I'm reading Stein and Shakarchi and on p. 232, they have for $0 < \alpha < 2$, $z = x\in \mathbb{R}^{-}$ the negative real numbers, that $f'(z) = \alpha |x|^{\alpha-1} e^{i\pi(\alpha-1)}$, and I don't see how they get this, where $f(z) = z^\alpha$
I'm trying to figure out some rules for the complex derivative of $z^\alpha$, where $z, \alpha \in \mathbb{C}$. I don't think it's true in general that it should be $\alpha z^{\alpha – 1}$, as in the case for real functions.
Here's what I have so far. Given any branch of the logarithm, the derivative of $\log(z)$ is $1/z$ by the inverse function theorem. So we wish to compute the derivative of $z^\alpha = e^{\alpha \log z}$. By the chain rule, this is
$$\frac{\alpha}{z} e^{\alpha \log z} = \frac{\alpha}{z} |z|^\alpha e^{\alpha i \arg \theta}.$$
Is this the most one can say about the derivative of $z^\alpha$, for arbitrary $z, \alpha$?
Also, they claim on p.231 that $z^\alpha = \alpha \int_0^z \zeta^{-\beta} \; d\zeta$, where $\alpha + \beta = 1$, and where the integral is taken along any path in the upper half-plane. They say "In fact, by continuity and Cauchy's theorem, we may take the path of integration to lie in the closure of the upper half-plane". I don't know what they mean by this last statement.
Best Answer
It is always true that if $f(z) = z^\alpha$ then $f'(z) = \frac \alpha z z^\alpha$, for any branch (and your calculation verifies this). One is sometimes careful about rewriting this as $\alpha z^{\alpha-1}$ since there are now two different powers and you may need to pay special attention to which branches they separately defined for.
As for what is meant by ``In fact, by continuity and Cauchy's theorem, we may take the path of integration to lie in the closure of the upper half-plane'' they are just claiming (I believe) that part of the path may lie on the real line (which is the boundary of the upper half plane).