Let be $L=\{(x,y)\in\mathbb{R²}\mid ((x-1)²+y²)((x+1)²+y²)=1 \}$. Let be path $\gamma$ the edge of $L$ so that the negative part goes in mathematical positive way and the positive part goes in mathematical negative way (like drawing the infinity symbol). Calculate $\int_{\gamma}\frac{dz}{1-z²}$ using Cauchy's integral formula.
I thought to first divide the path $\gamma$ to $\gamma_1$, which is on the left/negative side and $\gamma_2$, which is on the right/positive side. Taking $f(z)=\frac{1}{1+z}$ we get $\int_{\gamma}\frac{dz}{1-z²}=-\int_{\gamma}\frac{f(z)}{z-1}dz=\int_{\gamma_1}\frac{f(z)}{z-1}dz - \int_{\gamma_2}\frac{f(z)}{z-1}dz=0$. The reason I took the negative of the integral over the path $\gamma_1$ is because it goes clockweise and the Cauchy formula is for counterclockwise integrals.
Can someone say if this is right what I did here? I appreciate hints much more than solutions.