I want to calculate $$\displaystyle\int_{\partial B_2(0)}\underbrace{\frac{2z^2+7z+11}{z^3+4z^2-z-4}}_{=:f(z)}\;dz\tag{0}$$ Partial fraction decomposition yields $$f(z)=\underbrace{\frac{1}{z+4}}_{=:f_1(z)}-\underbrace{\frac{1}{z+1}}_{=:f_2(z)}+\underbrace{\frac{2}{z-1}}_{=:f_3(z)}\tag{1}$$ From this representation of $f$, it's easy to see that $-4$ and $\pm 1$ are poles of $f$. That means, that we can't take benefit from Cauchy's integral theorem, since $f$ is unbounded in a neighborhood of one of these poles.
However, since $f$ is holomorphic on $\mathbb{C}\setminus\left\{-4,\pm 1\right\}$ we can apply the residue theorem which states here $$\int_{\partial B_2(0)}f(z)\;dz=2\pi i\sum_{z_0\in\left\{-4,\pm 1\right\}}\text{res}(f,z_0)\;\text{ind}_{\partial B_2(0)}\text{ }z_0$$ The winding number of $-4$ is obvious equal to $0$ while that ones of $\pm 1$ are equal to $1$.
So, what would be smart to do now? Either we consider $f$ as a whole or as the sum of $f_1$, $f_2$ and $f_3$:
- In the first case, we would need to calculate the integrals $$\int_{\partial B_{\delta_\pm}(\pm 1)}f(z)\;dz$$ with $B_{\delta_\pm}(\pm 1)\subset B_2(0)$
- In the second case, we would need to determine the Laurent series expansion of $f_1$, $f_2$ and $f_3$ at $\pm 1$. We can take advantage of the fact, that $f_1$, $f_2$ and $f_3$ in $(1)$ are in their Laurent series form at $-4$, $-1$ and $1$, respectively.
What would be the easier way? Is there some rule of thumb in general?
It seems like in this case, both options are too complicated and it would be easier to calculate $(0)$ from the definition without the residue theorem. Or is there something what prevents me from doing this?
Notes:
- $B_r(z_0):=\left\{z\in\mathbb{C}:|z-z_0|<r\right\}$
- $A_{r,R}(z_0):=\left\{z\in\mathbb{C}:r<|z-z_0|<R\right\}$
- $\text{ind}_{\gamma}\text{ }z_0$ is the winding number of $z_0$ wrt $\gamma$
Notice that $f_1$ and $f_2$ are already in their laurent series form. You can just read off the residues.