Milnor Number of real and imaginary parts of holomorphic germs? By performing some computations using the Singular software, I've noticed the following pattern: if $\mu$ is the Milnor Number of a holomorphic germ $f\in \mathcal{O}_n$ at the origin, then the Milnor Number of its real and imaginary parts are equal and are $\mu^2$, that is,
$$
\mu_{(\mathcal{R}e(f),0)} = \mu_{f,0}^2  
$$
By Milnor number of the real part of $f$, I mean $u=\mathcal{R}e(f)$ as a germ of real analytic function of $2n$ real variables (the real and imaginary parts of each complex variable). If $\mathcal{A_{2n}}$ is the ring of germs of such real analytic functions, then
$$
\mu_{(\mathcal{R}e(f),0)}=\text{dim}_{\mathbb{R}} \dfrac{\mathcal{A}_{2n}}{\left<\frac{\partial u}{\partial x_1},..., \frac{\partial u}{\partial x_n}, \frac{\partial u}{\partial y_1},..., \frac{\partial u}{\partial y_n}\right> }
$$
I've noticed the same happens for the imaginary part of $f$.
I would like to know if there's any generalisation to this. I've tried using some direct sum properties on ideals but got nowhere. I suspect there might be some tensor products involved, but also got nowhere. 
 A: Let's try a partial solution, i.e. I'll prove the formula for the complex milnor number of $Re(f)$:
For two germs $f\in\mathcal{O}_{k,0}$, $g\in\mathcal{O}_{l,0}$ with isolated singularities we can define $f+g\in\mathcal{O}_{k+l,0}$ and we have $\mu_{f+g}=\mu_f\cdot \mu_g$. Outline of a proof: Take morsifications $f_\epsilon,g_\delta$. Then $f_\epsilon+g_\delta$ is a morsifications of $f+g$ and the non-degenerated singularities are given by the cartesian product of the singularities of $f_\epsilon$ and $g_\delta$.
Let $g\in\mathcal{O}_{2n,0}$ the complex extension of $Re(f)$, whereby $x,y$ denote the coordinates corresponding to the real/imaginary parts of $f$. Then we can write $2g= f(x+iy)+\tilde f(x-iy)$ ($\tilde f$ is the germ we get by conjugation of the coefficients). Applying the above mentions result we get $\mu_{Re(f)}=\mu_{f(x+iy)}\cdot\mu_{\tilde f(x-iy)} = \mu_f^2$.
Remark on coordinates. Define $u_k:=x_k+iy_k, v_k:=x_k-iy_k$. We get $2x_k=u_k+v_k$ and $2iy_k=u_k-v_k$. Therefore we have a change of coordinates and we get $2g(x,y)=f(u)+f(v)$, such that the formula for the Milnor number is applicable.
It remains to show $\mu^\mathbb{R}_{Re(f)} =\mu^\mathbb{C}_{Re(f)}$. Unfortunately we have only $\mathcal{A}_{2n}\otimes\mathbb {C}\subset \mathcal{O}_{2n}$, whereby $\mathcal{A}_{2n}\otimes\mathbb {C}$ is dense in $\mathcal{O}_{2n}$.
To show $\mu^\mathbb{R}_{Re(f)} =\mu^\mathbb{C}_{Re(f)}$ I would try
$$
\text{dim}_{\mathbb{R}} \dfrac{\mathcal{A}_{2n}}{\left<\frac{\partial u}{\partial x_1},..., \frac{\partial u}{\partial x_n}, \frac{\partial u}{\partial y_1},..., \frac{\partial u}{\partial y_n}\right> } = \\
\text{dim}_{\mathbb{C}} \dfrac{\mathcal{A}_{2n}\otimes\mathbb{C}}{\left<\frac{\partial g}{\partial x_1},..., \frac{\partial g}{\partial x_n}, \frac{\partial g}{\partial y_1},..., \frac{\partial g}{\partial y_n}\right> } = \\
\text{dim}_{\mathbb{C}} \dfrac{\mathcal{O}_{2n}}{\left<\frac{\partial g}{\partial x_1},..., \frac{\partial g}{\partial x_n}, \frac{\partial g}{\partial y_1},..., \frac{\partial g}{\partial y_n}\right> }
$$
The first equality follows from some properties on tensor product (Didn't checked it in detail).
For the second equality we need to do some more work. if we compare the rings $\mathcal{A}_{2n}\otimes\mathbb{C}\subset \mathcal{O}_{2n}$ we have for $f\in\mathcal{O}_{2n}$: $f\in \mathcal{A}_{2n}\otimes\mathbb{C}$ iff the coefficients of the power series splits into finite many subset such that the coefficients in a subset only differ by a real factor, i.e. we have a finite sum $f=\sum c_k f_k$ with $f_k\in\mathcal{A}_{2n}, c_k\in\mathbb{C}$.
The two rings are both local and the maximal ideals ($\mathfrak{m}$) are genereate by the same elements. Therefore we have $(\mathcal{A}_{2n}\otimes\mathbb{C})/\mathfrak{m}^k=\mathcal{O}_{2n}/\mathfrak{m}^k$ because they have finite $\mathbb{C}$-dimension. If we know $\mu^{\mathbb{R}}<\infty$ (*) we can choose $k\gg 1$ such that
$$ \dfrac{\mathcal{A}_{2n}\otimes\mathbb{C}}{\left<\frac{\partial g}{\partial x_1},..., \frac{\partial g}{\partial x_n}, \frac{\partial g}{\partial y_1},..., \frac{\partial g}{\partial y_n}\right> } = 
\dfrac{(\mathcal{A}_{2n}\otimes\mathbb{C})/\mathfrak{m}^k}{\left<\frac{\partial g}{\partial x_1},..., \frac{\partial g}{\partial x_n}, \frac{\partial g}{\partial y_1},..., \frac{\partial g}{\partial y_n}\right> }
$$
and the same for $\mathcal{O}_{2n}$ and we get $\mu^\mathbb{R}_{Re(f)} =\mu^\mathbb{C}_{Re(f)}$.
(*) If we can show $\mu^{\mathbb{C}}<\infty \Rightarrow \mu^{\mathbb{R}}<\infty$ the proof would be complete.
