Explaining an integral of $\mathrm{d}\theta$ Let $\gamma$ be a path (we can assume it is the straight line) in $\mathbb{C}$ from $v$ to $w$. Then, "morally", it seems that
$$\int_\gamma \mathrm{d}\theta=\mathrm{arg}w-\mathrm{arg}v.$$
It's been a while since I did much complex integration though, so I'm not sure how to formalise/prove this. Every reference I can find only talks about integrating $\mathrm{d}z$ along a path.
 A: This isn't so much a complex analysis problem as a multivariable calculus problem, so I'll translate everything in terms of calculus in $2$ variables. 
Let $U$ be an open subset of $\mathbb{R}^2 \setminus \{0\}$. One can define an argument function on $U$ to be a function $\arg : U \to \mathbb{R}$, such that
$$
 \forall (x,y) \in U, \quad (x,y) = (x^2+y^2)^{1/2}(\cos(\arg(x,y)),\sin(\arg(x,y))).
$$
Since
$$
 \forall (x,y) \in U \setminus \operatorname{Span}\{(1,0)\}, \quad \tan(\arg(x,y)) = \frac{y}{x},\\
\forall (x,y) \in U \setminus \operatorname{Span}\{(0,1)\}, \quad \cot(\arg(x,y)) = \frac{x}{y},
$$
we can apply the implicit function theorem locally to conclude that $\arg$ is continuously differentiable on $U$ with gradient
$$
 \nabla\arg(x,y) = \left(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2}\right),
$$
or equivalently, in terms of differential forms, with exterior derivative
$$
 d\arg = \frac{-y}{x^2+y^2}\,dx + \frac{x}{x^2+y^2}\,dy =: d\theta.
$$
As a result, if $\mathbf{v}, \mathbf{w} \in U$ and if $\gamma$ is a piecewise continuously differentiable curve in $U$ from $\mathbf{v}$ to $\mathbf{w}$, then by the fundamental theorem of calculus for line integrals,
$$
 \int_\gamma d\theta := \int_\gamma \nabla\arg \cdot \, d\mathbf{r} = \arg(\mathbf{w}) - \arg(\mathbf{v}).
$$
Of course, I've glossed over an extremely important question: does the function $\arg$ even exist on the domain $U$? It exists by the implicit function theorem on any small enough open neighbourhood of any point in $\mathbb{R}^2 \setminus \{0\}$. More generally, if $U$ is simply connected, then by Green's theorem, the irrotational vector field $\mathbf{F}(x,y) := \left(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2}\right)$ on $U$ is necessarily the gradient of an argument function on $U$, e.g.,
$$
 \arg_{\mathbf{r}_0}(\mathbf{r}) := \int_{\mathbf{r}_0}^\mathbf{r} \mathbf{F} \cdot \,d\mathbf{r},
$$
where $\mathbf{r}_0 \in U$ is some fixed point and where $\int_{\mathbf{r}_0}^\mathbf{r} \mathbf{F} \cdot \,d\mathbf{r}$ denotes the line integral of $\mathbf{F}$ along any piecewise $C^1$ path in $U$ from $\mathbf{r}_0$ to $\mathbf{r}$.
