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This is a question found in Theodore Gamelin's Complex Analysis, Chapter 3, Section 2.

We are given the differential $\frac{-ydx+xdy}{x^2+y^2}\quad \text{where }\ (x,y)\neq(0,0)$ The first part says to show that the differential is closed, which I've already done.

The second part says to show that the line integral in any annulus centered at $0$ is not independent of path. I need help for this second part.

I tried evaluating the line integral over the piecewise boundaries $|z|=R$ and $|z|=r$. When calculating the latter, I put a negative sign since the inner circle should have the opposite orientation. But both integrals came out to be $2\pi$ and hence the sum equaled $0$.

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    Just 'wiki'ed about winding numbers. Fascinating. Put a note next to the question, thank you sos440 :)2012-10-23

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As Alan Guo explained, the fact that $ \int_{|z|=r}\frac{-ydx+xdy}{x^2+y^2} = 2\pi $ implies that the integral is not path independent. Let $\gamma_1$ and $\gamma_2$ be the curves starting at $r$ and ending at $-r$, where $\gamma_1$ is upper semicircle and $\gamma_2$ is the lower semicircle. Then we have $ \int_{|z|=r}\frac{-ydx+xdy}{x^2+y^2} =\int_{\gamma_1}\frac{-ydx+xdy}{x^2+y^2} - \int_{\gamma_2}\frac{-ydx+xdy}{x^2+y^2} $ where the minus sign comes from $\gamma_2 $ being traveled in the opposite direction. So, the integrals over $\gamma_1$ and $\gamma_2$ are not equal, despite these paths having the same endpoints.