The problem is as follows:
For $|z| < 1,$ evaluate the integral $\int_{|\zeta|=1} \frac{1}{\zeta (\zeta - z)} d\zeta.$
I've been trying to utilize the Cauchy's theorem on annulus on the choice of $f(\zeta) = \frac{1}{\zeta},$ but so far this has led me to nowhere.
The problem additionally asks, for $|z_0| = 1,$ does $\lim_{z \to z_0, |z| < 1} \frac{1}{2\pi i}\int_{|\zeta|=1} \frac{1}{\zeta(\zeta -z)} d\zeta = \frac{1}{z_0}?$
So I guess I would have to use $\epsilon$ radius smaller annulus around $0$ and the unit disk as the larger annulus to apply Cauchy's theorem to compare $f(z_0)$ with, and if the integral on $\epsilon$ radius disk vanishes to 0, we get the equality on the fourth line. But this didn't work out as I thought.
Note that, unless $z=0$,$$\frac1{\zeta(\zeta-z)}=\frac1z\left(\frac1{\zeta-z}-\frac1\zeta\right).$$Therefore\begin{align}\int_{|\zeta|=1}\frac{\mathrm dz}{\zeta(\zeta-z)}&=\frac1z\left(\int_{|\zeta|=1}\frac{\mathrm dz}{\zeta-z}-\int_{|\zeta|=1}\frac{\mathrm dz}{\zeta}\right)\\&=\frac{2\pi i}z(1-1)\text{ (since $|z|<1$)}\\&=0.\end{align}And, if $z=0$, the function that you are integrating is $\frac1{\zeta^2}$, which has a primitive ($-\frac1\zeta$), and therefore the integral will be $0$ in that case as well.