Is $z/z$ holomorphic in $z=0$? To be holomorphic require to have derivative at $z\to0$ but $f(z)$ is undefined.
Does it mean $z/z$ is not holomorphic at $z=0$?
 A: I would say that no, the function is undefined. But you can set $f(0)=1$, and by Riemann's principle, the new $\tilde{f}$ is analytic everywhere, since $\lim_{z \to 0} z \tilde{f}(z)=0$.
A: Generally in complex analysis texts, after we are exposed to Riemann's Removable singularity theorem, every function written down that has a removable singularity is assumed to already have it removed, to avoid unnecessary extra clauses "and equal to $1$ at $z=0$" and the like. 
For example, $\displaystyle f(z) = \frac{\sin z}{z}$ may be written down and it may be written that $f(0)=1.$ Similarly, if we define $g(z)=z/z$ then without anything extra, one would normally go ahead and declare $g$ is holomorphic in $\mathbb{C}.$
A: I'd say for practical purposes the answer is simply yes. The function you are looking at is $g(z)=1$, which is so obviously extendable to $0$ that you don't have to apply anything like the Riemann principle and I usually would not even comment on it and I also would not write it down as $z/z$. I would also say that things are a bit different if you look at more complicated things, like the example Ragib proposed ($\sin(z)/z$), or more complicated functions. In that case I'd like to see a reason if you claim it can be extended to $0$ (which is, once you learned about the Riemann principle, of course easy to provide). 
For formal purposes it is sometimes useful to carefully observe the domain of definition of functions, in which case you are looking at a function which is simply not defined in $0$, so it simply does not make sense to ask whether it is continuous, differentiable or holomorphic there. In such cases the more elaborate way of speaking about extensibility is needed for different reasons.   
