Before introducing the problem, I need to introduce some notations and preliminary result. Let $f=(f_1,\ldots,f_n)$ be a homogeneous complex polynomial system in $n+1$ variables. This means each $f_i$ is a polynomial with degree $d_i\in\mathbb{N}$ such that all monomials of $f_i$ have the same degree $d_i$. The variables are denoted by $X=(X_0,X_1,\ldots, X_n)$ and we say that $f$ has degree $\textbf{d} = (d_1,\ldots,d_n)$.
Once $\textbf{d}$ is fixed, we denote by $\mathcal{H}_\textbf{d}$ the space of all homogeneous complex polynomial systems of degree $\textbf{d}$, as described above.
Fix $\zeta\in\mathbb{C}^n$ not null, then it's possible make a decomposition $\mathcal{H}_\textbf{d} = C_\zeta \oplus L_\zeta \oplus R_\zeta$ in the following way: first define $R_\zeta = \{h\in\mathcal{H}_\textbf{d}:\ h(\zeta)=0,\ Dh(\zeta)=0 \}$, ie, $R_\zeta$ is the space of all systems vanishing at $\zeta$ to higher order. Now define $L_\zeta$ as the subspace of $R_\zeta^\perp$ of the systems vanishing at $\zeta$ (it's not hard to check that $L_\zeta$ indeed is a well defined subspace). Finally, define $C_\zeta$ as the orthogonal complement of $L_\zeta$ in $R_\zeta^\perp$.
For any scalar $\lambda$ not null, we have that $C_{\lambda\zeta} = C_\zeta, L_{\lambda\zeta}=L_{\zeta}, R_{\lambda\zeta}=R_{\zeta}$. For any unitary matrix $U\in\mathbb{C}^{(n+1)\times (n+1)}$ and any subset $A\subset \mathcal{H}_\textbf{d}$, denote $UA=\{ f\circ U^\ast:\ f\in A \}$. We have that $UA$ is a subset of $\mathcal{H}_\textbf{d}$, also we have that $UC_\zeta = C_{U\zeta}, UL_\zeta = L_{U\zeta}, UR_\zeta = R_{U\zeta}$.
What I want to do is to find explicit formulas for the decomposition $$f = k + g + h \in C_\zeta\oplus L_\zeta\oplus R_\zeta. $$
I have a little hint for the prove:
Write $f_i$ expanding according to the powers of $X_0$ as $f_i = c_iX_0 + X_0^{d_i-1}\sqrt{d_i}\sum_{j=1}^n a_{ij}X_j + h_i$. My textbook considers important (for some reasons not so related to this problem in particular) to detach the terms $\sqrt{d_i}$, but this is not an issue. If we start with $\zeta = e_0$ we have $f_i(e_0) = c_i, \frac{\partial f_i(e_0)}{\partial X_0} = d_ic_i, \frac{\partial f_i(e_0)}{\partial X_j} = \sqrt{d_i}a_{ij}$, for $j\neq 0$. Note that $f\in R_{e_0} \iff c_i$ and $a_{ij} = 0$ for all $i,j$, which means $f_i=h_i$. Suppose now that $f\in R_{e_0}^\perp$ (so $h_i = 0$ for all $i$). In this case, we have $f\in L_{e_0}\iff c_i = 0$ for all $i$, which means $f_i = X_0^{d_i-1}\sqrt{d_i}\sum_{j=1}^n a_{ij}X_j$. Finally, we have $f\in C_{e_0} \iff a_{ij}=0$ for all $i,j$, which means $f_i = c_iX_0^{d_i}$. Putting all together, we conclude that $f_i=k_i+g_i+h_i$, where $$k_i = c_iX_0^{d_i} = f_i(e_0)X_0^{d_i}$$ and $$g_i = X_0^{d_i-1}\sqrt{d_i}\sum_{j=1}^n a_{ij}X_j = X_0^{d_i-1}\sqrt{d_i}\sum_{j=1}^n \frac{1}{\sqrt{d_i}}\frac{\partial f_i(e_0)}{\partial X_j} X_j.$$
The idea to find a formula for a general $\zeta$ is to start scaling $e_0$ by some factor $\lambda$, then using some unitary transformation $U$ such that $\zeta = \lambda Ue_0$.
Therefore, consider the decomposition $f\circ U = \tilde{f} = \tilde{k}+\tilde{g}+\tilde{h}\in C_{e_0}\oplus L_{e_0}\oplus R_{e_0}$. It follows that $f = \tilde{k}\circ U^\ast + \tilde{g}\circ U^\ast + \tilde{h}\circ U^\ast \in UC_{e_0}\oplus UL_{e_0}\oplus UR_{e_0} = UC_{\lambda e_0}\oplus UL_{\lambda e_0}\oplus UR_{\lambda e_0} =$ $ = C_\zeta\oplus L_\zeta \oplus R_\zeta$, so we have the decomposition. All that is left is to write $k = \tilde{k}\circ U^\ast$ and $g = \tilde{g}\circ U^\ast$ explicitly. I'm having trouble with what I found for $g$. That's where I hope you can help me. Of course, if there is any other problem please let me know.
My attempt: First, note that the $j-$th coordinate of $U^\ast X$ is $\langle X, U_j\rangle$, where $\langle, \rangle$ is the Hermitian inner product in $\mathbb{C}^{n+1}$ and $U_j$ is the $j-$th column of $U$.
