With regards to utilizing Cauchy's Integral Theorem for integration over closed contours:


In particular the result that $\int_\gamma f(z)\,dz = 0$ for closed paths $\gamma$ and holomorphic functions $f(z)$.

It is stated (and shown algebraically) in numerous tutorials and texts that $f(z)=z^{-1}$ violates the conditions for the above result to hold. But in all elaborations given to this statement it has not been clear to me why this is not also true for $z^{-2}$ (or indeed any $z^{n}$ for integers $n$ smaller than -1).

For example, in the Wikipedia article linked, it states:

"The Cauchy integral theorem does not apply here since $f(z)=\frac1 z$ is not defined (and certainly not holomorphic) at $z=0$."

Which I believe is also true for any other $f(z) = \frac 1 {z^n}$.

Alternatively, a set of notes from the Columbia's Complex Analysis course state:

$1/z$ is analytic in the region $\mathbb C − \{ 0 \} = \{z ∈ \mathbb C > : z \neq 0\}$, but this region is not simply connected.

Again believe this also applies to other $f(z) = \frac 1 {z^n}$.

Could someone please clarify what exact condition $\frac 1 z$ violates that $\frac 1 {z^n}$ does not?

P.S. Some background - I am not a mathematician by training but trying to teach myself for applied research, so answers that favor intuition are greatly appreciated!

  • 2
    $\begingroup$ The functions $f(z)=z^{-n}$ (where $n>0$ is an integer) all fail the hypothesis of the theorem. However, for the integers $n>1$, the function satisfies the conclusion. The reason for this lies beyond the scope of this one theorem. $\endgroup$ – Omnomnomnom Dec 26 '15 at 3:07
  • 2
    $\begingroup$ Note that the theorem is not an if and only if statement. $\endgroup$ – Omnomnomnom Dec 26 '15 at 3:09
  • $\begingroup$ Ah I see! Thank you. So just to be clear, all functions of the reciprocal form ($\frac 1 {z^n}$) violate the conditions. The fact that $\int_{\gamma} \frac 1 {z^n} dz = 0$ for n>1 is not an outcome of Cauchy's integral theorem but some other mechanism? $\endgroup$ – Sue Doh Nimh Dec 26 '15 at 3:11
  • 1
    $\begingroup$ Yes, exactly. The answer given explains that the mechanism at work here is the existence of an antiderivative. $\endgroup$ – Omnomnomnom Dec 26 '15 at 3:47

All the integral powers $z^n$ have an antiderivative $\frac{1}{n+1} z^{n+1}$ on $\mathbb{C}\backslash\{0\}$, except $z^{-1}$. Now, if $F'(z) = f(z) $ then $$\int_{\gamma} f(z) dz = F(\textrm{end point} ) - F(\textrm{initial point} )$$ so in particular, for a closed curve $$\int_{\gamma} f(z) dz = 0$$ if $f$ has an antiderivative on the domain.

  • $\begingroup$ As long as said closed curve doesn't have $z=0$ in its inside. $\endgroup$ – vonbrand Dec 26 '15 at 2:31
  • $\begingroup$ Thank you, but as far as I can see $z^{-1}$ also has an anti-derivative ($log(z)$) over the same space. Of course, I can see some complications arising because $log(z)$ is a multiple-valued function in the complex domain (and I imagine this is related to the answer I'm looking for), but like I mentioned I'm trying to understand which of the conditions of Cauchy's theorem $z^{-1}$ violates that parallel reciprocal functions do not. $\endgroup$ – Sue Doh Nimh Dec 26 '15 at 2:36
  • $\begingroup$ @vonbrand: The (complex) Leibniz-Newton works always, provided you have the antiderivative. If you don't ( like for $1/z$ ) but you still have it locally, then you consider $F$ the determination of the derivative along the curve $\gamma$. So yes, for $1/z$ around $0$ the $\log$ gathers multiples of $2 \pi i$. $\endgroup$ – orangeskid Dec 26 '15 at 2:37
  • $\begingroup$ @Sue Doh Nimh: Yes, locally you always have the antiderivative. But for $1/z$ it turns out to be multivalued. Of course you could consider functions of the form $z^s$ where $s$ is complex. Already the function is multivalued, but you could consider nevertheless a determination. The anti-derivative would exist again locally. I guess to be correct you have to say: Leibniz-Newton holds ( for the determination along the path.). if the functions is not multiply valued, then .... $\endgroup$ – orangeskid Dec 26 '15 at 2:42
  • $\begingroup$ @Sue Doh Nimh: the anti-derivatives of $z^n$ for $n\ne -1$ are uni-valued. Not so for $z^{-1}$. I guess that would also be a way to put it. $\endgroup$ – orangeskid Dec 26 '15 at 2:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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