If $f$ has no removable singularity at $0$, then either it has an essential singularity there or a pole. If it has an essential singularity at $0$, then $z^mf^{(m)}(z)$ also has an essential singularity at $0$. But it is also bounded near $0$, and this is impossible, by the Casorati-Weierstrass theorem.

And if it has a pole of order $k$ at $0$, then, near $0$, you have $f(z)=\sum_{n=-k}^\infty a_nz^n$. But then $f^{(m)}$ has a pole of order $k+m$ at $0$, and so $\lim_{z\to0}z^{m+k}f^{(m)}(z)$ exists in $\Bbb C\setminus\{0\}$. So, you cannot have $|f^{(m)}(z)|\leqslant\frac M{|z|^m}$, since this implies that $\lim_{z\to0}z^{m+k}f^{(m)}(z)=0$.

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 $\mathcal{O}_{\mathbb{C}^{n},0}$.

## Best Answer

We do not need the assumption $Z(g) \subseteq Z(f)$.

Let $w = (w_1,w_2) \in \mathbb{D}^2$. We want to show that $|f(w)| \leq |g(w)|$.

If $|w_1| = |w_2|$, then $w \in |w_1|\mathbb{T}^2$, and $|f(w)| \leq |g(w)|$ holds by hypothesis.

Thus, suppose $|w_1| \ne |w_2|$ and, wlog, $|w_1| < |w_2| =:r$. Let $F(z) = f(z,w_2)$ and $G(z) = g(z,w_2)$ (functions of one complex variable). For all $z \in \mathbb{C}$ with $|z|=r$, we have $(z,w_2) \in r\mathbb{T}^2$ and, hence, $|F(z)| \leq |G(z)|$. By the maximum principle, it follows that $|F(z)| \leq |G(z)|$ for all $z \in \mathbb{C}$ with $|z| \leq r$. In particular, this holds for $z=w_1$.