Integral of 1/z around the unit circle

Example

The circle integral of $f(z) = 1/z$ over the unit circle $|z| = 1$ is $$\oint_{|z| = 1} \frac{dz}{z} = 2 \pi i.$$ Computation: the path is $\gamma(\theta) = e^{i\theta}$ with $\theta \in [0, 2\pi]$, so $\gamma'(\theta) = i e^{i\theta}$. Then $$\oint_{|z| = 1} \frac{dz}{z} = \int_0^{2\pi} \frac{i e^{i\theta}}{e^{i\theta}} \, d\theta = \int_0^{2\pi} i \, d\theta = 2 \pi i.$$ More generally, let $c, w \in \mathbb{C}$ and $R > 0$. If $|w - c| < R$, then $\oint_{|z - c| = R} \frac{dz}{z - w} = 2 \pi i$. If $|w - c| > R$, the integral is $0$.

Remarks

For each integer $n \neq -1$, $\oint_{|z| = 1} z^n \, dz = 0$, because $z^n$ has a primitive on $\mathbb{C} \setminus \{0\}$. The value is not $0$, so $1/z$ has no primitive on $\mathbb{C} \setminus \{0\}$, and Cauchy's integral theorem does not apply on $\mathbb{C} \setminus \{0\}$. This integral is the base of Cauchy's integral formula, of the residue and of the winding number.

Explanations

L1 human

On the unit circle, $dz$ is $i z \, d\theta$, so $dz / z$ is just $i \, d\theta$. Going once around adds up $i \, d\theta$ over a full turn of $2\pi$, which gives $2 \pi i$.