Calculate the integrals below using the theory seen on Cauchy's integral formula. Do not use the formula that involves parameterization to $\gamma$ when the path is of class $C^1$ ny parts.
a)$\int_{\gamma} \frac{log z}{z^n}dz$, where $\gamma(t) = 1 + \frac{1}{2}e^{it}, 0 \le t \le 2\pi$ e $n \in \mathbb N$.
b)$\int_{\gamma} \frac{z^2 + 1}{z(z^2 + 9)}dz$, where $\gamma(t) = re^{it}, 0 \leq t \leq 2 \pi$, and $0 \lt r$ and $r \ne 3$.
a)Let $a =0, n=n,f(z)=log z$
$f'(1)= \frac{1}{zln10}$
$f^{(2)}= - \frac{1}{z^2 ln 10}$
$f^{(3)}= \frac{2}{z^3 ln 10}$
$f^{(4)}= - \frac{6}{z^4 ln 10}$
$f^{(5)}= \frac{24}{z^5 ln 10}$
$f^{(6)}= - \frac{120}{z^5 ln 10}$
$f^{(7)}= \frac{720}{z^5 ln 10}$
$f^{(n)}= (-1)^{(n+1)}\frac{(n-1)!}{z^5 ln 10} = \frac{n!}{2 \pi i} \int\frac{log z}{(z-0)^n} = (-1)^{(n+1)}\frac{2 \pi i}{z^5 ln 10 n} = \int\frac{log z}{(z-0)^n}$
b)
I think there's something missing for item a) but I'm not quite sure what. As for b) I'm kind of lost, would partial fractions work? seems quite complicated.
Since $\frac{z^2+1}{z(z^2+9)}=\frac{1/9}{z}+\frac{4/9}{z+3i}+\frac{4/9}{z-3i}$, we get that $$\int_{\gamma}\frac{z^2+1}{z(z^2+9)}dz=\int_{\gamma}\frac{1/9}{z}dz+\int_{\gamma}\frac{4/9}{z+3i}dz+\int_{\gamma}\frac{4/9}{z-3i}dz$$ Applying the cauchy's integral formula we obtain that
$$\int_{\gamma}\frac{z^2+1}{z(z^2+9)}dz=\begin{cases} \frac{2\pi i}{9} & 0<r<3 \\ 2\pi i & {r>3} \end{cases}$$
Note that using residues theorem is easier. Since the poles of the function are $z=0,z=\pm 3i$, which are all simple, we need to differenciate two cases.
If $0<r<3$
The onlye pole contained in the interior of our contour is $z=0$ so just applying the residues theorem we get $$\int_{\gamma}\frac{z^2+1}{z(z^2+9)}dz=2\pi i Res(f,0)=2\pi i \lim_{z\to 0}z\frac{z^2+1}{z(z^2+9)}=2\pi i \frac{1}{9}=\frac{2\pi i}{9}$$
If $3<r$
Now all the poles are contained in the interior of the contour so we just apply the residues theorem to all of them:
$$\int_{\gamma}\frac{z^2+1}{z(z^2+9)}dz=2\pi i (Res(f,0)+Res(f,3i)+Res(f,-3i))=2\pi i$$