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Complex analysis - Louvilles Theorem (I think)

  1. Oct 21, 2012 #1
    Hi I am doing a past exam for my complex analysis course and I should just mention right now that while it's a mix of pure/applied math I have never done a pure math unit before and i really really really suck at doing proofs and such..

    Given c>0 and f(z) is entire such that |f(z)| ≤ c|z| show that :

    f(z)=wz for some complex constant w.

    I just have no idea how to tackle problems like this whatsoever and have to turn to google and notes which I obviously wont be able to do in the exam. Also I have trouble remembering all these theorems, corollaries, propositions..

    My attempt at a solution was to state Louvilles theorem which is that if:

    |f(z)|≤M and f(z) is entire then f(z)=w.. I don't really know where to go from here..

    Any tips for tackling problems like this would be really appreciated!
  2. jcsd
  3. Oct 21, 2012 #2
    I think you are in the good path. Just apply Liouville's theorem to the auxiliary function $$g(z)=\frac{f(z)}{z}\text,$$ which is entire because ##f(z)=0## and we can continue ##g## to the origin as $$g(0)=\lim_{z\to 0}\frac{f(z)}{z}\text.$$
  4. Oct 21, 2012 #3


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    Alberto7: what do you mean f(z)=0? Maybe you mean to say that, _if_ f(z)=0 , then

    the discontinuity at z=0 can be removed (e.g., by the Riemann removable singularity

    theorem, since |f(z)/z|<c <oo)?

    Then, following (what I think was) Alberto7's idea ,

    |f(z)/z| itself is entire and bounded, so.....
  5. Oct 22, 2012 #4
    Sorry, I meant ##f(0)=0##.
    Last edited: Oct 22, 2012
  6. Oct 22, 2012 #5


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    jrp: you should have some theorem in your textbook saying that if f is analytic (holomorphic) and bounded in some punctured neigborhood of a point a (or someting simililar), then f can be extended to be analytic at a too. Apply this to f(z)/z and proceed as alberto7 and Bacle2 suggested.
    Last edited: Oct 22, 2012
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