Let $a \neq b\in \mathbb{C} $ and $U := \mathbb{C} -[a,b] $
Let $\Gamma$ be a cycle in $U$. The following equality is true?
$$\int_{\Gamma} \frac{1}{(z-a)(z-b)}dz=0$$
I saw it some notes of a friend. I know it's true for every cycle $\Gamma$ that is homologous to zero in $U$. But what about the others cycles?
Expanding the comment:
$$\frac{1}{(z-a)(z-b)} = \frac{1}{a-b}\left(\frac{1}{z-a}-\frac{1}{z-b}\right),$$
hence
$$\int_\Gamma \frac{1}{(z-a)(z-b)}\,dz = \frac{1}{a-b}\left(\int_\Gamma\frac{dz}{z-a} - \int_\Gamma \frac{dz}{z-b}\right) = \frac{2\pi i}{a-b} \left(n(\Gamma,a) - n(\Gamma,b)\right).$$
Since $a$ and $b$ belong to the same component of the complement of $U$, we have $n(\Gamma,a) = n(\Gamma,b)$ for every cycle $\Gamma$ in $U$.