What makes a holomorphic function non-analytic?

[Knissp]

Homework Statement

What is a real holomorphic function which is not analytic?

Homework Equations

Theorem from complex analysis: holomorphic functions and analytic functions are the same.
Definition 1: A holomorphic function is infinitely differentiable.
Definition 2: An analytic function is locally given by a convergent power series.

The Attempt at a Solution

I think one answer is e^{\frac{-1}{x^2}}. The function can be differentiated by the chain rule as many times as desired (so it is holomorphic) but has a Taylor series with all coefficients equal to zero (so it is not analytic).

However, I wonder about the complex-valued function e^{\frac{-1}{z^2}}. Does it not have the same problem? That is to say, how can we show that it is consistent with the holomorphic-analytic equivalence theorem if (1) we can differentiate it infinitely many times but (2) it's Taylor series has all coefficients equal to zero?

 

[Office_Shredder]

In the complex plane it's no longer continuous, let alone differentiable. Consider values near zero: for real values x that are near zero, your exponent is going to be close to negative infinity. Imagine for your value of z though you pick something like i/100000. Now your exponent is close to positive infinity, and we see that the function isn't even continuous

 

[Knissp]

I completely overlooked that, and it was really bothering me for a long time, but it makes a lot of sense now! Thank you so much office_shredder! :)

 

[gomunkul51]

yup, e^(-1/z^2) has an essential pole at z=0.
problem with the limit near the singularity. it does not behave well at that singularity at all !
it has a Laurent series representation - Analytic part (the Taylor series) and the Essential part.
so it is not a Entire function, it is not differentiable near z=0.