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.


To quote Wikipedia: "The sole existence of partial derivatives satisfying the Cauchy–Riemann equations is not enough to ensure complex differentiability at that point. It is necessary that u and v be real differentiable [...]".

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.

  • $\begingroup$ 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? $\endgroup$ – BowPark Oct 8 '14 at 14:08
  • 1
    $\begingroup$ 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, $\left[\frac{d}{dt} u(a+tv_1,b+tv_2)\right]_{t=0}=Av_1+Bv_2$. $\endgroup$ – Hans Lundmark Oct 8 '14 at 14:56

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