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: $$I=\int_{\gamma} \dfrac{1}{z}dz=\int_{0}^{2\pi}\dfrac{ire^{it}}{i+re^{it}}dt.$$
 A: 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.
A: Let us solve the following integral:
$$ \int \dfrac{ie^{it}}{a+e^{it}} \textrm{d}t $$
We may then use the substitutions $u=a+e^{it}$ and $\textrm{d}u=ie^{it}\textrm{d}t$ to reduce the integral to the following:
$$ \int \dfrac{\textrm{d}u}{u} = \ln(u) + C $$
Upon resubstituting, we get
$$ \ln(u) + C = \ln(a+e^{it}) + C $$
Notice that the original integral may be written as
$$ I = \int_0^{2\pi} \dfrac{ire^{it}}{r\frac{i}{r}+re^{it}} \textrm{d}t $$
If we cancel the $r$'s from both the top and the bottom and set $a=\dfrac{i}{r}$, we can simply say that the entire integral is just
$$ \ln(a+e^{2\pi i}) - \ln(a+e^{0}) = \ln(a) - \ln(a+1) = \ln\left(\dfrac{a}{a+1}\right) = \ln\left(\dfrac{\dfrac{i}{r}}{\dfrac{i}{r}+1}\right) = \ln\left(\dfrac{\dfrac{i}{r}}{\dfrac{i+r}{r}}\right) = \ln\left(\dfrac{{i}}{{i+r}}\right) $$
