Showing $\int_{\gamma}f(z)dz = \int_{\gamma_1}f(z)dz + \int_{\gamma_2}f(z)dz$ with non analytic points.

Suppose $f$ is analytic on the complex plane except at $z_1,z_2$, that $\gamma_1$ and $\gamma_2$ are simple closed curves with $z_1,z_2$ in their interiors and $\gamma_1$ and $\gamma_2$ are in the interior of $\gamma$. $\gamma$, $\gamma_1$, $\gamma_2$ are all traveling counterclockwise. Give an argument to show that $\int_{\gamma}f(z)dz = \int_{\gamma_1}f(z)dz + \int_{\gamma_2}f(z)dz$.

I assume that we will use the Cauchy Integral Theorem that if $f$ is analytic in a domain $D$, then any simple closed curve yields an integral of $0$ (very summarized), but we are unsure how to deal with the fact $f$ is not analytic at $z_1$ and $z_2$.

• This isn't true in general, you need to be carefull about the direction of $\gamma_1$, $\gamma_2$ and $\gamma$. For this formula to hold, all of the three paths need to have the same direction, i.e. all need to be either clockwise or counter-clockwise. – fgp Apr 7 '14 at 1:00
• There was a picture attached to the image, and they are all going counterclockwise. – Blaris Apr 7 '14 at 1:05
• We think we figured it out, we will use cancellation. Essentially, since we can draw a simple, closed curve around everything else that is not contained in $\gamma_1$ and $\gamma_2$, than the interior of THAT curve is 0, so the 'area' of $\gamma$ has to be partitioned by $\gamma_1$ and $\gamma_2$ and thus the 'area' of $\gamma =$ the 'area' of $\gamma_1$+$\gamma_2$. – Blaris Apr 7 '14 at 1:22
• I suspected there was a picture. You should have included that in your question, or at least explained how it looks like. Your solution sounds fine, btw. – fgp Apr 7 '14 at 9:58