d=2p$$. Take the curve $$F=(z_1^2+z_2^2+z_3^2)^p=0$$. It doesn't have real points at all. Taking a small perturbation $$F'$$ of the polynomial $$F$$ we get a smooth curve $$F'=0$$ also disjoint from the real plane.
Suppose $$d=2p+1$$. Take $$F=(z_1^2+z_2^2+z_3^2)^p(z_1+iz_2)$$. Note, the curve $$F=0$$, even though singular, intersects the real plane transversely at one point $$(0:0:1)$$. So, after a small perturbation, we get a curve intersecting the real plane in one point.
Lower bound. Let us now prove that if $$d=2p+1$$, there is always at least one point of intersection of the curve with the real plane. It is enough to prove this for a generic curve. So suppose $$F=0$$ is such a curve. Let $$\overline F$$ be the conjugated polynomial. Then since $$F$$ is generic, the curves $$F=0$$ and $$\overline F=0$$ intersect in $$(2p+1)^2$$ distinct points. Consider the action of conjugation on $$\mathbb CP^2$$ on this set. It is an involution, so it should have a fixed point since $$(2p+1)^2$$ is odd. Such a fixed point lies on the real plane.