When applying the Milne-Thomson Method to a function such as $$f(z)=u(x,y)+iv(x,y)$$ we take $z=z^¯$ which gives us:
$$f(z)=u(z,0)+iv(z,0)$$
I have not been able to find a thorough explanation as to why we can simply regard $x$ as equal to $z$ and $y$ equal to $0$. How would this logic be proven mathematically?
Assume $f(z) := (a+bi)(x+yi)^n = u(x,y) + iv(x,y)$ is a monomial in $\,z.\,$ When we replace $x$ with $z$ and $y$ with $0$ in $u$ and $v$ we get back $f(z).$ By linearity this holds for any polynomial $f(z)$ in $\,z.\,$ Now use the complex version of the Stone-Weierstrass theorem for approximation of analytic function by polynomials to conclude the result.