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For $C$ a simple closed contour in the counterclockwise direction and $C_1$, $C_2$, $C_3$, $C_4$ are subsets in $C$ all in the counterclockwise direction, use the Cauchy-Goursat theorem to prove that:

If $f(z)$ is holomorphic on $C$, $C_1$, $C_2$, $C_3$, $C_4$ and throughout the multiple connected domain consisting of all points inside $C$ and exterior to each $C_k$, $k =1, 2, 3, 4$ then $$\int_C f(z) dz=\int_{C_1} f(z) dz+\int_{C_2} f(z) dz+\int_{C_3} f(z) dz+\int_{C_4} f(z) dz$$

So far I know that if a function $f$ is holomorphic at all points interior and on a simple closed contour $C$, then $$\int_C f(z) dz=0$$ I do not know how to go about doing this graphical proof.

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  • $\begingroup$ Pick a point $p$ on $C$ and draw $4$ lines (not necessarily straight) connecting it to $C_1,C_2,C_3,C_4$. Integrate along the path that starts at $p$, goes around $C$ counterclockwise, then enters through one of those lines to $C_1$, goes around $C_1$ in the clockwise direction and goes back along the line back to $p$. Then it enters $C_2,C_3,C_4$ in a similar fashion. That integral should be $0$. But then split in into the integral along the different pieces involved. Each of the paths from $p$ to $C_1,C_2,C_,3,C_4$ were traveled twice in opposite directions. So, they cancel. $\endgroup$ – user647486 Mar 23 at 11:06
  • $\begingroup$ How would you integrate along each path? I'm unsure on how to go about this. Thank you $\endgroup$ – user504484 Mar 23 at 11:11
  • $\begingroup$ A picture says all $\endgroup$ – user647486 Mar 23 at 11:12
  • $\begingroup$ Thank you so much for your help! $\endgroup$ – user504484 Mar 23 at 11:16
  • $\begingroup$ Something is missing in the hypothesis. $\endgroup$ – Kavi Rama Murthy Mar 23 at 11:55

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