I have the problem, to use Cauchy's theorem or Cauchy's integral formula to evaluate $\oint_C \dfrac{\sin2z}{6z-\pi} dz$, where $C$ is the circle $\mod (z) = 3$.
I know that the denominator vanishes at z = $\dfrac{\pi}{6}$, which is within the contour, so the result of the integral is not $0$ (Cauchy's theorem). I have a singularity within the contour, then, and it is not analytic. How then can I use Cauchy's theorem or integral formula for this?
On
Apply the Residue Theorem. The integral has a singularity at $z=\frac{\pi}{6}$, which is inside the disk, so the value is $$2\pi i \text{Res} \left(\frac{1}{6}\frac{\sin (2z)}{z-\pi/6},\ z=\frac{\pi}{6}\right).$$
This will equal $$\frac{\pi i}{3}\sin\left(\frac{\pi}{3}\right)= \frac{\pi i}{2\sqrt{3}}.$$
No need for the residue theorem, Cauchy's integral formula gives the answer right away:
$$ \oint_C \dfrac{\sin2z}{6z-\pi}\,dz = \oint_C \dfrac{\frac{\sin2z}{6}}{z-\frac{\pi}{6}}\,dz = 2\pi i\,\frac{\sin \frac{\pi}3}{6} = \frac{\pi i\sqrt{3}}{6}. $$