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Math Help - Complex Analysis: Maximal Modulus Principle question

  1. #1
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    Complex Analysis: Maximal Modulus Principle question

    Hi all,

    I am stuck with this problem
    Complex Analysis: Maximal Modulus Principle question-image.jpg


    It is suggested that the Maximal Modulus Principle would help but in fact, I couldn't find a place to apply it! Anyone can suggest me how to approach this problem because I am really lost now. Thanks a lot.
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  2. #2
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    Re: Complex Analysis: Maximal Modulus Principle question

    I've been thinking of this problem but I just can't complete a proof, here's a possible route (hopefully) towards a solution: By a simple continuity argument there is an r>0 such that on 0\leq Im(z) \leq r we have that f does tend to zero. Now assume the following is true

    Claim: If f\to 0 on the line Im(z)=s with 0<s<1 then f\to 0 on 0\leq Im(z) \leq s

    then by an argument identical to the first f\to 0 on 0\leq Im(z) \leq s+r_1 so the set on which f\to 0 has to be D. I'm having a little trouble wih the claim though (particularly estimating f in the verticla boundary of the set 0\leq Im(z) \leq s, Re(z)>R for some R>0; this is enough by the MMP).

    Sorry I can't be of more help, if you find a proof please post it here.
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  3. #3
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    Re: Complex Analysis: Maximal Modulus Principle question

    There's probably a nicer way than this (i.e. a way that invokes maximum modulus at least). But for now I don't see why this doesn't work.

    Take the sequence of points \left\{a_n = n+i\left(\frac{1}{n}\right)\right\}_{n=2}^\infty. This sequence is in D.

    By continuity of f we can flex the limit in and out of the function:
    \lim_{n \to \infty} f(a_n) = f\left(\lim_{n \to \infty} a_n\right) = f\left(\lim_{n \to \infty} n + \lim_{n \to \infty} i\left(\frac{1}{n}\right)\right) = f\left(\lim_{n \to \infty} n\right)= f\left(\lim_{x \to \infty} x\right) = \lim_{x \to \infty} f(x) = A.

    Now f is holomorphic and bounded on D, so it has no singularity at infinity (in particular, no erratic essential singularity behavior). So
    \lim_{z \to \infty}f(z) exists. So all unbounded sequences in D tend to this limit. We just showed one such sequence tends to A.

    (So A is what all unbounded sequences in D tend to. In particular, any unbounded sequence with fixed imaginary component between 0 and 1 tends to A.)
    Last edited by gosuman; September 14th 2011 at 11:04 PM.
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  4. #4
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    Re: Complex Analysis: Maximal Modulus Principle question

    Quote Originally Posted by gosuman View Post
    Now f is holomorphic and bounded on D, so it has no singularity at infinity (in particular, no erratic essential singularity behavior). So
    \lim_{z \to \infty}f(z) exists.
    This I don't get, isn't this even stronger than the claim to be proved: We know that the function tends to a limit along the reals so by these two lines the problem becomes trivial.

    More to the point, can you prove your statement above?
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