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

For a polynomial $P\in\mathbb C[X,Y]$ define its complexification $P^{\mathbb C}(X,Y)=P\bigl(\tfrac{X+Y}{2},\tfrac{X-Y}{2i}\bigr)$ and its Cartesian substitution $P^{\mathrm{cart}}(X,Y)=P(X+iY,\,X-iY)$. The former rewrites a polynomial in the Cartesian coordinates $x=\tfrac{X+Y}{2}$, $y=\tfrac{X-Y}{2i}$ in terms of $X=z=x+iy$ and $Y=\bar z=x-iy$; the latter goes back. 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 $\bigl(P^{\mathrm{cart}}\bigr)^{\mathbb C}=P$, this makes complexification an automorphism of $\mathbb C[X,Y]$ whose inverse is the Cartesian substitution. In the note a real equation $f(x,y)$ is rewritten, just before equation (4), as $P(X,Y)=f\bigl((X+Y)/2,(X-Y)/(2i)\bigr)$; being an automorphism, this substitution preserves irreducibility, which Lemma 3 and Theorems 1 and 2 rely on.