TITLE: Mean value theorem for holomorphic functions
QUESTION [31 upvotes]: The mean value theorem for holomorphic functions states that if $f$ is analytic in $D$ and $a \in D$, then $f(a)$ equals the integral around any circle centered at $a$ divided by $2\pi$.  But if $f$ is analytic, then the line integral around any closed curve is 0, so $f(a) = 0$.  Why does the MVT not result in all holomorphic functions being identically zero?  There must be something I'm missing here.

REPLY [41 votes]: The difference is that for the mean value property, we consider the integral with the measure/form $d\varphi$, and for the integral theorem, the measure/form is $dz$.
The mean value property is actually the Cauchy integral formula for the centre of a disk:
$$\begin{align}
f(a) &= \frac{1}{2\pi i} \int_{\lvert z-a\rvert = r} \frac{f(z)}{z-a}\,dz\\
&= \frac{1}{2\pi i} \int_0^{2\pi} \frac{f(a+re^{i\varphi})}{(a+re^{i\varphi})-a}\, d(a+re^{i\varphi})\\
&= \frac{1}{2\pi i} \int_0^{2\pi} \frac{f(a+re^{i\varphi})}{re^{i\varphi}}rie^{i\varphi}\,d\varphi\\
&= \frac{1}{2\pi} \int_0^{2\pi} f(a+re^{i\varphi})\,d\varphi.
\end{align}$$
For a parametrised circle with centre $a$, we have $d\varphi = \dfrac{dz}{i(z-a)}$, so the integral of $d\varphi$ over a circle does not vanish, while the integral of $dz$ does.