Let $f: U \to \mathbb{C}$ be a holomorphic, with $U \subset \mathbb{C}$ open, and $a \in U$ with $f(a) = 0 ≠ f'(a)$. I then want to show that
$$∫_{\partial B_\epsilon(a)} \frac{1}{f(z)} dz = \frac{2πi}{f'(a)}$$
for sufficiently small $\epsilon > 0$, with $\partial B_\epsilon(a)$ being the edge of the circle with radius $\epsilon$ and center $a$.
I must admit that I so far didn't really know how to approach this problem. Of course we can plug in a parametrization like $\gamma(t) = \epsilon e^{t}, t \in [0, 2π]$ into the left integral, in order to get $∫_{\partial B_\epsilon(a)} \frac{1}{f(z)} dz = \int_0^{2π} \frac{1}{f(\epsilon e^{t})}\epsilon e^t$ if I'm not mistaken, but to me, that doesn't really seem to get anywhere.
Another thing that I'm confused about is: how do we even know that the integral on the left side is well-defined? Because it seems to me that no matter how small the circle around $a$, there will always be points on it's edge where $a$ is also $0$, hence, where $1/f(z)$ is not defined. (Which follows from the maximum principle, I guess. $a$ can neither be a strict maximum nor minimum of $|f(z)|$, hence, there are points $w, x \in U$ with $|f(w)| < 0 < |f(x)|$ around, and because of continuity also points with $f(z) = 0$.) Edit: $|f(w)| < 0 < |f(x)|$ of course doesn't make much sense since $|.| ≥ 0$. Seems I'm wrong in this paragraph and the integral is indeed well-defined on a small enough circle.
I'm sorry that I can't contribute more myself, but I couldn't really get anywhere so far. I suppose, rather than trying to evaluate the integral on the left "on foot", there's a clever theorem or so that we can use; but I couldn't find any so far that seemed to help.
The conditions $f(a) = 0 \ne f'(a)$ mean that $f$ has a simple zero at $a$. There is a nice formula for the residue of quotient $g/f$ at a point where $g$ is nonzero and $f$ has a simple zero: $$ \operatorname*{res}_{z=a}\frac{g(z)}{f(z)} = \frac{g(a)}{f'(a)} $$ This formula, and the residue theorem, imply the claim.