Looking for intuïtive explanation why contour integral of $\frac{dz}{z} $equals $2\pi i$ in complex analysis

Question

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

I've seen the derivation of it using the 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?

Answer 1

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 $\int dz = 0$?

Answer 2

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 $0$.