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Complex analysis prof

  1. Sep 8, 2009 #1
    1. The problem statement, all variables and given/known data

    suppose that f(z) is an analytic function on all of C, and suppose that, for all z in C, we have

    [tex] |f(z)| <= sqrt{|z|} [/tex]

    2. Relevant equations



    3. The attempt at a solution

    I'm unsure of how to start the proof. any help is greatly appreciated.
     
  2. jcsd
  3. Sep 8, 2009 #2

    gabbagabbahey

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    What are you supposed to prove? Your problem statement only lists the premises, and not the conclusion that you are meant to deduce.
     
  4. Sep 8, 2009 #3
    sorry, we are supposed to prove that f(z) must be a constant
     
  5. Sep 8, 2009 #4

    gabbagabbahey

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    There may be an easier way to do this, but I would start by writing [itex]z=x+iy[/itex] and [itex]f(z)=u(x,y)+iv(x,y)[/itex] and then seeing what restrictions the condition [itex]|f(z)|\leq\sqrt{|z|}[/itex] places on [itex]u(x,y)[/itex] and [itex]v(x,y)[/itex].
     
  6. Sep 8, 2009 #5
    my prof told us to look at the liouville theorem to do this proof

    i looked in my book and found this paragraph

    let f(z) be analytic inside and on a circle C of radius R centered about z_0. If |f(z)| <= M

    [tex] |f^n(z_o)| <= n! \frac {M}{R^n} [/tex]

    "This innocuous looking theorem actually places rather severe restrictions on the behaviour of analytic functions. Suppose, for instance, that f(z) is analytic and bounded by some number M over the whole plane. Then the conditions of the theorem hold for any z_o and for any R. Taking n = 1 in the above equations and letting R -> infinity, we counclude that f' vanishes everywhere; i.e., f must be constant"

    So, how can I apply this theorem over the whole complex plane instead of a circle centered at z_o?

    could I say that sqrt(|z|) = M for all z in C


    [tex] |f^n(z)| <= n! \frac {\sqrt{|z|}}{R^n} [/tex]

    and thus have the proof done by the argument in the above paragraph?
     
    Last edited: Sep 8, 2009
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