$$ \int_{\lvert z\rvert=1}(z-1)^2\lvert dz\rvert $$ This one is either simpler or I've have my experience jogged.
Let $z=e^{j\theta}$ for $\theta\in[0,2\pi]$ (this is the unit circle, $|z|=1$), then $\frac{dz}{d\theta}=je^{j\theta}$, so $dz=je^{j\theta}d\theta$
Clearly $\lvert dz\rvert=1\lvert d\theta\rvert = d\theta$ (as $\theta$ goes from 0 up to $2\pi$ - thiis is "A-level logic")
Anyway: $I=\int^{2\pi}_0(e^{j\theta}-1)^2d\theta=\int^{2\pi}_0(e^{2j\theta}-2e^{j\theta}+1)d\theta$
The integrals over $e^{j\theta}$ become zero,thus the answer is $2\pi$ I belive.
Yes, your solution looks good and is probably the easiest way to go about.