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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 $2 \pi i$?

Meow
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1 Answers1

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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:

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

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

DonAntonio
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  • 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? – Meow Oct 28 '12 at 13:31
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  • 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 $2 \pi i$? – Meow Oct 28 '12 at 14:23
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    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