As is well known a function $f:U\to \mathbb{C}$ defined on an open non-empty set $U\subseteq \mathbb{C}$ is called holomorphic on $U$ if it is (complex) differentiable at every $z\in U$ and the impressive theory of complex analysis is built on this very strong property. On the other hand, what can be inferred if the function is known to be (complex) differentiable at only one point $z_{0}\in U$? This is a very weak property as there are functions $f:\mathbb{C} \to \mathbb{C}$ which are complex differentiable at a single point $z_{0}$ but not even continuous at all $z\neq z_{0}$. Can you give an example?

Thus it only makes sense to define a function to be holomorphic on an open set, but not in a single point.

What I would still like to know is: if the function is know to be (complex) differentiable in a sequence of points in the open set having an accumulation point contained in this open set and it is also (complex) differentiable in the accumulation point, can it be concluded that the function is holomorphic in some open neighborhood of the accumulation point? Or can you give a counter-example for this?

  • $\begingroup$ Take the function $f(z)=x^2=(\Re z)^2$ and from the definition one can show that $f$ is complex differentiable on the imaginary axis only since there $f(iy)=0$ and if $z=\epsilon+iy$ we have $f(z)=\epsilon^2$ so $(f(z)-f(iy))(z-iy)=\epsilon \to 0$ as $z \to iy$ so indeed $f'(iy)=0$ and one can easily see that $f$ is non differentiable at any other point $\endgroup$
    – Conrad
    2 days ago

1 Answer 1


Suppose that $$ f(z)=\begin{cases} z^2&\text{if }\operatorname{Re}z,\operatorname{Im}z\in\mathbb{Q}\\ 0&\text{otherwise.} \end{cases} $$ Then $f(0)=0$ and so you have $$ \frac{f(z)-f(0)}{z}=\frac{f(z)}{z}= \begin{cases} z&\text{if }\operatorname{Re}z,\operatorname{Im}z\in\mathbb{Q}\\ 0&\text{otherwise.} \end{cases} $$ Therefore, $\lim_{z\to0}\frac{f(z)-f(0)}{z}=0$, which means that $f'(0)=0$. But $f$ is discontinuous at every complex number other than $0$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .