Complex derivative vs two-variable derivative: requiring same limit from different directions I am trying to understand how the CR equations encode the field properties present in $\mathbb{C}$ that aren't present in $\mathbb{R^2}$. Yes, I know their derivation involves multiplying by the inverse of a complex number, which you can't do for a vector, hence the different definition of a multivariable derivative. My question is about approaching along different paths: the crux of the CR derivation seems to involve forcing the limit of the difference quotient of two complex numbers to be the same along both the real and imaginary directions. My question is how this is different from the derivative of a function from $\mathbb{R^2}$ to $\mathbb{R^2}$: do such function not require limits along different paths to be the same? I get that the complex derivative, as a $2 \times 2$ matrix is a special case of a general $Df$ for $f: \mathbb{R^2} \to \mathbb{R^2}$ that enforces the CR condition amongst the entries of the matrix.
I have looked at this question, this question, and a few other questions, and it doesn't seem to answer the heart of my confusion: CR conditions do nothing more than enforce that limits of quotients along two orthogonal axes are the same. Are limits along different (orthogonal) paths allowed to be different in a well defined $Df$? Is this why complex derivatives are a stronger condition than multivariate real derivatives?
I also understand that complex functions can be thought of as functions in conservative vector fields, in some analogous sense (though they are not totally isomorphic).
 A: For uniformity of notation, let's say $f$ is defined in some non-empty open subset $G$ of the real plane (identified with the complex numbers via $(x, y) \leftrightarrow x + iy$), that $z_{0} = (x_{0}, y_{0})$ is a point of $G$.
Using the "linear approximation" definition, a mapping $f$ from $G$ to the plane is differentiable at $z_{0}$ if there exist:

*

*A linear mapping $L = Df(z_{0})$ from the plane to the plane, and

*A function $E$ such that $\frac{|E(h)|}{|h|} \to 0$ as $h \to 0$
such that
$$
f(z_{0} + h) = f(z_{0}) + L(h) + E(h).
$$
The function $f$ is complex differentiable at $z_{0}$ if and only if $Df(z_{0})$ is complex-linear, i.e., $L(ih) = iL(h)$ for all $h$. Not all real-linear transformations satisfy this condition. Under the identification $(x, y) \leftrightarrow x + iy$,

*

*A general real-linear transformation has matrix $\left[\begin{array}{@{}rr@{}}
    a & \phantom{-}c \\
    b & d \\
  \end{array}\right]$ for some real $a$, $b$, $c$, $d$;

*A general complex-linear transformation has matrix $\left[\begin{array}{@{}rr@{}}
    a & -b \\
    b &  a \\
  \end{array}\right]$ for some real $a$, $b$.

Complex-linearity of $Df(z_{0})$ is therefore equivalent to the Cauchy-Riemann equations for the components (real and imaginary parts) of $f$.
A: I hope I can clear up some of the confusion.
First, lets recap what it means complex derivative at a point. The derivative of $f(z)$ at $z_0$ is
$$\lim_{z\rightarrow z_0} \dfrac{f(z)-f(z_0)}{z-z_0}$$ provided this limit exists. For this limit to exist, you must approach $z_0$ from any possible direction in the complex plane, and the limit is always the same complex number (otherwise the limit doesn't exist).
Now, how do Cauchy-Riemann equations work? The most important hypothesis, is that the derivative of $f(z)$ at $z_0$ must exist. So, an assumption is that the limit I wrote before does exist, so I can see what happens from two particular directions and I know those limits are the same value. I could take any two paths I like, but for sake of simplicity, the paths chosen are parallel to the real and imaginary axes, and the rest of C-R equations is a known story.
Now, partial derivatives in $\mathbb{R^2}$ don't work the same way, because for the partial derivative to exist, you don't care what happens from any other direction. For example, consider $g=g(x,y)$, then $\dfrac{\partial g}{\partial x}$ is:
$$\lim_{h\rightarrow 0} \dfrac{g(x+h,y)-g(x,y)}{h}$$ provided this limit exist, BUT you don't need to see what happens from a path parallel to the $y$ axis.
A nice example of this is the so called "crossbar function" defined as:
$$f(x,y)=\left\{ \begin{array} 
.1 &\text{if } xy=0 \\ 
0 &\text{if } xy\neq0 
\end{array} \right.$$
So, its value is $1$ along the $x$ and $y$ axes, and $0$ everywhere else, ressembling the crossbar used to replace car wheels. At the origin, partial derivatives exits and the perpendicular direction doesn't matter. However, at the origin, this function isn't differentiable (I think we have a translation problem here, "derivable" and "diferenciable" are two different concepts that translate to differentiable). Let me point out what a differentiable in $\mathbb{R}^n$ is what I mean: https://en.wikipedia.org/wiki/Differentiable_function#Differentiability_in_higher_dimensions
So, even though Cauchy-Riemann equations only cumpute partial derivatives, those derivatives must be continuous at an open disk, implying that the real and imaginary parts of a complex function must be differentiable at the point we're looking at in order to $f'(z_0)$ exist.
The crossbar function couldn't be the real or imaginary part of a complex function that has complex derivative at $z_0$.
So, yes, in the end, complex derivative can be thought as a "stronger" version of partial derivative in $\mathbb{R}^2$ for its real and imaginary parts, because the former requieres differentiability while the latter doesn't.
I hope I understood your question correctly and that you may find this answer helpful.
