Just ask a very fundamental question.
I am reading the following textbook:
Ablowitz and Fokas: Complex Variables, Introduction and Applications, second edition
On p.83, it says
analytic only means $f'(z)$ exists, not that it is necessarily continuous
On p.38, it says
an analytic function has derivatives of all orders in the region of analyticity and that the real and imaginary parts have continuous derivatives of all orders as well
On p.37, Definition 2.1.1 says
$f(z)$ is said to be analytic at $z_0$ if $f(z)$ is differentiable in a neighborhood of $z_0$.
If $f(z)$ has derivatives of all orders, I think $f'(z)$ should be continuous, right? Otherwise $f(z)$ is not differentiable.
So I am confused about the three statements above.
Is there an example that $f(z)$ is analytic in $D$ but $f'(z)$ is not continuous in $D$?
Thanks so much!
Best Answer
It appears that the statements on pages 83 and 37 contradict each other.
The function $f(z) = |z|^2$ is differentiable only at $z=0.$ I would not call it analytic, nor holomorphic. There is a fastidious distinction that says "holomorphic" means (complex-)differentiable in an open set, whereas "analytic" means locally equal to the sum of a convergent power series. But in the context of functions from $\mathbb C$ to $\mathbb C,$ those two can be shown to be the same, and many authors will make no such distinction. But even those who make no such distinction I would expect not to consider $z\mapsto |z|^2$ to be $\text{“analytic at $0,$”}$ since there is no open neighborhood of $0$ within which it's differentiable.