In Rudin Example 10.4 it is said that $f(z) = \frac{1}{z}$ is holomorphic at $\mathbb{C} \setminus \{0\}$ (which i checked through Cauchy-Riemann equations).
What is going on at $0$? How can I establish conformity for $f$ at $0$? Is it just a convention?
On
$f$ is not defined at $z=0$. Moreover, the singularity at $z=0$ is not removable: there is no entire function $g$ extending $f$, in the sense that $g(z)=f(z)$ for $z\ne0$.
Contrast this with $h(z)=z^2/z$ or $h(z)=\sin(z)/z$, which are not defined at $z=0$ but can be extended to entire functions.
What is going on at $0$ is that it doesn't belong to the domain of $f$.