Polynomials that suffice $S = V(f,g)$ for $S$ finite set in $\mathbb{A}^2(\mathbb{C})$ Let $ S \subset \mathbb{A}^2(\mathbb{C})$  a finite set of points. Show that exists polynomials $f$ and $g$ such that S = V(f,g).

I know that if I am working in $\mathbb{A}^2(\mathbb{R})$ I can make horizontal lines and little circles tangent to those lines and make the polynomials $f$ and $g$ the multiplication of such lines and circles, (given that the radius of these circles is less then the minimum of the vertical distance between all points).

My problem is if we consider $\mathbb{A}^2(\mathbb{C})$, because then I lose control of other intersection points. How can I solve this prroblem considering in $\mathbb{A}^2(\mathbb{C})$?
Suppose $S = \{ (a_1, b_1), (a_2, b_2), \ldots, (a_n, b_n) \}$.  First, find a solution for the "generic" case in which all $a_i$ are distinct.  (Hint to get started: there exists a polynomial function $p$ such that $p(a_1) = b_1, \ldots, p(a_n) = b_n$.)

I will now give the idea of how to treat the general case.  The idea will be to find some perturbation of $\mathbb{A}^2(\mathbb{C})$ which maps our problem into a problem of the generic case which was solved in the previous paragraph.  As it turns out, instead of considering the whole class of automorphisms of $\mathbb{A}^2(\mathbb{C})$, it will suffice to consider a class of shear transformations $\phi_\lambda$, $(x, y) \mapsto (x + \lambda y, y)$.  So, what you will need to prove is:


There exists $\lambda \in \mathbb{C}$ such that $\phi_\lambda(S)$ has all $x$-coordinates distinct.
There exist $f,g$ such that $S = V(f,g)$ if and only if there exist $f', g'$ such that $\phi_\lambda(S) = V(f', g')$.
$S=\{(a_1,b_1),\dotsc,(a_n,b_n)\}$

$P=(z_1-a_1)\dotsb(z_n-a_n)$, $Q=(z_2-b_1)\dotsb(z-b_n)$.