# What does it mean for a complex function to be real-differentiable?

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.

• 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. – Angina Seng Jul 28 at 16:04
• 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. – 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 $$\mathbb R^2$$.