An analytic local map (or map germ) $f: (\mathbb{R}^n, 0) \to (\mathbb{R}^n, 0) $ can be considered as a vector field with zero at the origin. Assume that the origin is an isolated zero of $f$. How can we compute the index of the vector field $f$ at the origin?

The index is defined as follows. Let $ S^{n-1}_{\epsilon} \subset \mathbb{R}^n $ be a sphere around the origin with a small enough radius $ \epsilon$. That means there is no zero of $f$ inside $ S^{n-1}_{\epsilon} $ except at the origin. Then the index of $f$ at the origin is the degree of $ f/|f| : S^{n-1}_{\epsilon} \to S^{n-1}$.

Some examples and approaches:

If the Jacobian of $df_0$ is nondegenerate, then the index is $ \pm 1$, i.e. it is equal to $ \text{sign} (\det (df_0))$. In this case the origin is a nondegenerate zero. ( http://www.math.uchicago.edu/~may/VIGRE/VIGRE2009/REUPapers/Popa.pdf )

The index is the multiplicity of the zero at the origin. That is, if we take a small stable perturbation $ \tilde{f} $ of $f$, then $ \tilde{f} $ has nondegenerate zeros, and the sum of the indices of $ \tilde{f}$ is equal to the index of $f$ at the origin. (Do you know some reference for this statement?)

Let $f: \mathbb{C} \to \mathbb{C} $, $f(z)=z^n$. Then the index at the origin is equal to $n$. Let $f(z)= \bar{z}^n$, then the index is $-n$.

Let $f, g: (\mathbb{C}^2, 0) \to (\mathbb{C}, 0)$ be holomororhic germs. Assume that they are irreducible in the local ring $ \mathcal{O}_2$. Then the origin is an isolated zero of $(f,g): (\mathbb{C}^2, 0) \to (\mathbb{C}^2, 0)$, and its index is equal to the intersection multiplicity of the local curves $f^{-1}(0)$ and $g^{-1}(0)$. That is, equal to the complex dimension of $ \mathcal{O}_2 /(f,g) $, where $(f,g)$ denotes the generated ideal.

So the question is: can we calculate the index from the power series of $f$ in general in the real case, or for a holomorphic map (germ) $f: \mathbb{C}^n \to \mathbb{C}^n $?