# Degree of Zariski closure of curve parametrized by hypocycloids

I have a curve $(x(\theta),y(\theta))$ in $\mathbb{C}^2$, where $x(\theta)$ is described as $$x(\theta) = (k-1)\cos(\theta) + \cos((k-1)\theta) + i[(k-1)\sin(\theta)- \sin((k-1)\theta)]$$ and $y(\theta)$ is just the conjugate of $x(\theta)$.

This curve is a rational curve, so there is a polynomial $P(x,y)$ such that the polynomial is 0 on the curve. There is a minimal such polynomial w.r.t. to degree, that satisfies this, which is not the 0 polynomial.

How do I find this minimal degree?

For example, the curve has rotational symmetry $2\pi/k$ so this tells me that the degree must be a multiple of $k$, or something very similar, but I need something that gives me a lower bound.

EDIT: For each $k$ I have a candidate for $P$, but it is tedious to express this $P$. However, I do not think it is too hard to calculate the degree of each $P$. Now, if this degree matches the minimal degree above, then I know that $P$ is in fact the smallest polynomial. Thus, for a fixed $k$ I can of course use variable elimination, and compare the result to my candidate $P$, but how do I prove that these are the same for all $k$?

SOLUTION:

Turns out the solution was to start at the other end. I had a certain discriminant $P$ in mind, that was the Zariski closure to this hypocycloid. There was not a nice formula for the discriminant explicitly, but I managed to get a very nice parametrization of the set where the discriminant vanish. Restricting this discriminant to the subspace $x$ is the conjugate of $y$ gives exactly the hypocycloid above.

-

You write the cosines in terms of the variable $z\exp(i \theta)$ in the usual way, then write $x = f(z), y = g(z),$ and eliminate $z$ from this pair of equations by computing the resultant (the minimal polynomial will be a factor of the resultant, so you will need to factorize and check all the factors). For more on the subject, see