I have a small question about complex line integrals. I give an example to demonstrate what my question is. Consider the integral
$$\int_{|z|=2}\frac{1}{z-1}dz.$$ If I want to calculate this integral, can I take the path $f:[0,2\pi]\to \mathbb{C},\; f(t)=1+2e^{rit}$ to calculate the integral? I'm not sure if I could take this path, because it's not centered in zero, i.e. does it's not a parametrization of $|z|=2$. With this path I get:
$$\int_{|z|=2}\frac{1}{z-1}dz=\int_0^{2\pi}\frac{1}{2e^{rit}}2ire^{rit}dt=ir2\pi$$ if I did not make a mistake.
A comparison with the Cauchy-Integral-formula:
gives me $$\int_{|z|=2}\frac{1}{z-1}dz=\int_0^{2\pi}\frac{1}{2e^{rit}}2ire^{rit}dt=i2\pi$$, which isn't the same. But maybe I did a mistake in calculation. Is $f$ a possible choice for a path or has f to be centered in zero and has to be a parametrization of the boundary of $D=\{z\in\mathbb{C}: |z|\le 2\}$?
No, you haven't made a mistake, the contour you've "wraps" around the point $1$, $r$ times. The extra factor of $r$ is called the winding number.