Cauchy-Riemann conditions for complex differentiability

The Cauchy-Riemann conditions for the differentiability of $f(z) = f(x + iy) = u(x, y) + iv(x, y)$ in $(x_0, y_0)$ are

$$\displaystyle \frac{\partial u (x_0, y_0)}{\partial x} = \frac{\partial v (x_0, y_0)}{\partial y}$$

$$\displaystyle \frac{\partial u (x_0, y_0)}{\partial y} = - \frac{\partial v (x_0, y_0)}{\partial x}$$

These conditions are derived when $z = x + iy$ approaches $z_0 = x_0 + iy_0$ along the $x$-axis and the $y$-axis respectively. If the first derivative of $f(z)$ with respect to $z_0$ does exist, it must be not dependent on the pattern followed by $z$ approaching $z_0$.

So, how can the computation along just two possible patterns ($x$-axis and $y$-axis) be sufficient? How are (implicitly) considered all the other possible patterns while we obtain the Cauchy-Riemann conditions?

I know that they are sufficient conditions, but up to now, they appear to be not general and relative to only two specific cases of "approach" between infinite possibilities.

Assuming that a function $u(x,y)$ is real differentiable means assuming that that the graph of $u$ is "well approximated" by a tangent plane, whose slope is then determined by the partial derivatives; this means that $u$ has the property that its derivatives in all directions can be deduced from the derivatives in the $x$ and $y$ directions alone. (And similarly for $v$ of course.) And that's why it's sufficient to look at only two approach patterns.
• Thank you for your answer. Given the existence of the tangent plane, you state that "the derivatives in all directions can be deduced from the derivatives in the $x$ and $y$ directions alone". Is this based on the definition of directional derivative? – BowPark Oct 8 '14 at 14:08
• The definition of differentiability at $(x,y)=(a,b)$ is that $u(a+h,b+k)=u(a,b)+Ah+Bk+\sqrt{h^2+k^2} R(h,k)$ for some numbers $A$ and $B$, and some function $R(h,k)$ which is bounded near $(h,k)=(0,0)$. This implies that $u'_x(a,b)=A$ and $u'_y(a,b)=B$, and more generally, for the derivative in any direction, $\left[\frac{d}{dt} u(a+tv_1,b+tv_2)\right]_{t=0}=Av_1+Bv_2$. – Hans Lundmark Oct 8 '14 at 14:56