Finding algebraic curve satisfying given parameterization

Question

Is there an easy way to find an algebraic curve that satisfies a given parameterization? Specifically, I am talking about the following parameterization: $$ x=z(1-z),\hspace{10pt} y=\sum_{n=1}^r \frac{(-1)^{n+1}}{n} (z-1)^n, \hspace{10pt} z \in \mathbb{C}$$

What might make the above easier is that $\sum_{n=1}^r \frac{(-1)^{n+1}}{n} (z-1)^n$ is the truncation of the Taylor expansion of $log(z)$ around $z=1$. And so, when we take the limit $r \to \infty$ the unknown curve should be: $$ 1-e^y-xe^{-y}=0$$

Is it a matter of playing around until you find something? Or, is there some useful techniques?

Thank you.

Answer 1

This problem is called implicitization and can be solved systematically but not easily by hand.

In Mathematica (and Wolfram Alpha), you can use Eliminate.

For $n=2$, Eliminate[{x==z(1-z), y==(z-1)-(z-1)^2/2}, {z}] gives $$ (6-4 x) y+4 y^2 = (-x-6) x $$

For $n=3$, Eliminate[{x==z(1-z), y==(z-1)-(z-1)^2/2+(z-1)^3/3}, {z}] gives $$ (-36 y-66) y = 4 x^3-9 x^2+x (66-72 y) $$