How to calculate this complex integral (with poles on the contour) I've come up with calculating this complex integral:

Compute $\displaystyle\oint_C \dfrac{z}{z^2+4z+3} \mathrm{d}z$, where $C$ is the circle with center -1 and radius 2.

The function has a pole, namely $z=-3$, on the contour. So it seems that we cannot use the residue theorem (or digging holes in the region). Then how to calculate this integral?
Thank you all in advance!
 A: The integral does not exist in the usual sense of contour integrals, but we can find a principal value for the integral.
Consider the following lemma: if $g$ is analytic at $z_0$ then $$\lim_{\epsilon \to 0} \int_{\gamma_\epsilon} \frac{g(z)dz}{z - z_0} = ig(z_0)\theta$$ where $\gamma_\epsilon$ denotes a positively-oriented arc of radius $\epsilon$ and angle $\theta$ centred on $z_0$. This lemma can be proved by writing the integrand as a sum of an analytic part and a principal part, and observing that the integral of the analytic part vanishes in the limit. The integral of principal part can be computed by hand.
In this vein, consider the contour $C_\epsilon$ comprised of $C$ with a circular portion of radius $\epsilon$ cut away to avoid the pole at $z=-3$, and the corresponding negatively oriented circular arc of radius $\epsilon$ centred on $-3$. Call these two parts $C_\epsilon^1$ and $C_\epsilon^2$. The residue of $f(z) = \frac{z}{z^2 + 4z + 3}$ at $z=-1$ is $-\frac{1}{2}$ so by the residue theorem $$\int_{C'} fdz = - \pi i.$$ On the other hand $$\int_{C'} fdz = \int_{C_\epsilon^1} fdz + \int_{C_\epsilon^2}fdz.$$
If we take the limit as $\epsilon \to 0$ the second term on the right becomes $-\frac{3 \pi i}{2}$ using the lemma above. Therefore
$$ - \pi i = \left(\lim_{\epsilon \to 0}\int_{C_\epsilon^1} fdz \right) -\frac{3 \pi i}{2} \implies PV\int_{C} f dz = \frac{\pi i}{2}.$$
