If the complex function $f(z)$ is complex differentiable $\Rightarrow$ the Cauchy Riemann equations hold. $($This is because if $f'(z)$ is the same no matter in what direction $\delta z\rightarrow 0$. Choosing the special case of $\delta z\rightarrow 0$ along the real, then the imaginary, line yields the Cauchy Riemann equations, so it is obvious that any differentiable function will satisfy them$)$.
However, how is demonstrating that $f'(z)$ is the same for two perpendicular paths enough to show that $f'(z)$ is the same for all paths (i.e. how does one go from $\Rightarrow$ to $\Leftrightarrow$)? I have tried explicitly calculating the directional derivatives, but the mess that ensued did not enlighten whatsoever.
I do not know much about linear algebra if that helps in writing answers. Any geometric intuition would be greatly appreciated!
Best Answer
The directional derivative in any direction in $\mathbb{R}^n$ is completely determined by the directional derivative in $n$ linearly independent directions. This somewhat counterintuitive fact follows from linearity of the directional derivative: any vector $v$ can be expressed as $\sum \alpha_i b_i$ where $b_i$ are the linearly independent directions, and then $$Df(v) = \sum \alpha_i Df(b_i).$$
In the case of $\mathbb{R}^2$, if the directional derivative of two functions agree in two directions, they must agree in all directions.
This same phenomenon also explains why the curvature of a surface in $\mathbb{R}^3$ is completely determined by the two principle curvatures, why the bending stiffness of a rod with non-circular cross section is completely determined by the bending stiffness in two directions, etc.
EDIT: You asked me for geometry and I gave you a bunch of linear algebra, so let me say a bit more from a geometric perspective. A function $f:\mathbb{R}^2\to \mathbb{R}$ has its first derivatives at a point determined by its tangent plane at that point. If you know two lines contained in that plane (which is what the two directional derivatives tell you) then you know the entire plane, since it passes through the point in question, and has as its normal the cross product of those two lines.
If you think of a complex function as $f:\mathbb{R}^2\to \mathbb{R}^2$, nothing much changes, since the tangent-plane interpretation applies to each component of $f$.