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Let $\Omega \subseteq \mathbb{C}$ be open and $\gamma:[\alpha,\beta]\rightarrow \Omega$ be a piecewise continuously differentiable and closed path.

Why does $z^{-1}$ have no antiderivative on $\mathbb{C}\setminus0$?

Why is $\int\limits_\gamma z^{-1} dz = 2\pi i \cdot \text{ind}_\gamma (0)$, where $\text{ind}_\gamma$ denotes the winding number.

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    $\begingroup$ Your integral does not appear to be a function and so I can't make much sense out of your question. $\endgroup$
    – Seth
    Commented Jun 5, 2012 at 21:31
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    $\begingroup$ $\int_\gamma dz/z$ is a complex number, not a function... What do you mean ? $\endgroup$
    – Lierre
    Commented Jun 5, 2012 at 21:31
  • $\begingroup$ I think he wants to know why $1/z$ has no antiderivative on $\mathbb{C}-0$. $\endgroup$
    – Potato
    Commented Jun 5, 2012 at 21:37
  • $\begingroup$ Yes, Potato is right, sorry about that. $\endgroup$
    – Chris
    Commented Jun 5, 2012 at 21:41

3 Answers 3

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Potato's answer gives you all you need. But I'd like to mention not why these facts are true, but why they should be true. So here's how I like to think about it (all of which can be made precise):

We know that the derivative of the natural log function is $z^{-1}$, so locally the antiderivative of $z^{-1}$ is $\ln$. But since the exponential function is $2\pi i$-periodic, $\ln$ is not well-definied: if we start by defining, say, $\ln(1)$, we have to choose which inverse function to use: should that be $0$, or $2\pi i$, or $-38\pi i$? So we make a choice, and continuing along the unit circle, we define natural log to be the function whose anti-derivative is $z^{-1}$ which is consistent with our original choice. (This can be made precise, and is an example of what is called analytic continuation.) Since the imaginary part of $\ln$ measures argument (angle), an we have been moving in the positive direction the whole time, we see that our function is not consistent: when we get back to $1$ we are $2\pi i$ more than where we started! So although we can locally define $\ln$, we run into trouble trying to define the function on all of $\mathbb{C}\backslash \{0\}$. This same inconsistency, via the fundamental theorem of calculus, explains the residue formula as well: one bound of the integral is given by one choice of $\ln(1)$, and the other bound is evaluated using another choice.

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  • $\begingroup$ Thank you for the additional explanation, it is highly appreciated! $\endgroup$
    – Chris
    Commented Jun 6, 2012 at 15:57
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Suppose, for the sake of contradiction, that $1/z$ has an antiderivative $F$ on $\mathbb{C}-0$. Then if $\gamma:[a,b]\rightarrow \mathbb{C}-0$ is a path, the integral over this path is just $F(\gamma(a))-F(\gamma(b))$, by the fundamental theorem of calculus.

Now consider a closed loop that winds around 0. The integral should be $F(\gamma(a))-F(\gamma(a))=0$ because the starting and ending points are the same. But if you directly compute the integral of $1/z$ over the unit circle, you get $2\pi i$. This is a contradiction, and there is no antiderivative.

Edit: I see you have edited your question, so I will edit my answer. Your second equation, about winding numbers, is true because of the generalized residue theorem. You can find this in any good book on complex analysis.

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The last line of the question does not feel entirely valid, as it basically questions the definition of the winding number. Here is how the classical argument about the direct computation goes. Suppose $z=z(t)$ is the equation of $\gamma$ on $[\alpha, \beta]$. Consider the function: $$h(t)=\int_{\alpha}^t\frac{z'(t)}{z(t)-a}dt \tag{1}$$ Then $$h'(t)=\frac{z'(t)}{z(t)-a}$$ Now $$\left(e^{-h\left(t\right)}\left(z\left(t\right)-a\right)\right)'=-h'\left(t\right)e^{-h\left(t\right)}\left(z\left(t\right)-a\right)+e^{-h\left(t\right)}z'\left(t\right)=e^{-h\left(t\right)}\left(z'\left(t\right)-h'\left(t\right)\left(z\left(t\right)-a\right)\right)\equiv 0$$ Hence, $e^{-h\left(t\right)}\left(z\left(t\right)-a\right)$ reduces to a constant. By $(1)$ $h(\alpha)=0\implies e^{h(\alpha)}=1$ and we can write $$e^{h(\beta)}=\frac{z(t)-\alpha}{z(\alpha)-a}$$ Since $z(\beta)=z(\alpha)$ we obtain $e^{h(\beta)}=1$ and so $h(\beta)$ must be a multiple of $2\pi i$.

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