How to Integrate$\int_C\frac{Log~z}{(z-i)^2}$ where $C:|z-i|=1$?

Question

Let $C:|z-i|=\alpha$,where $0<\alpha<1$
I need to evaluate this expression

$~\int_C\frac{Log~z}{(z-i)^2}$

what I tried to do was representing $Log~z$ as a Laurent series on $|z-1|<1$.
But what I can think is only about $\frac{Log~z}{(z-1)^2}$ as a Laurent series. I don't know how to use what I know and what I don't know yet.

Any advice would be helpful.

Answer 1

Edit: Now that the OP has edited the question:
Recall Cauchy theorem: $$f(w) = \dfrac{1}{2\pi i} \int_\gamma \dfrac{f(z)}{z-w} dz$$ Differating in respect to $w$ we get: $$f'(w) = \dfrac{1}{2\pi i} \int_\gamma \dfrac{f(z)}{(z-w)^2} dz $$

Can you figure it out from here?