# 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

By Cauchy's Integral Theorem, $$\Gamma$$ is equivalent to the pair of contours $$\Gamma_0$$ and $$\Gamma_1$$, given by

The integrand can be brought to the following form via partial fraction decomposition,

$$\frac{}{}=\frac{3}{(z-1)^2} -\frac{1}{z-1} +\frac{1}{z}$$

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?

It must be from Cauchy's Integral Theorem, so maybe I really don't understand that theorem.

• They are just evaluating the residues. The term with the squared denominator does not contribute because it has an antiderivative defined along the whole path, namely $-3/(z-1)$. The other two don't because you wind up with logarithms.
– Ian
Nov 22, 2019 at 13:44
• Hmmm I will have to push forward in these notes first, haven't gotten to residues yet! Nov 22, 2019 at 14:46
• The residue theorem is technically not a new thing, it's just an application of Cauchy's integral formula essentially. The important thing is that $z^k$ for integer $k$ has a global antiderivative along paths that don't pass through $z=0$, except when $k=-1$. This means the squared term doesn't enter, and you're only left to work out how the other two terms enter (depending on the orientation of the path around the relevant poles).
– Ian
Nov 22, 2019 at 14:48

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$$\int_{\Gamma_1}\frac{\mathrm dz}{(z-1)^2}+\int_{\Gamma_1}-\frac{\mathrm dz}{z-1}+\int_{\Gamma_1}\frac{\mathrm dz}z=\int_{\Gamma_1}\frac{\mathrm dz}{(z-1)^2}+\int_{\Gamma_1}-\frac{\mathrm dz}{z-1}.$$