I wish to evaluate the following integral:
$$ \int_{\gamma} \frac{\log(z)}{z^n}$$ with $\gamma(t) = 1+\frac{1}{2}e^{it}, 0\le t \le 2\pi, n\ge 0.$
Now I tried to use the following Cauchy formula: $$ f^n(a) = \frac{n!}{2\pi i}\int_{\gamma} \frac{f(z)}{(z-a)^{n+1}} \mathrm{d}z. $$ where $f: G \to \mathbb{C}$ is analytic on $B(a;r) \subset G, \gamma(t) = a+re^{it}, 0\le t \le 2\pi.$
Since I need a $(z-1)^{n+1}$ in the denominator to use the above formula, I took $f(z) = \log(z)\frac{(z-1)^{n}}{z^n} = \log(z)\left(1-\frac{1}{z}\right)^n $ and then tried to apply the formula, but then I can't see a simple way to find the derivative. I need to use multiple times the product rule which becomes cumbersome. Is there some elegant way to do this calculation or I am totally wrong?
Assuming that, in this context, $\log$ is the main logarithm, then that integral is $0$, since $\frac{\log z}{z^n}$ has no singularities in the region bounded by $\gamma\bigl([0,2\pi]\bigr)$.