Using the Cauchy-Goursat theorem to prove a statement

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.

• 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. – user647486 Mar 23 at 11:06
• How would you integrate along each path? I'm unsure on how to go about this. Thank you – user504484 Mar 23 at 11:11
• A picture says all – user647486 Mar 23 at 11:12
• Thank you so much for your help! – user504484 Mar 23 at 11:16
• Something is missing in the hypothesis. – Kavi Rama Murthy Mar 23 at 11:55