There is a proposition I read which claims

Let $U$ be an open set. If $f(z)$ is real-differentiable on $U$ and satisfies Cauchy Riemann Equation then it is complex differentiable on $U$.

I am now confused as to what does it mean for a complex function to be real-differentiable? Does it simply mean if we write $f(z)=u(x,y)+iv(x,y)$ where $u(x,y)$ and $v(x,y)$ are real-valued functions, then $u,v$ are both differentiable in $\mathbb{R}^2$? However, that does not seem to justify the name real-differentiable at all.

Many thanks in advance!

  • $\begingroup$ Think of $\Bbb C$ as $\Bbb R^2$. Then $f$ is a map from an open subset of $\Bbb R^2$ to $\Bbb R^2$. We can ask whether this function of two real variables is differentiable. $\endgroup$ – Angina Seng Jul 28 at 16:04
  • $\begingroup$ Differentiable basically means finding a 'good' linear approximation $f(x+h)-f(x) = Df(x)h$. Real differentiable means the 'perturbations' $h$ are real, complex differentiable means the 'perturbations' $h$ are complex. It is 'more difficult' to be complex differentiable. $\endgroup$ – copper.hat Jul 28 at 16:07

You can see $f$ as a map

$$\begin{array}{l|rcl} f : & \mathbb R^2 & \longrightarrow & \mathbb R^2 \\ & (x,y) & \longmapsto & (u(x,y),v(x,y)) \end{array}$$

And $f$ is supposed to be differentiable. The term real-differentiable is used as $f$ is from $\mathbb R^2$ to $\mathbb R^2$.

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