Why does $f'(z)$ only involve partial derivatives on $x$ and not also on $y$? Let $f=u(x,y)+iv(x,y)$ be a complex function. Now:
$f'(z_0)=\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}=\lim_{z\to z_0}\frac{u(x,y)-u(x_0,y_0) + i(v(x,y)-v(x_0,y_0))}{x-x_0+i(y-y_0)}=\lim_{x\to x_0}\frac{u(x,y_0)-u(x_0,y_0) + i(v(x,y_0)-v(x_0,y_0))}{x-x_0}=\partial_x u + i \partial_x v(x,y)$
I don't get why we just fixed $y=y_0$. Doesn't ' usually describe the total derivative? I'd expect some kind of complex jacobina matrix or whatever.
Is this just a definition? If so, what's the motivation behind it? Why not fix $x=x_0$? [When we set $y=y_0$ what did we think?]
Is the only difference between $\partial_x f(z)$ and $f'(z)$ that in the later, we fixed $y=y_0$?
Also having the $f'(z)$ form above, we can now only talk about the x direction, no?
 A: This is only true when $f$ is a holomorphic function. If $f$ is not holomorphic, the derivative can be different depending on the direction of approach to the point $z_0$. When $f$ is holomorphic, which by definition means that the derivative is the same no matter what the direction of approach, we can derive a relationship between the $x$ and $y$ derivatives of $u$ and $v$:
Approaching along the direction where only $x$ varies gives
$$ f'(x_0+iy_0) = \lim_{x \to x_0} \frac{u(x,y_0)-u(x_0,y_0)+i(v(x,y_0)-v(x_0,y_0))}{x-x_0} = \partial_x u(x_0,y_0) + i \partial_x v(x_0,y_0). $$
Approaching along the direction where only $y$ varies gives
$$ f'(x_0+iy_0) = \lim_{y \to y_0} \frac{u(x_0,y)-u(x_0,y_0)+i(v(x_0,y)-v(x_0,y_0))}{i(y-y_0)} = -i\partial_y u(x_0,y_0) + \partial_y v(x_0,y_0). $$
Equating real and imaginary parts implies that
$$ \partial_x u = \partial_y v, \qquad \partial_x v = -\partial_y u. $$
So it is equally possible to write $f'(x_0+iy_0)$ in terms of partial derivatives with respect to $y$, or even
$$ \frac{1}{2}(\partial_x-i\partial_y) (u+iv) = \frac{1}{2} (\partial_x u -\partial_y v + i(\partial_x v - \partial_y u)) = f'(x_0+iy_0), $$
which is called a Wirtinger derivative, and is useful in higher complex analysis.
So if $f$ is holomorphic, because the derivative does not depend upon the direction of approach to $z_0$, we may choose the direction in which only $x$ varies to compute it. One can equally well choose the direction in which $y$ varies, or any other direction, or even a curve through $z_0$ that is not a straight line.
