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?

Question

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.

Answer 1

The answer is yes. Actualy Rouche's theorem is valid for smooth functions if instead of the number of zeros you count the signed number. (I suppose that this is what you want to count). That is the zero with positive Jacobian is counted with + and zero with negative Jacobian is counted with -. This is called the topological degree.