Let's start with $k_i$. Remember that $\tilde{k_i} = \tilde{f_i}(e_0)X_0^{d_i}$, therefore $$k_i = \tilde{f_i}(e_0)\langle X, U_0\rangle^{d_i} = f_i(Ue_0)\langle X, Ue_0\rangle^{d_i} = \frac{\lambda^{d_i}\overline{\lambda^{d_i}}}{\lambda^{d_i}\overline{\lambda^{d_i}}}f_i(Ue_0)\langle X, Ue_0\rangle^{d_i} =$$ $$ = \frac{f_i(\lambda Ue_0)}{|\lambda|^{2d_i}}\langle X, \lambda Ue_0 \rangle^{d_i} = \frac{f_i(\zeta)}{\|\zeta\|^{2d_i}} \langle X, \zeta \rangle^{d_i}.$$
For $g_i$ I want to do something similar, in particular, I don't want $\lambda$'s or terms of $U$ appearing. We can try starting like before. Remember that $\tilde{g_i} = X_0^{d_i-1}\sqrt{d_i}\sum_{j=1}^n \frac{1}{\sqrt{d_i}}\frac{\partial \tilde{f_i}(e_0)}{\partial X_j} X_j$, therefore $$g_i = \langle X,U_0\rangle^{d_i-1}\sqrt{d_i}\sum_{j=1}^n \frac{1}{\sqrt{d_i}}\frac{\partial \tilde{f_i}(e_0)}{\partial X_j} \langle X,U_j\rangle. $$
Note that $$\frac{\partial \tilde{f_i}(e_0)}{\partial X_j} = \frac{\partial (f_i\circ U)(e_0)}{\partial X_j} = Df_i(Ue_0)DU(e_0)e_j = Df_i(Ue_0)Ue_j = $$ $$= Df_i(Ue_0)U_j = \frac{\lambda^{d_i-1}}{\lambda^{d_i-1}}Df_i(Ue_0)U_j =\frac{Df_i(\lambda Ue_0)}{\lambda^{d_i-1}}U_j = \frac{Df_i(\zeta)}{\lambda^{d_i-1}}U_j.$$
Replacing above, we get $$g_i = \langle X,U_0\rangle^{d_i-1}\sqrt{d_i}\sum_{j=1}^n \frac{1}{\sqrt{d_i}}\frac{Df_i(\zeta)}{\lambda^{d_i-1}}U_j \langle X,U_j\rangle =$$ $$= \langle X,U_0\rangle^{d_i-1}\sqrt{d_i}\sum_{j=1}^n \frac{1}{\sqrt{d_i}\lambda^{d_i-1}}Df_i(\zeta)U_j\cdot\overline{U_j}^T X =$$ $$= \overline{\lambda^{d_i-1}}\langle X,U_0\rangle^{d_i-1}\sqrt{d_i}\sum_{j=1}^n \frac{1}{\sqrt{d_i}\overline{\lambda^{d_i-1}}\lambda^{d_i-1}}Df_i(\zeta)U_j\cdot\overline{U_j}^T X = $$ $$= \frac{\langle X,\zeta\rangle^{d_i-1}}{\|\zeta\|^{2(d_i-1)}}\sqrt{d_i}\sum_{j=1}^n \frac{1}{\sqrt{d_i}}Df_i(\zeta)U_j\cdot\overline{U_j}^T X. $$
Now I want to get get rid of the matrix $U_j\cdot\overline{U_j}^T$, to write it in terms of $d_i, \zeta, f_i$ and it's derivatives (assuming I didn't made any mistakes). I just don't know how to proceed.
EDIT: After some thought, I was able to make the term $U_j\cdot\overline{U_j}^T$ disappear, but this lead me to a contradiction. So...things are still not right and I still don't know what is wrong. Anyway, note that $$\sum_{j=1}^n Df_i(\zeta)U_j\overline{U_j}^T X = \sum_{j=1}^n\left( Df_i(\zeta) U_j\sum_{m=0}^nX_m\overline{U_{mj}} \right) = \sum_{m=0}^n\left( X_mDf_i(\zeta)\sum_{j=1}^n U_j\overline{U_{mj}} \right) = $$ $$= \sum_{m=0}^n\left( X_mDf_i(\zeta)U\cdot \left[\begin{array}{c} \overline{U_{m0}}\\ \vdots\\ \overline{U_{mn}} \end{array}\right]\right) = \sum_{m=0}^n \left( X_m Df_i(\zeta)UU^\ast e_m \right) = $$ $$ = \sum_{m=0}^n X_m Df_i(\zeta)e_m = \sum_{m=0}^n X_m \frac{\partial f_i(\zeta)}{\partial X_m}.$$
Therefore, $$g_i = \frac{\langle X,\zeta\rangle^{d_i-1}}{\|\zeta\|^{2(d_i-1)}}\sqrt{d_i} \sum_{j=0}^n \frac{1}{\sqrt{d_i}} X_j \frac{\partial f_i(\zeta)}{\partial X_j}.$$ This would be a nice solution, but it's not correct. In fact, we can consider that $\zeta$ is not a zero of $f_i$. By definition, we know that $g_i(\zeta) = 0$, for $g\in L_\zeta$. On the other hand, note that $$g_i(\zeta) = \frac{\langle \zeta,\zeta\rangle^{d_i-1}}{\|\zeta\|^{2(d_i-1)}}\sqrt{d_i} \sum_{j=0}^n \frac{1}{\sqrt{d_i}} \zeta_j \frac{\partial f_i(\zeta)}{\partial X_j} = \sum_{j=0}^n \zeta_j \frac{\partial f_i(\zeta)}{\partial X_j} = d_i f_i(\zeta) \neq 0.$$
The last equality comes from the Euler's Formula. This is a contradiction, for $g_i(\zeta) = 0$.