Take the 2-minute tour ×

To be holomorphic require to have derivative at $z\to0$ but $f(z)$ is undefined. Does it mean $z/z$ is not holomorphic at $z=0$?

share|improve this question
It doesn't make sense to talk about the derivative of a function at a point where that function is not defined. "Holomorphic at 0" means "differentiable in a neightbourhood of 0". –  mt_ Aug 18 '12 at 12:19
@mt_: I agree. That said, some authors would assume that a rational expression actually refers to its natural extension to include any removable singularities, e.g. when considering the field of Möbius functions. –  tomasz Aug 18 '12 at 12:24
I do not care what a function does at an isolated point ... –  Fabian Aug 18 '12 at 12:44
So, read the two answers. Then tell us why you ask the question. –  GEdgar Aug 18 '12 at 12:47

3 Answers 3

up vote 11 down vote accepted

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 $\mathbb{C}.$

share|improve this answer
I never understood why real analysis is taught so rigorously and complex analysis so loosely. Suppose that $f(z)=z/z$ and $z \in \mathbb{R}$: who would say, in 2012, that $f\equiv 1$? –  Siminore Aug 18 '12 at 13:49
@Siminore I think the difference is for two reasons: 1) Real Analysis often has to be learned first, and students have to learn how to be completely rigorous, filling in every detail. When we get to complex analysis, the ideas can be more free flowing as we are more experienced and the details can be just that - details. –  Ragib Zaman Aug 18 '12 at 14:19
2) We have to be more careful in Real Analysis, we almost begin to hate our intuitions for many things it wants us to believe are false. Even Gauss believed continuous functions had to be differentiable on "most" of their domain. On the other hand, Complex Analysis has many "Don't worry about it!" theorems - things are how you want them to be or even feel too good to be true. So there is slightly less need for utmost care in all the little details like where we are filling in a functions values, because these rarely cause errors later. –  Ragib Zaman Aug 18 '12 at 14:19

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 $\lim_{z \to 0} z \tilde{f}(z)=0$.

share|improve this answer

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.

share|improve this answer

Your Answer


