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 $Y$?
Thanks for reading.
 A: 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.
