Countour integration + partial fraction decompositon + poles Source: José Figueroa-O'Farrill's Mathematical Techniques III Lecture Notes Trying to evaluate the following integral: $\oint_{\text{ }\Gamma}\frac{2z+1}{z(z-1)^2}dz$ where the contour $\Gamma$ is given by ^2} -\frac{1}{z-1} +\frac{1}{z}$ Now, can someone explain to me when the contour integrals about $\Gamma_0$ and $\Gamma_1$ respectively are only non-zero for the following terms in the above decomposition? ![enter image description here]( It must be from Cauchy's Integral Theorem, so maybe I really don't understand that theorem.

Consider, for instance, $\frac1z$, whose domain is $\mathbb C\setminus\\{0\\}$. Then the loop $\Gamma_1$ is homotopically null (that is, it is homotopic to a _constant_ loop) in $\mathbb C\setminus\\{0\\}$, and therefore $\int_{\Gamma_1}\frac{\mathrm dz}z=0$. So$$\int_{\Gamma_1}\frac{\mathrm dz}{(z-1)^2}+\int_{\Gamma_1}-\frac{\mathrm dz}{z-1}+\int_{\Gamma_1}\frac{\mathrm dz}z=\int_{\Gamma_1}\frac{\mathrm dz}{(z-1)^2}+\int_{\Gamma_1}-\frac{\mathrm dz}{z-1}.$$