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I Analytic functions of analytic functions

  1. Jun 10, 2016 #1
    In our complex variables course we were told that an analytic function of an analytic function is itself analytic. i.e. For ##h(z)=g(f(z))## ##h(z)## is analytic.

    I was wondering is this is just a fact, or if it is possible to prove this statement. I did some googling and the best response I could find was :

    ##h'(z)= g'(f(z)) f'(z)##, which for ##f'(z)## and ##g'(f(z))## not equal to zero, describes an analytic function.. But I'm afraid this doesn't seem like much of a proof to me.

    Many thanks :)
     
  2. jcsd
  3. Jun 10, 2016 #2

    fresh_42

    Staff: Mentor

    What is your definition of an analytic function? Is it
    "We say that the function ##f## is analytic in a neighbourhood ##U## of ##z_0## if it is differentiable everywhere in ##U##." (http://www.maths.ed.ac.uk/~jmf/Teaching/MT3/ComplexAnalysis.pdf)
    or is it
    "In mathematics, an analytic function is a function that is locally given by a convergent power series."
    (https://en.wikipedia.org/wiki/Analytic_function)

    For the first one the differentiation you mentioned would be enough (plus that ##g(U)## is open).
    Then it remains to show that the two definitions are equivalent. (See chapter 2.3.3 in the first source I mentioned.)
     
  4. Jun 10, 2016 #3

    mathman

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  5. Jun 10, 2016 #4

    andrewkirk

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    I suggest attempting to prove it as follows.

    Choose an arbitrary point ##z## and a point ##z'## sufficiently close to ##z## that the power series ##\Big(S^{f,z}_n(z')\Big)_{n=1}^\infty## for ##f(z')## at ##z## converges to ##f(z')## and the power series ##\Big(S^{g,f(z)}_n(z')\Big)_{n=1}^\infty## for ##g(f(z'))## at ##f(z)## converges to ##g(f(z'))##. We know this can be done by the definition of analyticity of the two functions.

    Then take the power series for ##g(f(z'))## and replace all occurrences of ##f(z')## by the power series for ##f(z')##. If we write ##\Delta z'\equiv z'-z## then this gives us an infinite series whose terms are powers of an infinite series in ##\Delta z'##. By adding coefficients across the same powers of ##\Delta z'## we can find the coefficient of each power of ##\Delta z'## in the power series that is our candidate for the actual power series for ##h##. Write out the coefficients for the first few powers and a pattern should emerge that will allow writing a formula for the coefficient of ##(\Delta z')^n##. We now have a power series that we would hope will converge to ##h(z')## for ##z'## sufficiently close to ##z##.

    Now we need to use epsilon-delta arguments to show that there is some ##r## such that for ##|z'-z|<r##, the power series for ##g(f(z'))## does converge and is equal to ##g(f(z'))##.

    Both parts have their challenges, but it's an interesting yet eminently doable problem and should be enjoyable.
     
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