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Math Help - About analytic functions

  1. #1
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    About analytic functions

    How can i prove that an analytic, non constant function f(z) in the complex plane C, has a range dense in C ?
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  2. #2
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    Hi. It's an interesting problem and perhaps someone here can suggest a more elegant and non-overkill way to prove this than what I suggest here:

    I assume you mean "entire" function which can either be a polynomial or a non-polynomial. If it's a polynomial, then by the Fundamental Theorem of Algebra, the polynomial reaches all values in the complex plane exactly n times where n is the degree. If it's not a polynomial, then by Picard's First Theorem it has an essential singularity at infinity and by Casorati-Weierstrass, in any neighborhood of that singularity, the function comes arbitrarily close to any number in C, that is, it's dense in C.
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  3. #3
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    Sorry for my english,i'm greek and i try as much as i can to be clear.
    I can see what you mean. I was trying to use Picard's theorem too, but i found something else about the problem. I thought that i could consider that R is not dense in C. So there is an open disk of a point w in C, with no common points with R(range of f) and abs(f(z)-w)>e (e the radius of the disk). Then i consider g(z)=1/(f(z)-w), which is entire because f is entire. But absg(z)<1/e=d and then g(z) is entire and bounded in C, so g is constant(liouville theorem). But if g is constant then f is constant and this is impossible. So R is dense in the complex plane. This is my proof but i don't know if it's correct.
    Thanks for helping!!
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  4. #4
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    Quote Originally Posted by takis1881 View Post
    Sorry for my english,i'm greek and i try as much as i can to be clear.
    I can see what you mean. I was trying to use Picard's theorem too, but i found something else about the problem. I thought that i could consider that R is not dense in C. So there is an open disk of a point w in C, with no common points with R(range of f) and abs(f(z)-w)>e (e the radius of the disk). Then i consider g(z)=1/(f(z)-w), which is entire because f is entire. But absg(z)<1/e=d and then g(z) is entire and bounded in C, so g is constant(liouville theorem). But if g is constant then f is constant and this is impossible. So R is dense in the complex plane. This is my proof but i don't know if it's correct.
    Thanks for helping!!
    I'm not an analyst but that sounds really good to me and I can see nothing wrong with your logic.

    Thanks!
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  5. #5
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    I really appreciate your help! I wanted a second opinion about my proof and you were very helpful. I have to deliver an exercise about complex analysis till next week and i was stuck to this problem for a few days. Thank you again, my greetings from Greece and many wishes for a happy new year shawsend!
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