Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
add comment

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).

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.