Let's say I am integrating a function over $|z| = 1$ and $|z-1| = 1$, is there any difference? I think the answer for both cases will be same, as in both cases,
$$ z = \exp^{i\Theta} $$ and $$ dz = i \exp^{i\Theta} d\Theta $$
Please correct me if I am wrong.
$$ \int\frac{dz}{z+\frac 1 2} $$
Integrated over $|z-1|=1$, this is $0$, since it is holomorphic everywhere on and inside that circle. Integrated over $|z|=1$, running counterclockwise, this is $2\pi i$, because of its pole inside that circle.