A question about winding numbers and the Global Version of Cauchy's Theorem In his book, Introduction to Complex Analysis at a Graduate Level, Serge Lang starts the chapter on the Global Version of Cauchy's Theorem with the following discussion

However, I fail to follow what he means. He talks about how if a path winds around some point outside of $U$ then we will have an integral not equal to zero, but this can't be true. Take the integral along the unit circle of $\frac{1}{z-5}$. And furthermore, he mentions that the integrals of functions such as $f(z)=\frac{1}{z - \alpha}$ where $\alpha$ is a pint not in $U$ have an integral =/0 for closed paths in $U$, which is again false (for them to have values not equal to zero doesn't do point need to be in $U$? For example $f(z)=\frac{1}{z}$ along the unit circle has a value of $2\pi i$ and the point 0 and the closed path are both in the open set, say the circle of radius 2 centered at the origen.
I think I am probably missing something. Is whats written in the discussion by Serge Lang correct? And if yes could someone please help me understand where I have gone wrong?
All help is really appriciated.

Perhaps that you are missing that $U$ is the domain of $f$. So, in your example with $f(z)=\frac1z$, the natural domain of $f$ is $U=\Bbb C\setminus\{0\}$. And so the only point outside $U$ is $0$. And the integral of $f$ along $\gamma\colon[0,2\pi]\longrightarrow\Bbb C$ defined by $\gamma(t)=Re^{it}$ $(R>0$) is indeed $2\pi i$, which is different from $0$.
On the other hand, note that Lang wrote that if the path winds around some point outside of $U$, then we can find functions whose integral is not equal to $0$. He does not claim that the integral is always different from $0$.
