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Homework Help: Holomorphic = Analytic

  1. Aug 27, 2010 #1
    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 [tex]e^{\frac{-1}{x^2}}[/tex]. 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 [tex]e^{\frac{-1}{z^2}}[/tex]. 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. jcsd
  3. Aug 27, 2010 #2


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    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
  4. Aug 27, 2010 #3
    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! :)
  5. Aug 27, 2010 #4
    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.
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