TITLE: Why does the integral for the average value of a complex function on a circle normalize by $2\pi$, rather than $2\pi R$?
QUESTION [5 upvotes]: In Gamelin's Complex Analysis, the expression for the average value of a complex function on a circle is introduced before Cauchy's integral theorem.
It says that the average value of $h(z)$ on the circle $|z-z_0|=R$ is given by
$$
A(r) = \frac{1}{2 \pi}
       \int_0^{2 \pi} h\left(z_0 + re^{i \theta}\right) d \theta.
$$
I'm puzzled as to why we are normalizing by $2\pi$ instead of $2\pi R$.

REPLY [6 votes]: It is a matter of parametrization. In the integral you write, the circle is parametrized by$$\gamma:[0,2\pi]\to\mathbb{C},\quad \theta\mapsto z_0+re^{i\theta}.$$Then you divide by $2\pi$, which is the length of the interval $[0,2\pi]$.
I guess this puzzles you because this is not an arc-length parametrization. An equivalent way to calculate the average value is by$$A=\frac{1}{2\pi r}\int_0^{2\pi r}h\left(z_0+re^{it/r}\right)dt.$$This uses the parametrization$$\gamma:[0,2\pi r]\to\mathbb{C},\quad t\mapsto z_0+re^{it/r},$$which is arc-length.