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:

$\oint _{\partial D}{\frac {c}{\xi -z}}\,d\xi =\int _{0}^{2\pi }{\frac {c}{re^{it}}}(rie^{it}+r'(t)e^{it})\,dt=\int _{0}^{2\pi i}ci\,dt+\int _{0}^{2\pi }c{\frac {r'(t)}{r(t)}}\,dt$.

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:

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

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