Holomorphic function definition. Am I missing something very obvious?

Question

I'm reading a book of complex analysis in which the definition of holomorphic function is given as follows:

Definition: If $V$ is an open set of complex numbers, a function $f:V \to \mathbb C$ is called holomorphic if the first derivative $z \to f'(z)$ is defined and "continuous" as a function from $V$ to $\mathbb C$.

Can someone please illustrate why do we need the derivative map to be continuous?

I know this may be a easy doubt but I am unable to answer this.Thank you for your help !

Answer 1

I believe that a kind of classical approach of holomorphic function definition is following.

Definition 1 Function $f:D\to\mathbb C$ is called differentiable at $z_0$, if the limit exists $$ \lim\limits_{h\to 0}\frac{f(z_0+h)-f(z_0)}{h}. $$

And then

Definition 2 If $f$ is differentiable for all points in an open disk centered at $z_0$ then $f$ is called holomorphic at $z_0$. The function $f$ is holomorphic on the open set $D\subset\mathbb C$ if it is differentiable (and hence holomorphic) at every point in $D$.

Note, there is difference for function being differentiable and holomorphic at point.

Answer 2

The answer really is in LeBtz comment, but I would like to add something. The continuity assumption on $\frac{df}{dz}$, written $f\in C^1$, is superfluous but it might be useful pedagogically: it allows for the use of the basic theorems on differential forms to develop the theory. (So it is useful as long as one already knows something about differential forms).

Namely, if one writes $f=u+iv$ and computes formally \begin{equation} \begin{split} f\, dz&=(u+iv)(dx+idy) \\ &=(u\,dx-v\,dy)+i(u\,dy+v\,dx)\\ &\stackrel{\text{def}}{=}\omega+i\nu, \end{split} \end{equation} then one can prove the following unsurprising result: if $\gamma$ is a regular enough curve in the plane, one has $$ \int_{\gamma}f\, dz=\int_\gamma \omega + i\int_\gamma \nu. $$ (If the instructor wants to save time, she might even use this formula to define the left hand side, provided the audience already knows what the integral of a differential form is). The differential forms $\omega$ and $\nu$ are closed (meaning that they are $C^1$ and their exterior derivative vanishes, i.e., the cross derivatives of their Cartesian components are equal) precisely when $f$ satisfies Cauchy-Riemann equations.

If one is working in the $C^1$ setting, necessary in the theory of differential forms, one obtains the integrability theorem for free: every holomorphic function in a simply connected domain has a primitive. In particular, and that's very important, one has $$ \oint_\gamma f(z)\, dz =0 $$ for every closed curve $\gamma$ that does not enclose a singularity of $f$.

This last result is arguably the cornerstone of complex analysis. Note that we have obtained it quickly and easily by appealing to the language of differential forms. An instructor may choose to focus attention on this result and sweep the technical detail of the $C^1$ assumption under the rug. Later, one will discover that any holomorphic function is much smoother than $C^1$ anyway.