Cauchy integral formula for constant functions

Statement

Suppose $c\in \mathbb{C}$ is a complex number, and $U$ is a domain in $\mathbb{C}$. Then, if $D$ is a disk centered at $z_0$, and $z$ is any point in the interion of the disk, we have:

$c=\frac{1}{2\pi i}{\oint }_{\partial D}\frac{c}{\xi -z}d\xi$.

Proof

Let us parametrize $\xi$ as $z+r(t)e^{it}$, with $t\in [0,2\pi ]$. Then, the integral becomes:

${\oint }_{\partial D}\frac{c}{\xi -z}d\xi =\underset{0}{\overset{2\pi }{\int }}\frac{c}{r{e}^{it}}(ri{e}^{it}+r^{\prime }(t)e^{it})dt=\underset{0}{\overset{2\pi i}{\int }}cidt+\underset{0}{\overset{2\pi }{\int }}c\frac{r^{\prime }(t)}{r(t)}dt$.

The second integral is zero, because the expression being integrated is the differential of the function $\log (r(t))$, which has the same values at limits $0$ and $2\pi$. Thus, we get:

${\oint }_{\partial D}\frac{c}{\xi -z}d\xi =ci\underset{0}{\overset{2\pi }{\int }}dt=2\pi ic$.

Rearranging this gives the statement of the Cauchy integral formula for constant functions.