Solve $\int _{|z-1|=1} \frac{\log{z}}{z^{4}} dz $

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this problem came in my exam today and I couldn't solve it. This is my failed attempt, using Cauchy's Integral Formula for derivatives.

I think that the theorem tells me that if $z_ {0} \in $ my curve or on the curve itself, I can apply this formula, where $$\int _{|z-1|=1} \frac{\log{z}}{(z+0)^{4}} dz $$ $$f(z)=\log{z} , \text{main branch} $$ $$z_0=0, n+1=4$$enter image description here

Since z_0 is on the curve

$$f'''(0)=\int _{|z-1|=1} \frac{3!}{2\pi i}\frac{\log{z}}{(z+0)^{4}} dz$$

Now, My problem is that the third derivative of $f (z) $ is $2/z^3$

But, I still have an indeterminacy and I did not know how to advance, am I using the theorem wrong?

A huge apology and thank you so much for the help