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I understand that the Cauchy Riemann condition for complex function starts by defining differentiable complex functions as those whose derivative exists and is the same regardless of whether we go along the real axis or the imaginary.

Let $z=x+iy$. The function $f(z)$ is differentiable if $\frac{f(x+\Delta x+i y) - f(x+i y)}{\Delta x} = \frac{f(x+i(y+\Delta y)) - f(x+i y)}{i\Delta y}$

As a definition and what follows, it's fine but can anyone explain (to a physicist) why derivatives should be direction independent? I imagine there are plenty of functions that have direction dependent derivatives but I'd like to understand the motivation behind this way of defining differentiability.

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The whole point is that we do not want to deal with general functions of two variables, but rather with functions of the single complex variable $z$, with respect to which we want the derivative to exist, if we get different results depending on the direction then we don't have a well-defined function $f'(z)$.

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Even for functions on $\mathbb R$ we already care about direction independence: consider $f(x) = |x|$, which has different left and right directional derivatives at $x=0$, and is therefore not considered differentiable there, since there is no single value that one can ascribe as the derivative of $f$ at $0$.

It's only natural that when defining the complex derivative at a given point, we continue to require that the derivative is independent of a particular direction. The definition of differentiable, as implicit in Shahar's answer, is that the derivative exists.

Certainly there are more general notions of differentiability that admit set-valued or direction-specific derivatives, but I see no natural reason to use the term "differentiable" (without any qualifiers) in those contexts.

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The claim is that IF $f$ is differentiable, then the Cauchy-Riemann eqns are satisfied. This is so because as you point out the result must be independent of choice of direction of the limit. The converse is NOT true however. It may happen the the C-R eqns are satisfied but $f$ is not differentiable. Take $f$ to be a function that is $1$ everywhere except on the $x,y$ axes, where we define it to be $0$. Clearly $f$ not continuous at $(0,0)$ but the partials exist. For the converse, you need the extra condition that $f$ has continuous partial derivatives at $ z_0$. In any case, the best way to view differentiation of complex-valued functions of a cmplex variable, for theoretical reasons, is using series.

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  • $\begingroup$ Could you elaborate the last part, which is the main thrust of my question? Why (from a series pov) should I care about the direction independence of differentiability? $\endgroup$ – user1936752 Sep 7 '15 at 15:12

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