# Homework Help: Holomorphic = Analytic

1. Aug 27, 2010

### Knissp

1. The problem statement, all variables and given/known data
What is a real holomorphic function which is not analytic?

2. Relevant 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.

3. 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?

2. Aug 27, 2010

### Office_Shredder

Staff Emeritus
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

3. Aug 27, 2010

### 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! :)

4. Aug 27, 2010

### 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.