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$$\oint \frac{dz}z = 2\pi i$$

I've seen the derivation of it using the enter image description here parametrisation. Since this result is used all the time in my complex analysis course, i'd like to understand this intuïtively. I tried making sense of it like this:

$$\oint \frac{dz}{z} = \mathrm{mean}\left(\frac{1}{z}\right) \cdot \oint dz = \mathrm{mean}\left(\frac 1 z\right)\cdot 2\pi$$

Or by thinking of it like a 2D version of the flux integral of a field through a sphere. Then the singularity in the origin would act like a source of field lines.

Is there another way to think about this?

BigM
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  • What do you mean by "mean"? The association between means and integrals doesn't apply in contour integrals. – Thomas Andrews Nov 01 '14 at 14:09
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    Why do you think $\oint dz = 2\pi$, for example? It is actually $0$. You first have to back away and understand what the contour integral is and isn't. – Thomas Andrews Nov 01 '14 at 14:12

2 Answers2

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Letting $\gamma(t)=e^{it}$ be your curve, with $t\in[0,2\pi]$, the integral is:

$$\int_{0}^{2\pi} \frac{\gamma'(t)dt}{\gamma(t)}$$ But $\gamma'(t)=i\gamma(t)$, so this is just $$\int_0^{2\pi} i dt = 2\pi i$$

Essentially, the tangent to the circle is at a right angle to the circle, and $i$ acts like a 90 degree rotation under multipication.

Your question, at heart, shows a misunderstanding of what a contour integral is. $\oint dz =0$, for example. While in real integrals, there is a relationship between integrals and the mean, that problem is that the contour integral depends on the curve.

You are forgetting that $\gamma'(t)$ is part of the integral. That is what makes contour integrals different from mere integrals.

Now, it turn out that if you have a curve from $1$ to $z_0$ then $\oint_{\gamma} \frac{dz}{z} = \ln z_0$ is true, but in complex numbers, $\ln z_0$ is a multi-valued function, and which value we get depends on the curve we take to $z_0$. What happens is that the integral "counts" how far the curve winds around $0$.

Thomas Andrews
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At a given complex number $z = e^{i\theta}$ on the unit circle, the effect of multiplying by $z$ on another point $p$ is to rotate $p$ through an angle of $\theta$. The effect of multiplying by $\frac{1}{z} = e^{-i\theta}$ is to rotate through an angle of $-\theta$.

At the point $z$, $dz$ is a small complex number which, as a geometric vector, is tangent to the unit circle. Since the tangent line and radius of a circle are perpendicular you can see (by geometry) that $\frac{1}{z}dz$ should point straight up. In other words $\frac{1}{z}dz = ids$, where $ds$ is a little bit of the arclength of the circle.

Summing all of these, we get $i$ times the circumference of the circle.

Can you see intuitively, using the same method, why $\int dz = 0$?

Steven Gubkin
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  • I also speak a bit more about this perspective here: http://math.stackexchange.com/questions/110334/line-integration-in-complex-analysis/914273#914273 – Steven Gubkin Nov 01 '14 at 14:24