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Bounded complex valued function

  1. Jun 5, 2010 #1
    1. The problem statement, all variables and given/known data
    1. f(z) is a function that is analytic on all of the complex plane, and mod(f)<=mod(z). Prove that f=cz.
    2. f(z) is analytic on all of the complex plane, and mod(f)<= sqrt(mod(z)). Prove that f is constant


    2. Relevant equations
    Liouvilles thm: the only bounded entire functions are constant
    Maximum modulus thm: If g is analytic on a domain D and achieves a maximum modulus on the interior of D, then g is constant


    3. The attempt at a solution
    1. I think I have this one. So, |f(z)/z|<=1, and analytic enerywhere except at z=0. If z=0 is an essential singularity or a pole, f(z)/z is not bounded near z, which is a contradiction. Thus, z=0 is a removable singularity or f(z)/z. So, There is an entire function h(z) where h(z)=f(z)/z when z is not 0, and h(z)=c when z=0 for some value c. Then, |h(z)| is bounded by max(1,c), and is constant. Then, f(z)/z=c for all z not equal to zero, so that f=cz. At zero, f(z)=0 by continuity because f is analytic everywhere on C. thus, f(z)=cz on the complex plane.
    2. I don't know where to begin here. I would like to show that f is bounded, and then use Liouville's thm, but I don't know if that's the way to go.
     
  2. jcsd
  3. Jun 6, 2010 #2

    Dick

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    Start from |f(z)^2/z|<=1.
     
  4. Jun 13, 2010 #3
    In case of 2), you can use that the inequality implies that f(0) = 0. Then the fact that f is analytic and thus complex differentiable at zero implies that the limit of f(z)/z for z to zero must exist as it is the derivative of f at z = 0. Then you can complete the proof by splitting the complex plane in some small neighborhood of zero and the rest of the complex plane and considering the behavior of f(z)/z.

    You will be led to a stronger conclusion than asked in the problem: f(z) is actually zero.
     
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