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Let $f: U \subset \mathbb{C}$ be analytic and $\gamma_1, \gamma_2$ be two arbitrary closed paths. Prove that:

$ \int_{\gamma} f(z) \ dz = 0$ for any closed path $ \iff\int_{\gamma_1} f(z) \ dz = \int_{\gamma_2} f(z) \ dz $.

It's easy to see the $\implies$ direction, but I'm having a harder time with the other.

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  • $\begingroup$ Take the $\gamma_1$ and $\gamma_2$ from the right side. Can you see that $\gamma_1 - \gamma_2$ forms a closed path? $\endgroup$ – Kezer Sep 14 '18 at 16:44
  • $\begingroup$ @Kezer, yes, but can any closed path be written like that? $\endgroup$ – Matheus Andrade Sep 14 '18 at 16:45
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    $\begingroup$ I'm sorry, I misread, I didn't see that $\gamma_1$ and $\gamma_2$ are arbitrary closed paths. I assume your $U$ is supposed to be open? Then, choose some circle with its disc fully contained in $U$ - Cauchy makes the contour integral $0$. It's a closed path, so all integrals for closed paths equal $0$. $\endgroup$ – Kezer Sep 14 '18 at 16:54
  • $\begingroup$ Thanks a lot, @Kezer! $\endgroup$ – Matheus Andrade Sep 14 '18 at 19:45
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I am assuming that the question actually is: to prove that $\displaystyle\int_\gamma f(z)\,\mathrm dz=0$ for any closed path if and only if for any two paths $\gamma_1$ and $\gamma_2$ with the same initial points and the same final points, $\displaystyle\int_{\gamma_1}f(z)\,\mathrm dz=\int_{\gamma_2}f(z)\,\mathrm dz$.

If the second condition holds, let $\gamma$ be a closed path. And let $\gamma^\star\colon[0,1]\longrightarrow\mathbb C$ be the constant path such that $\gamma^\star(t)$ is equal to the initial (and final) point of $\gamma$. Then$$\int_\gamma f(z)\,\mathrm dz=\int_{\gamma^*}f(z)\,\mathrm dz=0.$$

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  • $\begingroup$ Why am I allowed to take the path $\gamma^{*}$? $\endgroup$ – Matheus Andrade Sep 14 '18 at 21:32
  • $\begingroup$ @MatheusAndrade I don't understand your question. If $[a,b]$ is the domain of $\gamma$ and if $\gamma(a)=\gamma(b)=z$, then I defined $\gamma^\star (t)=z$. That's all. $\endgroup$ – José Carlos Santos Sep 14 '18 at 22:44
  • $\begingroup$ It was a stupid question. Thanks anyway. I see what I missed now. $\endgroup$ – Matheus Andrade Sep 14 '18 at 22:51

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