0
$\begingroup$

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?

Thanks

  • 1
    @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.2012-03-13

1 Answers 1

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 $h \to 0$: $A=\lim_{h\to 0}\frac{f(z+h) - f(z)}{h}$