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](https://mathoverflow.net/questions/85490/implicit-function-theorem-at-a-singular-point) 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](https://arxiv.org/pdf/1311.0088.pdf) but there is too much algebra for me to understand even the statements. Maybe someone could confirm whether they can be applied or not?



Implicit function theorem with singularities of any order
[Rouché's theorem](https://en.wikipedia.org/wiki/Rouch%C3%A9%27s_theorem) provides a simple solution.


Let $f\_1(x,z)=z^k+x^h$ and $f\_2(x,z)=h.o.t.$. For each $x$, the functions $f\_1$ and $f\_2$ are holomorphic in $z$. We know $f\_2(x,z)=o(z^k)+o(x^h)$, so there exist $z\_0,x\_0>0$ such that for $|z|\leq z\_0$ and $|x|\leq x\_0$ we have
$$ |f\_2(x,z)|\leq \frac{1}{4}(|z|^k+|x|^h).$$
(Note that this is weaker than the type of inequality discussed in the comments).


Fix now any $|x|\leq x\_0$ small enough so that
$$r:= (2|x|^h)^{1/k}\leq z\_0. $$
For $|z|=r$ we then have
$$|f\_2(x,z)|\leq \frac{1}{4}\left( |z|^k+|x|^h\right)= \frac{3}{8}r^k< \frac{1}{2}r^k= |z|^k-|x|^h\leq |f\_1(x,z)|. $$
Therefore by Rouché's theorem, we deduce that $f\_1(x,\cdot)$ and $f(x,\cdot)=f\_1(x,\cdot)+f\_2(x,\cdot)$ have the same number of zeroes in $B\_r(0)$. Since $f\_1(x,\cdot)$ has $k$ zeroes, thus $f(x,\cdot)$ has $k$ zeroes $z\_1(x),\dots,z\_k(x)$ in $B\_r(0)$. In particular, we have
$$\lim\_{x\to 0}|z\_j(x)|\leq \lim\_{x\to 0}(2|x|^h)^{1/k}=0,\qquad j=1,\dots, k. $$
This proves that the $z\_j$ are at least continuous at the origin. In order to extend this to a neighborhood it should suffice to apply Rouche's theorem to balls centered on the individual roots of $f\_1$, but I am content with this. Besides, remarkably, the above proof does not require any regularity in $x$.


