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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.

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In your first second, I don't know what you mean by $z=x\in{\bf R}^-$, and I don't see how anyone gets a formula for $f'(z)$ without saying what $f$ is. –  Gerry Myerson Apr 13 '12 at 5:49

2 Answers 2

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).

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I think all the information you need is in Example 2 on page 210. The key here I think is that the boundary behaviour of $f$ is such that if $x$ travels from $-\infty$ to $0$ on the real line then $f(x)$ travels from $ \infty e^{i \alpha \pi} $ to $0$ determined by $arg(z) = \alpha \pi$.

The RHS of the derivative you wrote down (via the chain rule) may exhibit your answer now by considering $z=-x$ and noting that in the final stages of simplification replace $-1$ by $e^{i \pi}$. A few lines of algebra may then present yo with you answer.

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