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Let's $$f:\mathbb{C}\to\mathbb{R}$$ is holomorphic. In all lectures notes that I've read, said that $$f'(z)=\lim_{h\to 0}\frac{f(z+h) - f(z)}{h}.$$ Is it true. Or correctly next definition $$f'(z)=\lim_{h\to 0}\frac{f(z+h) - f(z)}{|h|}.$$ On the other hands, is it true that $f'(z)$ is complex?

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I'm certain that my complex analysis is very rusty but, how do you define a real valued holomorphic function? –  Martin Argerami Mar 13 '12 at 3:23
    
@anon: so you allow a complex derivative of a real-valued function? –  Martin Argerami Mar 13 '12 at 3:24
    
@Martin: I didn't pay attention to the $\mathbb{C}\to\mathbb{R}$ clearly in the question! Aspirin: There's no such thing as a (nonconstant) real-valued holomorphic function. The first definition of derivative is correct, and applies to $\mathbb{C}\to\mathbb{C}$ functions. What are you talking about? –  anon Mar 13 '12 at 3:28
    
@anon: I am curious and my complex analysis is rusty, too. Why is $f$ constant? –  shuhalo Mar 13 '12 at 6:22
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@Martin: Obviously a constant function is holomorphic, so picking a real constant obtains a trivial real-valued holomorphic function. Conversely, a real-valued function has constant zero imaginary part, so work with the Cauchy-Riemann equations to find that the function must be constant on the whole. –  anon Mar 13 '12 at 6:30

1 Answer 1

Setting aside the codomain issue (it should be $\mathbb C$ instead of $\mathbb R$ if we want nonconstant holomorphic functions), the answer is that $$\lim_{h\to 0}\frac{f(z+h) - f(z)}{|h|}$$ is not a suitable limit for definition of derivative. While the difference-quotient-limit formalism is a convenient tool in some cases, the deeper meaning of the derivative is linear approximation to $f$. The existence of derivative at $z$ means that $f(z+h)-f(z) \approx \ell(h)$ on small scales, where $\ell$ is a linear function of $h$. In the complex domain we can write $\ell(z) = Az$ for some complex number $A$. This number is called the derivative $f'(z)$. It can be calculated by dividing the relation $f(z+h)-f(z) \approx Ah$ by $h$ (which I decided not to make precise by including the error term) and then letting $h \to 0$: $$A=\lim_{h\to 0}\frac{f(z+h) - f(z)}{h}$$

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