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Math Help - Complex Analysis proof

  1. #1
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    Complex Analysis proof

    Hey guys, this question was on my final last week and I couldn't get it. If anyone could enlighten me as to what the solution is I'd really appreciate it. Thanks.

    Let f(z) be entire (i.e. analytic on all of C) and let |f(z)| > 1. Prove that f is constant.

    I know there is a theorem that says the only bounded entire functions are constant functions, so my thought was to show that f is bounded since we assumed it is entire. However, we also assumed that |f(z)| > 1 so it's clearly unbounded. This is where I'm stuck; the only solution I can think of is that my teacher made a typo (and the inequality should read the other way). Any help is appreciated.
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  2. #2
    Senior Member Shanks's Avatar
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    consider the function \frac{1}{f(z)}: it is bounded and analytic, thus it is a constant.
    further, the range of a nonconstant entire function is the complex field C.
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  3. #3
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    Quote Originally Posted by Shanks View Post
    consider the function \frac{1}{f(z)}: it is bounded and analytic, thus it is a constant.
    further, the range of a nonconstant entire function is the complex field ...or the whole complex plane minus one single point, according to (little) Picard's Theorem
    Tonio
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  4. #4
    Senior Member Shanks's Avatar
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    Tonio, THX very much!
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