Contour integration

Question

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 $2 \pi i$?

Answer 1

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:

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

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