Mean value theorem for holomorphic functions

Question

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.

Answer 1

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.