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Let $f$ be a holomorphic function defined in $U\in \mathbb{C}^n$, and $f=0$ gives an analytic variety. Suppose $f$ is irreducible as a germ of holomorphic function in $\mathcal{A}_z$, here $z$ is a fixed point in $U$, $\mathcal{A_z}$ denotes ring of germs of holomorphic function defined near $z$.

Question: Can we find a reparametrisation of $f(z)$ to be a polynomial?

More precisely, can we find a biholomorphic map $g:V\rightarrow B(z_,\epsilon)$ for some open set $V\subset \mathbb{C}^n$ and some $\epsilon>0$ such that $f\circ g:V\rightarrow \mathbb{C}$ is a polynomial function?

When $n=1$, we can always find such $g$ since every holomorphic function $f$ of one variable can be written as $f(z)=f(z_0)+a(z-z_0)^m+$Higher Order Terms ($a\neq0$), then we can find a single-value branch $h(z)$ of the multi-value function $(f(z)-f(z_0))^{1/m}$ near $z_0$, and we always have $h'(z_0)=a^{1/m}\neq 0$, hence $f(z)=f(z_0)+h(z)^m$, hence $g=h^{-1}$ gives a reparametrisation of $f$ as $(f\circ g) (w)=f(z_0)+w^m$.

Also, when $n\geq 2$ and $df_{z_0}\neq 0$, then $g$ always exists by implicit function theorem. But I don't know whether such $g$ exists when $z_0$ is a singular point of $f=0$.

Any comments are welcome.

Remark: I found a closed related question here:

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