Weierstrass Preparation Theorem, simple exercise I am working with a specific version of the Weierstrass preparation theorem, trying to understand it with an example. I'll cite the theorem before stating the question:

Theorem
If $f:U\subset \mathbb{C}^{n+1}\rightarrow \mathbb{C}$ is a holomorphic function in the variables $(z,w_1,\dots,w_n)$ such that $f(z,0,\dots,0)$ is not identically zero near $z=0$, then there exist

*

*a holomorphic function $h:B \subset \mathbb{C}^{n+1}\rightarrow \mathbb{C}$ with $h(0,\dots,0)\neq 0$,

*a Weierstrass polynomial of the form
$$
p(z,w_1,\dots,w_n) = z^d + \alpha_1(w_1,\dots,w_n) z^{d-1} + \dots + \alpha_d(w_1,\dots,w_n)
$$
where the coefficients are holomorphic functions in $(w_1,\dots,w_n)$ vanishing at the origin,

such that locally near the origin in $\mathbb{C}^{n+1}$ we have $f=h\cdot p$. The polynomial $p$ is unique.

Question: I am trying to get the Weierstrass  polynomial for the following function $f:\mathbb{C}\rightarrow \mathbb{C}$ given by
$$
f(z_1,z_2) = z_1^3z_2 + z_1 z_2 + z_1^2z_2² + z_2^2 + z_1 z_2^3
$$
Obviously $f(z_1,0)$ vanishes identically, so the variable that I need to work on is $z_2$. By differentiating $f=h\cdot p$ with respect to $z_2$ I have been able to show that the degree $d$ is necessarily 2, i.e.
$$
p(z_1,z_2) = z_2^2+\alpha(z_1)z_2 + \beta(z_1),
$$
for holomorphic coefficients $\alpha,\beta$ vanishing at $z_1=0$. However, I am stuck since I have too many degrees of freedom: $h,\alpha,\beta$. How can I construct a method for finding this coefficients?
 A: I have found a solution to the problem by reading the proof of the WPT (as in Proposition 1.1.6 of Huybrechts' Complex Geometry).
We first note that the sections $f(z_1,0) \equiv 0$ and $f(0,z_2)= z_2^2$ so we can only single out the coordinate $z_2$ as "the good coordinate". In order to construct the Weierstrass polynomial, the proof of theorem considers
$$
p_{z_1}(z_2) = \prod_{i=1}^{d} (z_2 - a_i(z_1)),
$$
where $a_i(z_1)$ are the roots of $f_{z_1}(z_2)$ that restrict to the original root $z_2=0$ when the parameter $z_1=0$. The proof of the theorem gives an argument that shows that this is a Weierstrass polynomial in which the coefficients $\alpha_j$ are holomorphic, and written in terms of sums of the form
$$
\sum_{i=1}^{k}a_i(z_2)^k.
$$
which then proves that are holomorphic by using the residue theorem.
In my particular case I can calculate the zeros of $f_{z_1}(z_2)$ by the quadratic formula. Factoring out $z_2$:
$$
f_{z_1}(z_2) = z_2(z_1^3 + z_1 + (z_1^2+z_1)z_2 + z_1 z_2^2)
$$
and by the quadratic formula I get
\begin{align*}
  a_0(z_1) & = 0,\\
  a_1(z_1) &= \frac{-1-z_1^2 + \sqrt{-3 z_1^4 - 2 z_1^2 + 1}}{2 z_1},\\
  a_2(z_1) & = \frac{-1-z_1^2 - \sqrt{-3 z_1^4 - 2 z_1^2 + 1}}{2 z_1}.
 \end{align*}
It is easy to see that $a_2(z_1)$ is not well defined for $z_1\rightarrow 0$, and that $a_1(z_1)\rightarrow 0$ for $z_1\rightarrow 0$. We consider then the Weierstrass polynomial
$$
p_{z_1}(z_2) = z_2 \left( z_2 - \frac{-1-z_1^2 + \sqrt{-3 z_1^4 - 2 z_1^2 + 1}}{2 z_1}\right).
$$
This implies that the holomorphic function $h(z_1,z_2)$ is precisely
\begin{align*}
h(z_1,z_2) & = f(z_1,z_2)/p_{z_1}(z_2) = \frac{z_1 z_2 (z_2-a_1(z_1))(z_2-a_2(z_1))}{z_2(z_2-a_1(z_2))} \\
& = z_1 (z_2 -a_2(z_1)) =
z_1z_2 + \frac{1+z_1^2 + \sqrt{-3 z_1^4 - 2 z_1^2 + 1}}{2} 
\end{align*}
which, note, does not vanish at the origin $h(0,0)= 1$ and is thus a unit in the ring of germs $\mathcal{O}_{\mathbb{C}^{n},0}$.
