Domain of a complex-valued function Let $f:\mathbb{R}^2 \to \mathbb{C}$ be function from the real plane to a complex plane which we denote by $f(x,y)$. If we introduce the new variables $z = x + iy$ and $\bar{z} = x - iy$, what is the domain of the new function $g(z,\bar{z}) = f(x,y)$? Is it $g:\mathbb{C} \to \mathbb{C}$ or $g:\mathbb{C}^2 \to \mathbb{C}$?
 A: Inverting the definition of $z$ and $\bar z$ you have $x={z+\bar z\over2}$, $y={z-\bar z\over 2i}$. The correct definition of $g$ would therefore have to be
$$g(z,\bar z):=f\left({z+\bar z\over2},{z-\bar z\over 2i}\right)\ .\tag{1}$$
If $f$ is only defined for real $x$ and $y$ then the RHS of $(1)$ is undefined unless $\bar z$ happens to be the complex conjugate of $z$. In other words: You cannot consider $g$ as a function of two independent complex variables $z$ and $\bar z$.
It is another thing if $$f:\quad{\mathbb R}\times{\mathbb R}\to{\mathbb C},\quad (x,y)\mapsto f(x,y)$$
has a natural extension to the domain ${\mathbb C}\times{\mathbb C}$. You can then make  use of this extension in $(1)$, but have to say so.
A: The function $g(a, b)$, as you define it, is a function of 2 complex variables $\mathbb{C}^2$, though you may define only at the line where $b = \bar{a}$ wich is a 1-dimensional complex mainfold $\subset \mathbb{C}^2$.
The function $h(z) = g(z, \bar{z})$ has a domain $\mathbb{C}$.
A: If your function takes two variables (i.e $f(z,w)=zw^2$), then the function is indeed $f:\mathbb C^2 \to \mathbb C$. 
Now if you're only using the conjugate of the input variable and you wanted a function like $g(z)=f(z,\overline z)=z\cdot\overline z^2$, then this is $g:\mathbb C \to \mathbb C$.
