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Math Help - Complex Analysis - Liouville's Theorem

  1. #1
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    Complex Analysis - Liouville's Theorem

    Hello,
    I am asked, whether for every analytic non-constant function f there exist z such as \Re f(z) > |f(z)|^2.
    I am pretty sure it is true because I read about Picard's theorems, but I cannot use that. I can use Liouville's Theorem and if needed Cauchy.
    I've been trying to assume that for every z, \Re f(z) \leq |f(z)|^2 and then find a function g(f(z)) such as the condition bounds g, therefore g is constant and so is f, but with no luck so far.


    Thank you!
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  2. #2
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    Re: Complex Analysis - Liouville's Theorem

    Note that every point in the interval (0,1) of the real axis satisfies your inequation. So you need only to show that the domain of f intersects with this interval.
    If f is a polynomial, since every polynomial( except the constants) has a zero point, a neighborhood of 0 is covered by the domain of f, we're done.
    Otherwise the power series of f has infinitly many terms, thus \infty is an essential singular point of f. According to big picard, we're done.
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  3. #3
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    Re: Complex Analysis - Liouville's Theorem

    Thank you for your prompt reply.

    Firstly, it is known that f is analytic everywhere, so we don't even have to use Big Picard.
    Secondly, as I said, I cannot use too advanced theorems, I can use Liouville and Cauchy.
    However, if I can proof that indeed if f has a root then in its neighbourhood I can find z which satisfies \Re f(z) > |f(z)|^2, then I'm done: let f be a non-constant analytic everywhere function. If f has a root, then according to that, I'm done. Otherwise, we can look at the function e^{\frac{1}{f(z)}: suppose \Re f(z) \leq |f(z)|^2, or in other words: \Re ( \frac{1}{f(z)} ) \leq 1. So | e^{\frac{1}{f(z)}} | = e^{\Re ( \frac{1}{f(z)}  )} \leq e. So it's bounded, ergo f is constant.
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