I know well the Cauchy integral formula, and I am interested in a precise consequence. It states that for any holomorphic function defined on a open disc $D$, having only one zero $z_0$, then $$\frac{1}{2i\pi} \int_{\partial D}\frac{f'(z)}{f(z)}\mathrm{d}z = \mathrm{ord}_{z_0}(f).$$
I often see that there is a "partial" version of it, stating that if you take a collapsing arc $\gamma$ of the $\partial D(0, \varepsilon)$, say of angular deviation $\alpha$, then $$\frac{1}{2i\pi} \int_{\gamma}\frac{f'(z)}{f(z)}\mathrm{d}z \longrightarrow_{\varepsilon \to 0} \frac{\alpha}{2\pi} \mathrm{ord}_{z_0}(f).$$
Is that true in general? And why is that so?
A precise reference is Proposition 1.2 in these notes.
If $f(z_0) = 0$ with multiplicity $k$ then in a neighborhood $U$ of $z_0$ $$ f(z) = (z-z_0)^k g(z) $$ where $g$ is holomorphic in $U$ and $g(z_0) \ne 0$.
Now fix $t_0 < t_1$ and let $\gamma_\epsilon : [t_0, t_1] \to \Bbb C$, $\gamma_\epsilon(t) = z_0 + \epsilon e^{it}$ be an arc on $D(z_0, \epsilon)$ of angle $t_1 - t_0$. Then for (sufficiently small) $\epsilon > 0$ $$ \frac{1}{2 \pi i}\int_{\gamma_\epsilon} \frac{f'(z)}{f(z)} \, dz = \frac{k}{2 \pi i}\int_{\gamma_\epsilon} \frac{dz}{z - z_0} + \frac{1}{2 \pi i}\int_{\gamma_\epsilon} \frac{g'(z)}{g(z)} \, dz \, . $$ The first integral can be computed explicitly: $$ \frac{k}{2 \pi i}\int_{\gamma_\epsilon}\frac{dz}{z - z_0} = \frac{k (t_1 - t_0)}{2 \pi} \, , $$ and $$ \frac{1}{2 \pi i}\int_{\gamma_\epsilon}\frac{g'(z)}{g(z)} \, dz \to 0 $$ for $\epsilon \to 0$ because $g'/g$ is bounded in a neighborhood of $z_0$.