Suppose $\gamma$ is a piecewise differentiable closed curve that does not pass through the point $a \in \mathbb{C}$. I'm reading a proof in Ahlfors that shows under this condition we will obtain
$$ \int_\gamma {dz \over z - a} = k 2 \pi i \text{ for some }k \in \mathbb{N} $$
However, in the beginning of his proof, Ahlfors remarks that we may write
$$ \int_\gamma {dz \over z-a} = \int_\gamma d\ \log(z-a) = \int_\gamma d\ \log |z-a| + i \int_\gamma d\ \arg(z-a) $$
My question is what does $d\ \log(z-a)$ in the integrands above denote? In particular, what does the "$d$" denote? As written, it's very unclear to me where this equation came from or what it denotes.
Formally, $df(z) = f'(z)\,dz$. The concept behind this notation is a differential form; operator $d$ acts on such forms, turning, in particular, $0$-forms into $1$-forms. But this terminology is not really needed to understand the computation. Saying that $$\int_\gamma z^2 \,dz = \int_\gamma d(z^3/3)$$ asserts that we can evaluate the integral of $z^2$ by taking the difference of $z^3/3$ at the endpoints, because this is what the Fundamental Theorem of Calculus (for line integrals) says. It's a bit more subtle with $$\int_\gamma \frac{1}{z} \,dz = \int_\gamma d\log(z)$$ because $\log (z)$ is a multi-valued function. Here it is understood that we pick some value of $\log z$ at the beginning of $\gamma$, and use analytic continuation along $\gamma$ to evaluate it at the end.