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!
Best Answer
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 $\mathbb R^2$.