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My question is about the derivative of holomorphic complex functions.

Assume there is a function $f(x) := \mathbb{R} \rightarrow \mathbb{R}$ , and a function $f(z) := \mathbb{C} \rightarrow \mathbb{C}$. Further assume that both functions have a "counterpart", both in the complex and real domain (e.g. $\sin(x)$, $e^x$, $|.|$ or polynomials).

When $f(z)$ is holomorphic on $\mathbb{C}$ can we then conclude that $f'(x) = f'(z)$?

From my understanding the second Wirtinger derivative must be zero for any holomorphic function $f(z)$.

$$\frac{\partial f(z)}{\partial z*} = \frac{1}{2}\left(\frac{\partial f(z)}{\partial a}+i\frac{\partial f(z)}{\partial b}\right) = 0$$

Which somehow leads me to believe that $\frac{\partial f(z_0)}{\partial z} = f'(x_0)$ assuming $x_0 = z_0$. However, as this is only an "Intuition" from my side, and I am by no means a professional Mathematician. Maybe one of you can construct a counterexample to my observation.

Alain Remillard
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Bastian
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    It could happen that the real derivative exists but the complex derivative does not exist. However, in the question it is assumed that the complex derivative exists. – GEdgar May 16 '24 at 19:32
  • To amend @GEdgar's comment, if you keep studying complex analysis you'll learn the very surprising theorem that if you start with a differentiable function on $\mathbb R$, then there are infinitely many ways to extend that to a function on $\mathbb C$, but at most one of them will be holomorphic. – Torsten Schoeneberg May 17 '24 at 15:06
  • That's indeed very interesting! Thanks for sharing. – Bastian May 19 '24 at 21:01

2 Answers2

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Differentiability in $\mathbb{C}$ is a much stronger condition than differentiability in $\mathbb{R}$ (because there are more "paths of approach" for evaluating the limit of the difference quotient).

In particular (regarding $\mathbb{R}$ as $\{z\in\mathbb{C} : \operatorname{Im}(z) = 0\}$), if $f$ is differentiable as a function $\mathbb{C}\to\mathbb{C}$, then it is differentiable as a function $\mathbb{R}\to\mathbb{R}$ and necessarily $f|_{\mathbb{R}}'(x) = f'(x)$.

MPW
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I'll temporarily denote the real derivative as $f'_\mathbb{R}$ and the complex derivative as $f'_\mathbb{C}$.
Let $x_0 \in \mathbb{R}$, and assume that $f$ is real-analytic on a neighbourhood of $x_0$ in $\mathbb{R}$, so that it is holomorphic in a neighbourhood of $x_0$ in $\mathbb{C}$.
Then, pretty much by definition, we have: $$f'_\mathbb{C}(x_0) = \lim_{h \to 0 \atop h \in \mathbb{C}} \frac{f(x_0 + h) - f(x_0)}{h} = \lim_{h \to 0 \atop h \in \mathbb{R}} \frac{f(x_0 + h) - f(x_0)}{h} = f'_\mathbb{R}(x_0)$$

Bruno B
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