3
$\begingroup$

I just read that a sufficient condition for a function $f:A \rightarrow \mathbb{C},f(z) = u(z)+ i v(z)$ to be holomorphic is:

  • $A$ open.
  • f is $\mathbb{R}$ - differentiable in $A$.
  • The Cauchy Riemann equations hold in $A$.

In the book i' m reading $\mathbb{R}$ - differentiability is defined as:

$f: D \rightarrow \mathbb{C}$ is $\mathbb{R}$ - differentiable in $z_0$ if there exists an $\mathbb{R}$ - linear map $T$ such that

$$\lim_{z \rightarrow z_0}\frac{f(z)-f(z_0)-T(z-z_0)}{| z-z_0 |}=0$$

Is this equivalent to saying that $Re(z)$ and $Im(z)$ (which are real valued) are differentiable in $A$?

$\endgroup$
  • $\begingroup$ yes, it is the same $\endgroup$ – Masacroso Apr 15 at 10:41
1
$\begingroup$

You can look at a complex-valued function as being represented by two real-valued functions, say $f(z) = f(x,y) = u(x,y) + iv(x,y)$ for $z = x + iy$. Now we want $f$ to be real-differentiable before even considering the Cauchy-Riemann equations (in fact, we need both to be complex-differentiable). What does it mean for a function to be real-differentiable? This is simple in one variable, but you might want to look up what it means for a function to be differentiable in multiple variables and think about what the differential actually is (spoiler: it is basically the best linear approximation of our function at a point).

$\endgroup$

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.