Integral $\int_{|z-a|=r} \frac{dz}{z}$ How am I allowed to modify the integral
$$\int_{|z-a|=r} \frac{dz}{z}$$
?
It looks reminiscent of the Cauchy integral formula, but is it related?
 A: If you don't have access to the residue theorem (or Cauchy's integral formula) just yet, but have seen Cacuchy's integral theorem, you can argue in the following way:
First, if $z=0$ is outside the disc $|z-a| < r$ (i.e. if $r < |a|$), the integrand is holomorphic on and inside the curve. Cauchy's integral theorem then shows that the integral is $0$.
If $r > |a|$, the problematic point $z=0$ is inside the curve, but Cauchy's integral theorem shows that the integral you are looking for is equal to 
$$
\int_{|z|=\varepsilon} \frac{dz}{z}
$$
for some suitably small $\varepsilon$. (The integrand is holomorphic on the region between the two circles.) This new integral is easy to compute via parametrization.
Finally, if $r = |a|$ you are in trouble, since the integrand has a (non-integrable) singularity on the curve.
A: Assuming that the origin does not lie on the circle, the integral is equal to $2\pi i$ or $0$ depending on whether the origin is inside the circle or not. This is a consequence of the residue theorem.
