Suppose we have an analytic function $f(z)=\frac1{z^2+az+b}$. Let $f=u+iv$ where $u,v$ are real valued functions on $\mathbb{R}^2$. Also, let $C$ be the closed disk which contains one of the roots of the denomiantor of $f$ in the complex plane. Then, how can we evaluate the line integral $$\int\limits_{C} vdx+udy $$?
How could we use Cauchy's integral theorem or formula in this case? If integrand were $vdx-udy$, then probably the Cauchy integral theorem would have worked, but it is not the case here. Again, if we have to evaluate using the laborious version of separating the real and imaginary parts of the function and then setting the limits of the disk separately, it would take a lot of time. Is the green theorem/ stokes theorem not applicable here? Any hints? Thanks beforehand.
The integral was quite easy, thanks to @DanielFischer. We have the integral $\int_Cf(z) dz=\int_C(u+iv)(dx+idy)=\int_C(udx-vdy)+i(vdx+udy)$. Therefore our integral is $$\int_CIm(f(z)dz)=Im\int_Cf(z)dz=2\pi i[Res_{r}f(z)]$$, by residue theorem, where $r$ is the root inside the curve $C$ of the denominator of $f(z)$.