I'm reading about holomorphic functions in section 1.2 of Complex Analysis by Stein and Shakarchi, and I am pretty confused about the derivative notation that the authors employ. In this section the authors derive the Cauchy-Riemann equations for a complex-valued function $f: \Omega \to \mathbb{R}$, where $\Omega$ is an open subset of $\mathbb{C}$. The following passage from page 11 has me confused:
...consider the limit in (1) [$\lim_{h \to 0} \frac{f(z_0+h) - f(z_0)}{h}$] when $h$ is first real, say $h = h_1 + ih_2$ with $h_2 = 0$. Then, if we write $z = x + iy, \, z_0 = x_0 + iy_0$, and $f(z) = f(x,y)$, we find that \begin{align*} f'(z_0) &= \lim_{h_1 \to 0} \frac{f(x_0 + h_1, y_0) - f(x_0,y_0)}{h_1} \\[5pt] &= \frac{\partial f}{\partial x}(z_0), \end{align*}
where $\partial/\partial x$ denotes the usual partial derivative in the $x$ variable. (We fix $y_0$ and think of $f$ as a complex-valued function of the single real variable $x$.) Now taking $h$ purely imaginary, say $h = ih_2$, a similar argument yields \begin{align*} f'(z_0) &= \lim_{h_2 \to 0} \frac{f(x_0, y_0 + h_2) - f(x_0,y_0)}{ih_2} \\[5pt] &= \frac{1}{i} \frac{\partial f}{\partial y}(z_0), \end{align*}
where $\partial/\partial y$ is partial differentiation in the $y$ variable.
What exactly is the author's definition of $\frac{\partial f}{\partial x}(z_0)$ and $\frac{\partial f}{\partial y}(z_0)$? I find this very confusing as the author seems to be using $f$ to represent two different functions, one with domain $\Omega$ and the other with domain $\{(x,y): x + iy \in \Omega \}$...
On page 11 of my copy, the authors of speak of "associating each complex-valued function $f=u+iv$ with the mapping $F(x,y)=(u(x,y),v(x,y))$ from $\mathbb R^2$ to $\mathbb R^2$". This suggests that when they use notation like $\frac{\partial f}{\partial x}(z_0)$ for functions $\mathbb C\to\mathbb C$, a string of natural identifications are taking place in the background.
If we write $z_0=x_0+iy_0$ with $x_0,y_0$ real, then the corresponding vector in $\mathbb R^2$ is $(x_0,y_0)$. The notation $\frac{\partial f}{\partial x}(z_0)$ thus actually means "the complex number identified with $\frac{\partial F}{\partial x}(x_0,y_0)$". Explicitly, $\frac{\partial f}{\partial x}(z_0)$ is the complex number identified with the following vector in $\mathbb R^2$: $$ \frac{\partial F}{\partial x}(x_0,y_0)=\lim_{h\to 0}\frac{F(x_0+h,y_0)-F(x_0,y_0)}{h} \, . $$ If you want a more direct definition of $\frac{\partial f}{\partial x}(z_0)$, it is $$ \lim_{h\to 0}\frac{f(z_0+h)-f(z_0)}{h} \, , $$ where $h$ is a real number tending to zero. Similarly, $$ \frac{\partial f}{\partial y}(z_0)=\lim_{h\to 0}\frac{f(z_0+hi)-f(z_0)}{h} \, . $$ Remark: this means that if $f'(z)$ exists, then $f'(z)=\frac{\partial f}{\partial x}(z_0)$; however, the existence of $f'(z)$ is a much stronger condition than the existence of $\frac{\partial f}{\partial x}(z_0)$. Intuitively speaking, this is because in the definition of $f'(z)$, the number $h$ is complex, and so we are mandating that $\frac{f(z+h)-f(z)}{h}$ approaches the same limit regardless of the path in the complex plane that $h$ takes as it approaches $0$. Indeed, if $f'(z)$ exists for all $z$ on some open set $U$, then $f$ is infinitely differentiable (even analytic) on $U$, and the partial derivatives of all orders exist; this is one of the miracles of complex analysis.
NB Given that $\mathbb C$ is often formally defined as $\mathbb R^2$ together with the operations $(a,b)+(c,d)=(a+c,b+d)$ and $(a,b)\cdot(c,d)=(ac-bd,ad+bc)$, one shouldn't worry too much about these "identifications" seeming unrigorous.