Let $f:(\mathbb{R}^n, 0) \to (\mathbb{R}^n, 0)$ be an analytic germ. Assume that it has isolated zero at 0, that is, $f^{-1}(0)=\{0\}$, what is more, assume that the dimension of the local algebra* $Q(f)$ of $f$ is finite.
$\dim Q(f)$ is equal to the local degree $\deg_0 f_{\mathbb{C}} $ of the complexified germ $f_{\mathbb{C}}: (\mathbb{C}^n, 0) \to (\mathbb{C}^n, 0)$, see Eisenbud-Levine or Mond-Ballesteros. The number of zeros of any generic perturbation of $f_{\mathbb{C}}$ is equal to $\dim Q(f)=\dim_{\mathbb{C}} Q(f_{\mathbb{C}})$.
On the other hand, the number of zeros counted with signs of any generic perturbation of $f$ is equal to the local degree $\deg_0 f$.
Hence the number $k$ of zeros of a generic perturbation of $f$ satisfies $$ |\deg_0 f | \leq k \leq \deg_0 f_{\mathbb{C}} =\dim Q(f) . \ \ **$$ Question: Which are the possible values of $k$? Do you know some results about it?
Specially, can we always reach that all complex zeros of a suitable perturbation are real, implying $k=\dim Q(f)$?
Can we always reach that all zeros of a suitable perturbation have the same sign, implying $k=\deg f$?
*The local algebra of $f$ is the quotient $\mathcal{C}_n/(f)$, where $\mathcal{C}_n$ is the algebra of the analytic germs $(\mathbb{R}^n, 0) \to (\mathbb{R}, 0)$, and $(f)$ is the ideal generated by the components of $f$.
** For a constant ''shift'' perturbation $f_q=f-q$ with a regular value $q \in \mathbb{R}^n$ close to $0$ the inequality (**) holds by definition: $\deg_0 f$ is the sum of the signs of the points of $f_q^{-1}(0)=f^{-1}(q)$, $k$ is the cardinality of $f_q^{-1}(0)$, and $\dim Q(f)$ is the cardinality of $f^{-1}_{\mathbb{C}}(q)$.
