Say I have a two variable polynomial
$$f(x,y)=xy(1-x-y)$$
where $x$ and $y$ are real.
The solutions to $f(x,y)=0$ are the three lines $x=0$, $y=0$ and $x=1-y$.
I am interested in the "double zeroes" of this function (not sure if this is the appropriate terminology), which I call the points where these three lines intersect. These are namely $(0,0)$, $(0,1)$ and $(1,0)$.
The fact that these points are pairs of real values $(a,b)$ made me wonder if there was a better way to encapsulate the double zeroes of this function in terms of complex numbers. For example the function
$$g(z)=z(z-1)(z-i)$$
with $z$ complex, has zeroes at the same points as the "double zeroes" of $f(x,y)$ if one associates $C\to R^2$. $g$ is clearly not unique, but for a holomorphic polynomial I believe that the most general form would have to be of the form
$$g(z)=a\cdot z^{n_1}(z-1)^{n_2}(z-i)^{n_3}$$
with $n_i$ integers.
My question is, is it possible to obtain a polynomial $g(z)$ from the polynomial $f(x,y)$, without knowing the locations of the "double zeroes" of $f$?. For the example above finding the roots of $f$ was easy, but I am working with complicated polynomials for which finding the "double zeroes" directly to construct a function $g(z)$ is very computationally intensive and must be avoided at all costs.
Thank you very much in advance for any help.
What you're talking about are singular points on a plane curve.
These satisfy at least the condition $f=\partial f/\partial x=\partial f/\partial y=0$.
"Double zero" refers, I think, to crunodes.
At any rate, I can't imagine how you would get such a complex polynomial in an easy way from the coefficients of $f$.
I've done a few searches on Google and cannot find any way to relate complex polynomials to the singular points of an affine real plane curve. All I see are articles on the ways to compute singular points.
A quote from "Singular Points of Algebraic Curves" by Sakkalis and Farouki: