1
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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 $\mathbb R^2$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.