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Entire function

  1. May 24, 2010 #1
    1. The problem statement, all variables and given/known data
    suppose that f is an entire function and that lim z->inf. f(z)=inf.
    show that the image of f is all of C


    2. Relevant equations
    I know that f(z) must have isolated values


    3. The attempt at a solution
    I think the approach should be to pick an arbitrary value w in C and show that there is a z such that f(z)=w... but I don't know where to go from there
     
  2. jcsd
  3. May 24, 2010 #2

    Dick

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    Suppose f(z) is never equal to w. Then what kinds of things can you say about g(z)=1/(f(z)-w)?
     
  4. May 24, 2010 #3
    lets see, if f(z) never equals w, then g(z) is analytic everywhere on C. Then, as z approaches infinity, g(z) approaches 0 because f(z) approaches infinity. I'm not sure why this is a contradiction.
     
  5. May 24, 2010 #4

    Dick

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    Think about whether g(z) might be bounded.
     
  6. May 24, 2010 #5
    Lets see... the modulus of f is definately bounded from below by 0. If I could get it to be bounded from above that would be a contradiction b/c of the maximum modulus principle. I'm not sure how to do that though. You could say that if z if large, f is above a certain value so that g is below a certain value. Then, f has a minimum value on the remaining compact set, so that g has a maximum on that compact set. Then the max of the two would be the bound for g. Is that right?
     
  7. May 24, 2010 #6

    Dick

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    You are garbling that a bit. But you've got the basic idea right. Yes, it's compactness and |f(z)|->infinity. Try and state it a little more carefully and then I'll believe you.
     
  8. May 24, 2010 #7
    Alright, so pick an N. Then, when |z|>M for some value M, |f(z)|>N. Thus, when |z|>M, |g(z)|<1/|N-w|. When |z|<=N, then |g(z)| attains a maximum (on the boundary) because g is continuous and the disc |z|<=N is compact. Call this maximum P. Then, g(z) is bounded by max(P, 1/|N-w|). But, g is entire, so by the max-modulus principle g is constant. Then, f must also be constant because w is fixed. This is a contradiction. Does that sound better?
     
  9. May 24, 2010 #8

    Dick

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    Does |f(z)|>N really imply that 1/|f(z)-w|<1/|N-w|? Suppose N=1, f(z)=1.01*i and w=i. You are just being really sloppy. Shouldn't your choice of N somehow involve |w|?
     
  10. May 25, 2010 #9
    Lets see, if |f|>N, then |g|=1/|f-w|. By the triangle inequality, |f-w|>||f|-|w||>||N|-|w||
    I could let N=S|w|, so that then |f-w|>||N|-|w||=||S||w|-|w||=|S-1|
    Then, |g(z)|=1/|f-w|<1/(S-1). I don't get why having |w| in the bound of g(z) is a problem.
     
  11. May 25, 2010 #10
    Oh, maybe I need that N>|w|?
     
  12. May 25, 2010 #11

    Dick

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    N>2|w| will do it, won't it? Then |g(z)|<1/|w|.
     
  13. May 25, 2010 #12
    Okay, so if I let N>2|w|, then my argument will work?
     
  14. May 25, 2010 #13

    Dick

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    Sure.
     
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