I have some questions on the following exercises
Calculate the following path integrals using the Cauchy formula:
a) $\oint_{|z|=1}(\frac{\exp(z)}{z}+1)dz=2\pi i\dot 2 =4\pi i$
Here I just take as $f(z)=e^z+1$ and as $\xi=0$ and I use the cauchy formula $f(\xi)=\frac{1}{2\pi i}\oint_{∂B}\frac{f(z)}{z-\xi}dz$, right?
I know that there exists another part of the Cauchy's theorem that says:
*:$U \subset \mathbb{C}$ open, $f: U \rightarrow \mathbb{C}$ holomorphic. Then the following applies: $\oint_{\gamma}f(z)dz=0$ for all $\gamma$ closed integration path with $|\gamma| \subset U$ and $int(\gamma) \subset U$
I still haven't fully understood this part of the theorem and when to apply it and when not. Because for me if I try to use (*) for a) I have: $\oint_{|z|=1}(\frac{\exp(z)}{z}+1)dz=\oint_{|z|=1}\frac{\exp(z)}{z}dz+\oint_{|z|=1}1dz$
Now we know that
$f(z):\mathbb{C}\rightarrow \mathbb{C}$, $f(z)=1$ is holomorph and $|\gamma| \subset \mathbb{C}$ and $int(\gamma) \subset \mathbb{C}$ with $\gamma(t)=e^{it}$
Therefore $\oint_{|z|=1}1dz=0 \Rightarrow \oint_{|z|=1}\frac{\exp(z)}{z}dz+\oint_{|z|=1}1dz=2\pi i$
Now I can't understand which of the two results is correct because both make sense to me.
b) $\oint_{|z|=3}\frac{\cos(\pi z)}{z^2+1}dz=\oint_{|z|=3}=\oint_{|z|=3}\frac{}{z+i}+\frac{}{z-i}dz$, but from here I dont know how to proceed. Can someone help me?
c) $\oint_{|z-i|=1}\frac{\exp(\exp(z))}{(z-i)^3}dz=\frac{2 \pi i}{2} f^{(2)}(i)=\pi if^{(2)}(i)$.
For (b) with Cauchy's Integral Formula:
$$\oint_{|z|=3}\frac{\cos\pi z}{z^2+1}dz=\frac1{2i}\oint_{|z|=3}\left[\frac{\cos\pi z}{z-i}-\frac{\cos\pi z}{z+i}\right]dz=$$
$$=\frac1{2i}2\pi i\left[\cos\pi i-\cos(-\pi i)\right]=\ldots$$
and now complete the easy end.
As for (c) and with the same formula as above (for derivatives, of course):
$$\oint_{|z-i|=1}\frac{e^{e^z}}{(z-i)^3}=\frac{2\pi i}{2!}\left(e^{e^z}\right)_{z=i}''=\ldots$$
and thus you are right in your comment.