I'm trying to find $$\int \dfrac{dz}{(z-1)^3}$$ on the circle $|z|=2$.
I'm pretty new to this, so my instincts were to use Cauchy's theorem for this, namely that the integral of an analytic function on the boundary of a region of this type is zero. However, the function isn't analytic, at $z=1$.
I know that we can modify Cauchy's theorem for when there is only one non analytic point, but I'm pretty sure we can only do that when $\lim_{z\rightarrow z_0}(z-z_0)f(z)$ equals zero, and here I don't think it does, as it becomes $$\lim\frac{(z-1)}{(z-1)^3}=\lim \frac{1}{(1-z)^2}$$ which isn't zero. So I'm pretty stuck. I know that residue is a thing that might be applicable, but I havent learned it very much.
Can it be done without it, or if residue is needed, how would I do it? Any help is welcome, thank you!