Let $(f_1,f_2)$ be a continuously differentiable vector field in $\mathbb{R}^2\setminus\left\{0\right\}$, which fullfills the integrability condition $\frac{\partial f_1}{\partial y}=\frac{\partial f_2}{\partial x}$. Show that it exists a $c\in\mathbb{R}$ so that the vector field $$ g(x,y):=\left(f_1(x,y)-c\frac{y}{x^2+y^2},f_2(x,y)+c\frac{x}{x^2+y^2}\right) $$ is conservative.
To my opinion conservative means here, that the integral of the vector field over each closed curve is 0. This works for each closed curve that does not contain the zero point, because such curves are homotopic to a point curve and therefore vanish. Only closed curves around the zero point are problematic. And to my opinion there exists a lemma which says:
It is enough to find one closed curve around the zero point over which the integral over the vector field vanishes - then the integral vanishes for all closed curves around the zero point.
First of all $g$ fullfills the integrability condition. Then, according to the lemma above, I chose an arbitrary curve around the zero point; my decision was
$$ \gamma\colon t\mapsto (\cos t,\sin t), t\in [0,2\pi]. $$
Then I calculated the integral over $\gamma$ and identified it with $0$, getting
$$ \int_{\gamma}g(x,y)\, d(x,y)=\int_0^{2\pi}\langle g(\gamma(t)),\dot{\gamma}(t)\rangle\, dt\\=2\pi c\int_0^{2\pi}\cos t f_2(\cos t,\sin t)-\sin t f_1(\cos t,\sin t)\, dt=0\\ \Leftrightarrow c=-\frac{1}{2\pi}\int_0^{2\pi}\cos t f_2(\cos t,\sin t)-\sin t f_1(\cos t,\sin t)\, dt. $$
So my question to you is, if
$$ c=-\frac{1}{2\pi}\int_0^{2\pi}\cos t f_2(\cos t,\sin t)-\sin t f_1(\cos t,\sin t)\, dt $$ is the searched real number and if everything is fine with my appliance of the lemma.
\With regards\