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let $\gamma$ be a closed continuosly differentiable path in the upper half plane not passing through $i$. What are the possible values of the integral $$\frac{1}{2\pi i}\int_{\gamma}\frac{2i}{z^2+1}dz$$ well the integral can be broken like $$\frac{1}{2\pi i}\int_{\gamma}\frac{2i}{(z+i)(z-i)}dz=$$

$$\frac{1}{2\pi i}\int_{\gamma}\frac{dz}{z-i}dz-\frac{1}{2\pi i}\int_{\gamma}\frac{dz}{z+i}dz=$$ by Cauchy Integral Formuale $$f(i)-f(-i)$$,so the second integral is $0$ as it is analytic in the upper half plane, but the first integral iss just $n(\gamma,i)$, which iss the windding number of $\gamma$ around $i$ what more I can say? Thank you.

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up vote 3 down vote accepted

The answer depends on two things: (1) if $\gamma$ is a Jordan curve (what is its winding number?); (2) does $i$ lie in the bounded region surrounded by $\gamma$?

It seems to me that the point $-i$ is completely useless, since you assume that $\gamma$ is in the upper half-plane. The answer seems to be: the possible values of $$ \frac{1}{2\pi i}\int_\gamma \frac{dz}{z-i}. $$ I think there is nothing more to say, as you noticed.

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