Q: (complex variables) Calculate the integrals below using the theory seen on Cauchy's integral formula. Do not use the formula that involves ... 
Calculate the integrals below using the theory seen on Cauchy's integral formula. Do not use the formula that involves parameterization.


a)$\int_{\gamma} \frac{e^{iz}}{z^2}dz$, where $\gamma(t) = e^{it}, 0 \le t \le 2\pi$.


b)$\int_{\gamma} \frac{log z}{(z-1)^n}dz$, where $\gamma(t) = 1 + \frac{1}{2}e^{it} , 0 \le t \le 2\pi$ e $n \in \mathbb N$.

a)Let $a=0,n=1,f(z) = e^{iz} \to f'(z) = ie^{iz}$ so $f(z)$ is analitic on $\mathbb C$ and $B(0,1) \subset \mathbb C$. From

$$ f^{n(n)}(a) = \frac{n!}{2 \pi i} \int_{\gamma} \frac{f(w)}{(w-a)^{n+1}}dw$$ where $\gamma (t) = a + re^{it}, 0 \le t \le 2 \pi$.

We have
$f'(0) = \frac{1!}{2 \pi i} \int_{\gamma} \frac{e^{iz}}{(z-0)^2} dz \to i = \frac{1}{2 \pi i} \int_{\gamma} \frac{e^{iz}}{z^2} dz \to \int_{\gamma} \frac{e^{iz}}{z^2} dz = - 2 \pi$.
b)Let $a =-1, n=n,f(z)=log z \to f'(1)= \frac{1}{zln10} \to f^{2}= - \frac{1}{z^2 ln 10} \to f^{3}= \frac{2}{z^3 ln 10} \to f^{4}= - \frac{6}{z^4 ln 10} \to f^{5}=  \frac{24}{z^5 ln 10}$
I think a) is correct but I'm stuck on b).
Thanks.

A: Hint
For $b)$, you can see that $f^{(n)}(1)=(-1)^{n+1}(n-1)!$.
There's no $\ln10$, because $\log z$ means the principal branch of log.

A: Cauchy integral formula says:
$\oint \frac {f(z)}{z-a}^n \ dz = 2\pi i \frac {f^{(n-1)}(a)}{(n-1)!}$
But, calculating those derivatives can be a pain in the butt.  Here is a shortcut.
Remember that that the Taylor series of a function is:
$f(x) = f(a) + f'(a)(z-a) + \cdots +\frac {f^{(n)}(a)}{n!}(z-a)^n + \cdots$
Those coefficients are exactly what we are looking for.  If we know the Talor series of our function, all we have to do is find the right coefficient.
$\log z = \sum_\limits{n=1}^\infty \frac {(-1)^{n+1}}{n} (z-1)^n$
(in complex analysis and every math class thereafter "log" should be assumed to be $\ln$)
When $n>1$
$\oint \frac {\log z}{(z-1)^n} \ dz = 2\pi i\frac {(-1)^{n}}{n-1}$
