Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In Penrose's 'Road to Reality', he states that for any integer n, $n \ne 1$, $ \oint z^n dz=0$. Qualitatively, why is this so, given that for any negative n poles in the complex graph exist (namely at z=0): integrating around them would surely yield $2 \pi i$?

share|cite|improve this question
$n \neq -1 \implies \oint z^n dz=0$ – Mhenni Benghorbal Oct 28 '12 at 12:27
What's the Latex for that? – Alyosha Oct 28 '12 at 12:27
up vote 1 down vote accepted

First approach: For any $\, -1\neq n<0\,$ , the function $\,z^n\,$ has a pole of order $\,n\,$ at zero with residue zero, thus by Cauchy's Theorem we get that (assuming the closed path $\,C\,$ encloses zero, otherwise it is trivial)

$$\oint_C z^ndz=2\pi i\cdot\operatorname{Res}_{z=0}(z^n)=0$$

Of curse, if $\,n\geq0\,$ the function $\,z^n\,$ is analytic everywhere so we trivially get zero, too.

Second approach: Assuming we've the integration path

$$C:=\{(r\cos t\,,\,r\sin t)\;\;;\;\;r>0\;,\;\;0\leq t\leq 2\pi\}=\{z\in\Bbb C\;;\;\;|z|=r\}=\{z=re^{it}\}$$

We get, doing the complex line integral:

$$z=re^{it}\Longrightarrow dz=rie^{it}dt\Longrightarrow$$

$$\Longrightarrow \oint_Cz^ndz=r^{n+1}i\int_0^{2\pi}e^{(n+1)it}dt=\left.\frac{r^{n+1}i}{(n+1)i}e^{(n+1)it}\right|_0^{2\pi}=0$$

share|cite|improve this answer
1- How is Res_{z=1}(z^n) actually calculated (if it's not too complex)? 2- Shouldn't it be $e^{i(n+1)t}$? 3- Given that $e^{ik}=e^{ik \pm i2 \pi m}$, why is the integral not equal to 0 when n=-1? – Alyosha Oct 28 '12 at 13:31
1) Use Laurent series: $$-1\neq n<0\Longrightarrow z^n=\frac{1}{z^{-n}}+0\cdot\frac{1}{z}+...$$ 2) You're right and I'll edit my answer thought it doesn't change the final result. 3) When $\,n=-1\,$ we have residue $\,=1\,$ – DonAntonio Oct 28 '12 at 14:03
Thanks for that. By which process do we get the n=-1 integral is actually $2 \pi i$? – Alyosha Oct 28 '12 at 14:23
Perhaps the most direct one is using branch theory for the complex logarithmic function, yet I think that by complex line integration we can do it way more basically and pretty simple, too. Using the same $\,C\,$ as in my answer, with $\,r=1\,$ , we get: $$\oint_C\frac{dz}{z}=\int_0^{2\pi}i\,dt=2\pi i$$ – DonAntonio Oct 28 '12 at 14:28

Your Answer


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.