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In the paper by Fedotov and Dragomir the formula to calculate the Index of a vector field $\mathbf{F} = (f(x,y), g(x,y))^T$ is given as $$ I \equiv \frac{1}{2 \pi} \oint_C \frac{f(x,y) \, \mathrm{d} g(x,y) - g(x,y) \, \mathrm{d}f(x,y)}{f^2(x,y) + g^2(x,y)} \tag{1} $$ where C is a closed smooth curve in $\mathbb{R}^2$, i.e. the unit circle. On the bottom of page 1 they state: "Formula (1) allows to evaluate the rotation of a field $\mathbf{F}$ with respect to the curve."

From this I understand that (1) must be connected to the following integral $$ I_{rot} \propto \oint_C \nabla \times \mathbf{F} \cdot \mathrm{d} \mathbf{l} \tag{2} $$ However I don't see how these are connected.

Update I have added some own calculations that suggest that $(1)$ and $(2)$ dont share any connection.

Let us assume $C$ to be the unit circle centered at $(0, 0)$. Then $(1)$ can be written as $$ I = \frac{1}{2 \pi} \int_{0}^{2 \pi} \left[ \frac{f(x, y)}{{f^2(x,y) + g^2(x,y)}} \left( \frac{\partial g}{\partial x} \dot{x} + \frac{\partial g}{\partial y} \dot{y}\right) + \frac{g(x, y)}{{f^2(x,y) + g^2(x,y)}} \left( \frac{\partial f}{\partial x} \dot{x} + \frac{\partial f}{\partial y} \dot{y} \right) \right] \, \mathrm{d}t \tag{3} $$ with $x \equiv x(t) = \cos(t)$ and $y \equiv y(t) = sin(t)$. For $(2)$ we use that $$ \nabla \times \mathbf{F} = \frac{ \partial g}{\partial x} - \frac{\partial f}{\partial y} $$ for $\mathbf{F}: \mathbb{R}^2 \rightarrow \mathbb{R}^2$. Hence $\nabla \times \mathbf{F}: \mathbb{R}^2 \rightarrow \mathbb{R}$, for which the line integral is given as $$ \int_0^{2 \pi} \left( \frac{ \partial g}{\partial x} - \frac{\partial f}{\partial y} \right) \underbrace{\lvert \dot{x}^2 + \dot{y}^2 \rvert}_{=1} \, \mathrm{d} t \tag{4} $$

To me integrals $(3)$ and $(4)$ don't seem to have any connection for arbitary $f(x, y), \; g(x, p)$. Is there some error in my calculation or some identiy that would connect both integrals?

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