The Cartesian substitution followed by complexification is the identity of $\mathbb C[X,Y]$

For a polynomial $P\in\mathbb C[X,Y]$ let $P^{\mathrm{cart}}(X,Y)=P(X+iY,\,X-iY)$ be its Cartesian substitution and $P^{\mathbb C}(X,Y)=P\bigl(\tfrac{X+Y}{2},\tfrac{X-Y}{2i}\bigr)$ its complexification, which rewrites a polynomial in $x=\tfrac{X+Y}{2}$, $y=\tfrac{X-Y}{2i}$ in terms of $X=z=x+iy$ and $Y=\bar z=x-iy$. Both are $\mathbb C$-algebra endomorphisms of $\mathbb C[X,Y]$.

Then for every $P\in\mathbb C[X,Y]$,

Together with the reverse identity $\bigl(P^{\mathbb C}\bigr)^{\mathrm{cart}}=P$, it shows that complexification, the passage made just before equation (4) of the note from a real equation $f(x,y)$ to $P(X,Y)=f\bigl((X+Y)/2,(X-Y)/(2i)\bigr)$, is an automorphism of $\mathbb C[X,Y]$. This direction is what allows an equation given in the coordinates $z,\bar z$ to be converted back into a Cartesian one, as in the descent to real Cartesian equations used for the examples of Theorems 1 and 2.