Path complex integral of $\dfrac{1}{z}$ along a ball not centered at $0$ I have some troubles to solve the integral $$I=\int_{\gamma} \dfrac{1}{z}dz,$$ with the parametrisation of a circumference with radius $r$ and centre $i$: $$\gamma(t)=i+re^{it}.$$
I know if $\gamma$ would centered at $0$, then $I=2\pi i$. But in this particular case i have some troubles with the algebra since i don’t know how to deal with the denominator: $$I=\int_{\gamma} \dfrac{1}{z}dz=\int_{0}^{2\pi}\dfrac{ire^{it}}{i+re^{it}}dt.$$

If $r\lt1,$ then the integral is $0$ by Cauchy's integral theorem. If $r=0,$ then the integral does not exist. If $r\gt0,$ then the integral is $2\pi{i}$ by Cauchy's residue theorem. There is no need to actually concern yourself with the parametrization with the contour and convert the integral into a Riemann integral.
