# If you perturb a polynomial by a smooth function, then is the signed number of small zeros of the perturbed equation the same as the lowest non zero derivative?

Let $f: \mathbb{C} \rightarrow \mathbb{C}$ be a function of the form

$$f(z) = z^n + z^{n+ 1} g(z)$$ where $g$ is a $\textbf{smooth}$ function (not necessarily holomorphic).

Is it true that the number of solutions counted with a sign near'' the origin for the equation $$f(z)-\nu =0$$ is $n$, where $\nu$ is a small perturbation? More precisely, this is what I mean:

Let $\nu : \mathbb{C} \rightarrow \mathbb{C}$ be a smooth function that is zero outside a compact set. Let the supremum norm (which makes sense) be less than $\epsilon$. Consider the equation $$f(z) -\nu =0$$ restricted to an open ball of radius $R$. Also assume that restricted to this open ball zero is a regular value of the function $f(z) - \nu(z)$. (I think one can show such a $\nu$ exists, in fact any generic $\nu$ will satisfy that). Hence now we can ask how many solutions does the equation $$f(z) - \nu(z) =0$$ have inside a ball of radius $R$. My question is that if $\epsilon$ and $R$ are sufficiently small then is it going to be $n$? Its certainly true if $g$ was holomorphic.

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