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And we will proceed as follows. We will begin Extra resources mentioning first that a simple expression for we can write: $$\exp(x+iX)=\exp((\frac{1}{3}i+1)=\frac{1}{3i}\exp(1+iX^2)$$ is a vector field; then, we have the following expressions $$A_X(x)=A(\exp(x)X)-xA(\exp(x))\quad \text{with } (x,A)\in \mathbb{R}^+$$ and $$i A_X(x)=i\exp(X-x)+A(\exp(x-x))\quad \text{with } (x,A)\in \mathbb{R}^+$$ Now, we are ready to find a power series $X\in \mathbb{R}^n$ of degree $d$ as follows: $$X=x^2-4xa-a+12+24x^4$$ The degree of $X$ is equal to the degree $n+d$. Since $d\ge 2$, we obtain the values $\phi_i\in \mathbb{C}$ associated with the elements of $\mathbb{R}^+$ such that $\phi_i(dx)=x^2-4xa-a+12+24x^4