Let $\mathcal{U}\subset \mathbb{R}\times \mathbb{C}$ a neighborhood of $(0,0)$, and $f:\mathcal{U}\to \mathbb{C}$ differentiable in the first variable and holomorphic in the second variable, with $f(0,0)=0$. I want to locally express the zeros of $f$ as one or more curves $z=z(x)$ with $z(0)=0$. The hypothesis I want to eliminate is that $\partial_z f(0,0)\neq 0 $. Instead, suppose that for some $k,h\in \mathbb{N}$ with $k\geq 2$, as $(x,z) \to (0,0)$ we have $$ f(x,z)= z^k+ x^h+ h.o.t.,$$ where $h.o.t.$ stands for higher order terms in $x$ or $z$, i.e. they are $o(z^k)$ or $o(x^h)$ as $(x,z)\to (0,0)$. I wish to prove that there is at least one continuous (not necessarily differentiable) curve $z(x)$ on $\mathcal{U}$ such that $f(x,z(x))=0$ in a neighborhood of $x=0$. Stronger statement: there are $k$ curves $z_j(x)$ such that in a neighborhood of $(0,0)$ we have $f(x,z)=0\iff z=z_j(x)$ for some $j=0,\dots, k-1$.

Could someone point at a specific result from the literature which implies the above (or explain why it doesn't hold)?

Example: At least if $f$ is a polynomial in $z$ then the result must hold. If $n\geq k$ is the degree then the polynomial has $n$ roots but at $(0,0)$ the root only has multiplicity $k$. Because the roots of polynomials depend continuously on the coefficients, which in turn are continuous with respect to $x$, this gives rise to exactly $k$ continuous roots $z_1(x),\dots, z_k(x)$ with $z_j(0)=0$.

Attempts at proof.

The issue is to generalize the result to functions which aren't just polynomials in $z$. My efforts at trying to transform $f$ so that the standard IFT can be applied haven't been successful:

• Define $g(x,z):=f(x,z)^{1/k}_j$, then $g$ is no longer differentiable in $(0,0)$ and hence does not satisfy the assumptions of the IFT. Also, the limit of $f$ to $(0,0)$ does not exist.
• One can apply the IFT to $\partial_z^{k-1}f$ but this is useless. We could even assume by induction that the statement holds for $k'<k$, but the curves where $\partial_z^{k'}f$ vanishes are different for each $k'=1,\dots, k-1$.
• Dividing by $(z)^{k-1}$ doesn't work as the limit of $f$ in $(0,0)$ no longer exists.

The problem is also equivalent to finding a (continuous, differentiable for $x\neq 0$) solution $z(x)$ to the I.V.P. $$z'(x)\partial_z f(x,z(x))=-\partial_x f (x,z(x)),\qquad z(0)=0.$$ In standard IFT, we can divide by $\partial_z f$ and apply the standard local existence and uniqueness result. Here we cannot because $\partial_z f(0,0)=0$. I do not know any existence theorem that would apply in this case. We could perturbate the initial condition by a $w\in \mathbb{C}$, obtaining a curve $z_w(x)$ such that for $w$ small enough, $$ |f(x,z_w(x))|=|f(0,w)|\leq 2|C_1| |w|^k. $$ Here the map $w\mapsto z_w(x)$ is continuous on $\mathbb{C}\setminus \{{0\}}$, but this is not enough to conclude that the limit as $w\to 0$ exists.

This question provides a positive answer to the case $k=2$ but the proof does not extend to higher $k$. I suspect the result might be a special case of this paper but there is too much algebra for me to understand even the statements. Maybe someone could confirm whether they can be applied or not?